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Sorting is a very classic problem of reordering items (that can be compared, e.g., integers, floating-point numbers, strings, etc) of an array (or a list) in a certain order (increasing, non-decreasing (increasing or flat), decreasing, non-increasing (decreasing or flat), lexicographical, etc).

There are many different sorting algorithms, each has its own advantages and limitations.

Sorting is commonly used as the introductory problem in various Computer Science classes to showcase a range of algorithmic ideas.

Without loss of generality, we assume that we will sort only Integers, not necessarily distinct, in non-decreasing order in this visualization. Try clicking Bubble Sort for a sample animation of sorting the list of 5 jumbled integers (with duplicate) above.

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If you are an NUS student and a repeat visitor, please login.

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Sorting problem has a variety of interesting algorithmic solutions that embody many Computer Science ideas:

1. Comparison versus non-comparison based strategies,
2. Iterative versus Recursive implementation,
3. Divide-and-Conquer paradigm (e.g., Merge Sort or Quick Sort),
4. Best/Worst/Average-case Time Complexity analysis,
5. Randomized Algorithms, etc.

Pro-tip 1: Since you are not logged-in, you may be a first time visitor (or not an NUS student) who are not aware of the following keyboard shortcuts to navigate this e-Lecture mode: [PageDown]/[PageUp] to go to the next/previous slide, respectively, (and if the drop-down box is highlighted, you can also use [→ or ↓/← or ↑] to do the same),and [Esc] to toggle between this e-Lecture mode and exploration mode.

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When an (integer) array A is sorted, many problems involving A become easy (or easier):

1. Searching for a specific value v in array A,
2. Finding the min/max or the k-th smallest/largest value in (static) array A,
3. Testing for uniqueness and deleting duplicates in array A,
4. Counting how many time a specific value v appear in array A,
5. Set intersection/union between array A and another sorted array B,
6. Finding a target pair xA and yA such that x+y equals to a target z,
7. Counting how many values in array A is inside range [lo..hi], etc.

Discussion: In real-life classes, the instructor may elaborate more on these applications.

Pro-tip 2: We designed this visualization and this e-Lecture mode to look good on 1366x768 resolution or larger (typical modern laptop resolution in 2021). We recommend using Google Chrome to access VisuAlgo. Go to full screen mode (F11) to enjoy this setup. However, you can use zoom-in (Ctrl +) or zoom-out (Ctrl -) to calibrate this.

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

Pro-tip 3: Other than using the typical media UI at the bottom of the page, you can also control the animation playback using keyboard shortcuts (in Exploration Mode): Spacebar to play/pause/replay the animation, / to step the animation backwards/forwards, respectively, and -/+ to decrease/increase the animation speed, respectively.

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There are two actions that you can do in this visualization.

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The first action is about defining your own input, an array/a list A that is:

1. Totally random,
2. Random but sorted (in non-decreasing or non-increasing order),
3. Random but nearly sorted (in non-decreasing or non-increasing order),
4. Random and contain many duplicates (thus small range of integers), or
5. Defined solely by yourself.

In Exploration mode, you can experiment with various sorting algorithms provided in this visualization to figure out their best and worst case inputs.

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The second action is the most important one: Execute the active sorting algorithm by clicking the "Sort" button.

Remember that you can switch active algorithm by clicking the respective abbreviation on the top side of this visualization page.

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View the visualisation/animation of the chosen sorting algorithm here.

Without loss of generality, we only show Integers in this visualization and our objective is to sort them from the initial state into non-decreasing order state. Remember, non-decreasing means mostly ascending (or increasing) order, but because there can be duplicates, there can be flat/equal line between two adjacent equal integers.

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At the top, you will see the list of commonly taught sorting algorithms in Computer Science classes. To activate each algorithm, select the abbreviation of respective algorithm name before clicking "Sort".

To facilitate more diversity, we randomize the active algorithm upon each page load.

The first six algorithms in this module are comparison-based sorting algorithms while the last two are not. We will discuss this idea midway through this e-Lecture.

The middle three algorithms are recursive sorting algorithms while the rest are usually implemented iteratively.

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To save screen space, we abbreviate algorithm names into three characters each:

1. Comparison-based Sorting Algorithms:
1. BUB - Bubble Sort,
2. SEL - Selection Sort,
3. INS - Insertion Sort,
4. MER - Merge Sort (recursive implementation),
5. QUI - Quick Sort (recursive implementation),
6. R-Q - Random Quick Sort (recursive implementation).
2. Not Comparison-based Sorting Algorithms:
1. COU - Counting Sort,
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We will discuss three comparison-based sorting algorithms in the next few slides:

They are called comparison-based as they compare pairs of elements of the array and decide whether to swap them or not.

These three sorting algorithms are the easiest to implement but also not the most efficient, as they run in O(N2).

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Before we start with the discussion of various sorting algorithms, it may be a good idea to discuss the basics of asymptotic algorithm analysis, so that you can follow the discussions of the various O(N^2), O(N log N), and special O(N) sorting algorithms later.

This section can be skipped if you already know this topic.

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You need to already understand/remember all these:
-. Logarithm and Exponentiation, e.g., log2(1024) = 10, 210 = 1024
-. Arithmetic progression, e.g., 1+2+3+4+…+10 = 10*11/2 = 55
-. Geometric progression, e.g., 1+2+4+8+..+1024 = 1*(1-211)/(1-2) = 2047
-. Linear/Quadratic/Cubic function, e.g., f1(x) = x+2, f2(x) = x2+x-1, f3(x) = x3+2x2-x+7
-. Ceiling, Floor, and Absolute function, e.g., ceil(3.1) = 4, floor(3.1) = 3, abs(-7) = 7

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Analysis of Algorithm is a process to evaluate rigorously the resources (time and space) needed by an algorithm and represent the result of the evaluation with a (simple) formula.

The time/space requirement of an algorithm is also called the time/space complexity of the algorithm, respectively.

For this module, we focus more on time requirement of various sorting algorithms.

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We can measure the actual running time of a program by using wall clock time or by inserting timing-measurement code into our program, e.g., see the code shown in SpeedTest.cpp | py | java.

However, actual running time is not meaningful when comparing two algorithms as they are possibly coded in different languages, using different data sets, or running on different computers.

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Instead of measuring the actual timing, we count the # of operations (arithmetic, assignment, comparison, etc). This is a way to assess its efficiency as an algorithm's execution time is correlated to the # of operations that it requires.

See the code shown in SpeedTest.cpp | py | java and the comments (especially on how to get the final value of variable counter).

Knowing the (precise) number of operations required by the algorithm, we can state something like this: Algorithm X takes 2n2 + 100n operations to solve problem of size n.

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If the time t needed for one operation is known, then we can state that algorithm X takes (2n2 + 100n)t time units to solve problem of size n.

However, time t is dependent on the factors mentioned earlier, e.g., different languages, compilers and computers, etc.

Therefore, instead of tying the analysis to actual time t, we can state that algorithm X takes time that is proportional to 2n2 + 100n to solving problem of size n.

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Asymptotic analysis is an analysis of algorithms that focuses on analyzing problems of large input size n, considers only the leading term of the formula, and ignores the coefficient of the leading term.

We choose the leading term because the lower order terms contribute lesser to the overall cost as the input grows larger, e.g., for f(n) = 2n2 + 100n, we have:
f(1000) = 2*10002 + 100*1000 = 2.1M, vs
f(100000) = 2*1000002 + 100*100000 = 20010M.
(notice that the lower order term 100n has lesser contribution).

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Suppose two algorithms have 2n2 and 30n2 as the leading terms, respectively.

Although actual time will be different due to the different constants, the growth rates of the running time are the same.

Compared with another algorithm with leading term of n3, the difference in growth rate is a much more dominating factor.

Hence, we can drop the coefficient of leading term when studying algorithm complexity.

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If algorithm A requires time proportional to f(n), we say that algorithm A is of the order of f(n).

We write that algorithm A has time complexity of O(f(n)), where f(n) is the growth rate function for algorithm A.

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Mathematically, an algorithm A is of O(f(n)) if there exist a constant k and a positive integer n0 such that algorithm A requires no more than k*f(n) time units to solve a problem of size n ≥ n0, i.e., when the problem size is larger than n0, then algorithm A is (always) bounded from above by this simple formula k*f(n).

Note that: n0 and k are not unique and there can be many possible valid f(n).

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In asymptotic analysis, a formula can be simplified to a single term with coefficient 1.

Such a term is called a growth term (rate of growth, order of growth, order of magnitude).

The most common growth terms can be ordered from fastest to slowest as follows:
O(1)/constant time < O(log n)/logarithmic time < O(n)/linear time <
O(n log n)/quasilinear time < O(n2)/quadratic time < O(n3)/cubic time <
O(2n)/exponential time < O(n!)/also-exponential time < ∞ (e.g., an infinite loop).

Note that a few other common time complexities are not shown (also see the visualization in the next slide).

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We will see three different growth rates O(n2), O(n log n), and O(n) throughout the remainder of this sorting module.

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Given an array of N elements, Bubble Sort will:

1. Compare a pair of adjacent items (a, b),
2. Swap that pair if the items are out of order (in this case, when a > b),
3. Repeat Step 1 and 2 until we reach the end of array
(the last pair is the (N-2)-th and (N-1)-th items as we use 0-based indexing),
4. By now, the largest item will be at the last position.
We then reduce N by 1 and repeat Step 1 until we have N = 1.

Without further ado, let's try Bubble Sort on the small example array [29, 10, 14, 37, 14].

You should see a 'bubble-like' animation if you imagine the larger items 'bubble up' (actually 'float to the right side of the array').

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`void bubbleSort(int a[], int N) { // the standard version  for (; N > 0; --N) // N iterations    for (int i = 0; i < N-1; ++i) // except the last, O(N)      if (a[i] > a[i+1]) // not in non-decreasing order        swap(a[i], a[i+1]); // swap in O(1)}`

Comparison and swap require time that is bounded by a constant, let's call it c. Then, there are two nested loops in (the standard) Bubble Sort. The outer loop runs for exactly N iterations. But the inner loop runs get shorter and shorter:

1. When i=0, (N−1) iterations (of comparisons and possibly swaps),
2. When i=1, (N−2) iterations,
...,
3. When i=(N−2), 1 iteration,
4. When i=(N−1), 0 iteration.

Thus, the total number of iterations = (N−1)+(N−2)+...+1+0 = N*(N−1)/2 (derivation).

Total time = c*N*(N−1)/2 = O(N^2).

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Bubble Sort is actually inefficient with its O(N^2) time complexity. Imagine that we have N = 105 numbers. Even if our computer is super fast and can compute 108 operations in 1 second, Bubble Sort will need about 100 seconds to complete.

However, it can be terminated early, e.g. on the small sorted ascending example shown above [3, 6, 11, 25, 39], Bubble Sort can terminates in O(N) time.

The improvement idea is simple: If we go through the inner loop with no swapping at all, it means that the array is already sorted and we can stop Bubble Sort at that point.

Discussion: Although it makes Bubble Sort runs faster in general cases, this improvement idea does not change O(N^2) time complexity of Bubble Sort... Why?

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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Given an array of N items and L = 0, Selection Sort will:

1. Find the position X of the smallest item in the range of [L...N−1],
2. Swap X-th item with the L-th item,
3. Increase the lower-bound L by 1 and repeat Step 1 until L = N-2.

Let's try Selection Sort on the same small example array [29, 10, 14, 37, 13].

Without loss of generality, we can also implement Selection Sort in reverse:
Find the position of the largest item Y and swap it with the last item.

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`void selectionSort(int a[], int N) {  for (int L = 0; L <= N-2; ++L) { // O(N)    int X = min_element(a+L, a+N) - a; // O(N)    swap(a[X], a[L]); // O(1), X may be equal to L (no actual swap)  }}`

Total: O(N2) — To be precise, it is similar to Bubble Sort analysis.

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Quiz: How many (real) swaps are required to sort [29, 10, 14, 37, 13] by Selection Sort?

3
2
1
4
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Insertion sort is similar to how most people arrange a hand of poker cards.

2. Pick the next card and insert it into its proper sorted order,
3. Repeat previous step for all cards.

Let's try Insertion Sort on the small example array [40, 13, 20, 8].

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`void insertionSort(int a[], int N) {  for (int i = 1; i < N; ++i) { // O(N)    int X = a[i]; // X is the item to be inserted    int j = i-1;    for (; j >= 0 && a[j] > X; --j) // can be fast or slow      a[j+1] = a[j]; // make a place for X    a[j+1] = X; // index j+1 is the insertion point  }}`
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The outer loop executes N−1 times, that's quite clear.

But the number of times the inner-loop is executed depends on the input:

1. In best-case scenario, the array is already sorted and (a[j] > X) is always false
So no shifting of data is necessary and the inner loop runs in O(1),
2. In worst-case scenario, the array is reverse sorted and (a[j] > X) is always true
Insertion always occur at the front of the array and the inner loop runs in O(N).

Thus, the best-case time is O(N × 1) = O(N) and the worst-case time is O(N × N) = O(N2).

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Quiz: What is the complexity of Insertion Sort on any input array?

O(N log N)
O(1)
O(N)
O(N^2)

Ask your instructor if you are not clear on this or read similar remarks on this slide.

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We will discuss two (and a half) comparison-based sorting algorithms soon:

1. Merge Sort,
2. Quick Sort and its Randomized version (which only has one change).

These sorting algorithms are usually implemented recursively, use Divide and Conquer problem solving paradigm, and run in O(N log N) time for Merge Sort and O(N log N) time in expectation for Randomized Quick Sort.

PS: The non-randomized version of Quick Sort runs in O(N2) though.

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Given an array of N items, Merge Sort will:

1. Merge each pair of individual element (which is by default, sorted) into sorted arrays of 2 elements,
2. Merge each pair of sorted arrays of 2 elements into sorted arrays of 4 elements,
Repeat the process...,
3. Final step: Merge 2 sorted arrays of N/2 elements (for simplicity of this discussion, we assume that N is even) to obtain a fully sorted array of N elements.

This is just the general idea and we need a few more details before we can discuss the true form of Merge Sort.

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We will dissect this Merge Sort algorithm by first discussing its most important sub-routine: The O(N) merge.

Given two sorted array, A and B, of size N1 and N2, we can efficiently merge them into one larger combined sorted array of size N = N1+N2, in O(N) time.

This is achieved by simply comparing the front of the two arrays and take the smaller of the two at all times. However, this simple but fast O(N) merge sub-routine will need additional array to do this merging correctly.

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`void merge(int a[], int low, int mid, int high) {  // subarray1 = a[low..mid], subarray2 = a[mid+1..high], both sorted  int N = high-low+1;  int b[N]; // discuss: why do we need a temporary array b?  int left = low, right = mid+1, bIdx = 0;  while (left <= mid && right <= high) // the merging    b[bIdx++] = (a[left] <= a[right]) ? a[left++] : a[right++];  while (left <= mid) b[bIdx++] = a[left++]; // leftover, if any  while (right <= high) b[bIdx++] = a[right++]; // leftover, if any  for (int k = 0; k < N; ++k) a[low+k] = b[k]; // copy back}`

Try Merge Sort on the example array [1, 5, 19, 20, 2, 11, 15, 17] that have its first half already sorted [1, 5, 19, 20] and its second half also already sorted [2, 11, 15, 17]. Concentrate on the last merge of the Merge Sort algorithm.

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Before we continue, let's talk about Divide and Conquer (abbreviated as D&C), a powerful problem solving paradigm.

Divide and Conquer algorithm solves (certain kind of) problem — like our sorting problem — in the following steps:

1. Divide step: Divide the large, original problem into smaller sub-problems and recursively solve the smaller sub-problems,
2. Conquer step: Combine the results of the smaller sub-problems to produce the result of the larger, original problem.
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Merge Sort is a Divide and Conquer sorting algorithm.

The divide step is simple: Divide the current array into two halves (perfectly equal if N is even or one side is slightly greater by one element if N is odd) and then recursively sort the two halves.

The conquer step is the one that does the most work: Merge the two (sorted) halves to form a sorted array, using the merge sub-routine discussed earlier.

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`void mergeSort(int a[], int low, int high) {  // the array to be sorted is a[low..high]  if (low < high) { // base case: low >= high (0 or 1 item)    int mid = (low+high) / 2;	    mergeSort(a, low  , mid ); // divide into two halves    mergeSort(a, mid+1, high); // then recursively sort them    merge(a, low, mid, high); // conquer: the merge subroutine  }}`
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Contrary to what many other CS printed textbooks usually show (as textbooks are static), the actual execution of Merge Sort does not split to two subarrays level by level, but it will recursively sort the left subarray first before dealing with the right subarray.

That's it, running Merge Sort on the example array [7, 2, 6, 3, 8, 4, 5], it will recurse to [7, 2, 6, 3], then [7, 2], then [7] (a single element, sorted by default), backtrack, recurse to [2] (sorted), backtrack, then finally merge [7, 2] into [2, 7], before it continue processing [6, 3] and so on.

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In Merge Sort, the bulk of work is done in the conquer/merge step as the divide step does not really do anything (treated as O(1)).

When we call merge(a, low, mid, high), we process k = (high-low+1) items.
There will be at most k-1 comparisons.
There are k moves from original array a to temporary array b and another k moves back.
In total, number of operations inside merge sub-routine is < 3k-1 = O(k).

The important question is how many times this merge sub-routine is called?

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Level 1: 2^0=1 calls to merge() with N/2^1 items each, O(2^0 x 2 x N/2^1) = O(N)
Level 2: 2^1=2 calls to merge() with N/2^2 items each, O(2^1 x 2 x N/2^2) = O(N)
Level 3: 2^2=4 calls to merge() with N/2^3 items each, O(2^2 x 2 x N/2^3) = O(N)
...
Level (log N): 2^(log N-1) (or N/2) calls to merge() with N/2^log N (or 1) item each, O(N)

There are log N levels and in each level, we perform O(N) work, thus the overall time complexity is O(N log N). We will later see that this is an optimal (comparison-based) sorting algorithm, i.e., we cannot do better than this.

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The most important good part of Merge Sort is its O(N log N) performance guarantee, regardless of the original ordering of the input. That's it, there is no adversary test case that can make Merge Sort runs longer than O(N log N) for any array of N elements.

Merge Sort is therefore very suitable to sort extremely large number of inputs as O(N log N) grows much slower than the O(N2) sorting algorithms that we have discussed earlier.

There are however, several not-so-good parts of Merge Sort. First, it is actually not easy to implement from scratch (but we don't have to). Second, it requires additional O(N) storage during merging operation, thus not really memory efficient and not in-place. Btw, if you are interested to see what have been done to address these (classic) Merge Sort not-so-good parts, you can read this.

Merge Sort is also a stable sort algorithm. Discussion: Why?

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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Quick Sort is another Divide and Conquer sorting algorithm (the other one discussed in this visualization page is Merge Sort).

We will see that this deterministic, non randomized version of Quick Sort can have bad time complexity of O(N2) on adversary input before continuing with the randomized and usable version later.

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Divide step: Choose an item p (known as the pivot)
Then partition the items of a[i..j] into three parts: a[i..m-1], a[m], and a[m+1..j].
a[i..m-1] (possibly empty) contains items that are smaller than (or equal to) p.
a[m] = p, i.e., index m is the correct position for p in the sorted order of array a.
a[m+1..j] (possibly empty) contains items that are greater than (or equal to) p.
Then, recursively sort the two parts.

Conquer step: Don't be surprised... We do nothing :O!

If you compare this with Merge Sort, you will see that Quick Sort D&C steps are totally opposite with Merge Sort.

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We will dissect this Quick Sort algorithm by first discussing its most important sub-routine: The O(N) partition (classic version).

To partition a[i..j], we first choose a[i] as the pivot p.

The remaining items (i.e., a[i+1..j]) are divided into 3 regions:

1. S1 = a[i+1..m] where items are ≤ p,
2. S2 = a[m+1..k-1] where items are ≥ p, and
3. Unknown = a[k..j], where items are yet to be assigned to either S1 or S2.

Discussion: Why do we choose p = a[i]? Are there other choices?

Harder Discussion: If a[k] == p, should we put it in region S1 or S2?

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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Initially, both S1 and S2 regions are empty, i.e., all items excluding the designated pivot p are in the unknown region.

Then, for each item a[k] in the unknown region, we compare a[k] with p and decide one of the three cases:

1. If a[k] > p, put a[k] into S2,
2. If a[k] < p, put a[k] into S1,
3. If a[k] == p, throw a coin and put a[k] into S1/S2 if it lands head/tail, respectively.

These three cases are elaborated in the next two slides.

Lastly, we swap a[i] and a[m] to put pivot p right in the middle of S1 and S2.

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`int partition(int a[], int i, int j) {  int p = a[i]; // p is the pivot  int m = i; // S1 and S2 are initially empty  for (int k = i+1; k <= j; ++k) { // explore the unknown region    if ((a[k] < p) || ((a[k] == p) && (rand()%2 == 0)))  { // case 2+3      ++m;      swap(a[k], a[m]); // C++ STL algorithm std::swap    } // notice that we do nothing in case 1: a[k] > p  }  swap(a[i], a[m]); // final step, swap pivot with a[m]  return m; // return the index of pivot}`
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`void quickSort(int a[], int low, int high) {  if (low < high) {    int m = partition(a, low, high); // O(N)    // a[low..high] ~> a[low..m–1], pivot, a[m+1..high]    quickSort(a, low, m-1); // recursively sort left subarray    // a[m] = pivot is already sorted after partition    quickSort(a, m+1, high); // then sort right subarray  }}`
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Try Quick Sort on example array [27, 38, 12, 39, 29, 16]. We shall elaborate the first partition step as follows:
We set p = a[0] = 27.
We set a[1] = 38 as part of S2 so S1 = {} and S2 = {38}.
We swap a[1] = 38 with a[2] = 12 so S1 = {12} and S2 = {38}.
We set a[3] = 39 and later a[4] = 29 as part of S2 so S1 = {12} and S2 = {38,39,29}.
We swap a[2] = 38 with a[5] = 16 so S1 = {12,16} and S2 = {39,29,38}.
We swap p = a[0] = 27 with a[2] = 16 so S1 = {16,12}, p = {27}, and S2 = {39,29,38}.

After this, a[2] = 27 is guaranteed to be sorted and now Quick Sort recursively sorts the left side a[0..1] first and later recursively sorts the right side a[3..5].

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First, we analyze the cost of one call of partition.

Inside partition(a, i, j), there is only a single for-loop that iterates through (j-i) times. As j can be as big as N-1 and i can be as low as 0, then the time complexity of partition is O(N).

Similar to Merge Sort analysis, the time complexity of Quick Sort is then dependent on the number of times partition(a, i, j) is called.

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When the array a is already in ascending order, e.g., a = [5, 18, 23, 39, 44, 50], Quick Sort will set p = a[0] = 5, and will return m = 0, thereby making S1 region empty and S2 region: Everything else other than the pivot (N-1 items).

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On such worst case input scenario, this is what happens:

The first partition takes O(N) time, splits a into 0, 1, N-1 items, then recurse right.
The second one takes O(N-1) time, splits a into 0, 1, N-2 items, then recurse right again.
...
Until the last, N-th partition splits a into 0, 1, 1 item, and Quick Sort recursion stops.

This is the classic N+(N-1)+(N-2)+...+1 pattern, which is O(N2), similar analysis as the one in this Bubble Sort analysis slide...

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The best case scenario of Quick Sort occurs when partition always splits the array into two equal halves, like Merge Sort.

When that happens, the depth of recursion is only O(log N).

As each level takes O(N) comparisons, the time complexity is O(N log N).

Try Quick Sort on this hand-crafted example input array [4, 1, 3, 2, 6, 5, 7].
In practice, this is rare, thus we need to devise a better way: Randomized Quick Sort.

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Same as Quick Sort except just before executing the partition algorithm, it randomly select the pivot between a[i..j] instead of always choosing a[i] (or any other fixed index between [i..j]) deterministically.

Mini exercise: Implement the idea above to the implementation shown in this slide!

Running Random Quick Sort on this large and somewhat random example array a = [3,44,38,5,47,15,36,26,27,2,46,4,19,50,48] feels fast.

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It will take about 1 hour lecture to properly explain why this randomized version of Quick Sort has expected time complexity of O(N log N) on any input array of N elements.

In this e-Lecture, we will assume that it is true.

If you need non formal explanation: Just imagine that on randomized version of Quick Sort that randomizes the pivot selection, we will not always get extremely bad split of 0 (empty), 1 (pivot), and N-1 other items. This combination of lucky (half-pivot-half), somewhat lucky, somewhat unlucky, and extremely unlucky (empty, pivot, the rest) yields an average time complexity of O(N log N).

Discussion: For the implementation of Partition, what happen if a[k] == p, we always put a[k] on either side (S1 or S2) deterministically?

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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We will discuss two non comparison-based sorting algorithms in the next few slides:

These sorting algorithms can be faster than the lower bound of comparison-based sorting algorithm of Ω(N log N) by not comparing the items of the array.

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It is known (also not proven in this visualization as it will take another 1 hour lecture to do so) that all comparison-based sorting algorithms have a lower bound time complexity of Ω(N log N).

Thus, any comparison-based sorting algorithm with worst-case complexity O(N log N), like Merge Sort is considered an optimal algorithm, i.e., we cannot do better than that.

However, we can achieve faster sorting algorithm — i.e., in O(N) — if certain assumptions of the input array exist and thus we can avoid comparing the items to determine the sorted order.

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Assumption: If the items to be sorted are Integers with small range, we can count the frequency of occurrence of each Integer (in that small range) and then loop through that small range to output the items in sorted order.

Try Counting Sort on the example array above where all Integers are within [1..9], thus we just need to count how many times Integer 1 appears, Integer 2 appears, ..., Integer 9 appears, and then loop through 1 to 9 to print out x copies of Integer y if frequency[y] = x.

The time complexity is O(N) to count the frequencies and O(N+k) to print out the output in sorted order where k is the range of the input Integers, which is 9-1+1 = 9 in this example. The time complexity of Counting Sort is thus O(N+k), which is O(N) if k is small.

We will not be able to do the counting part of Counting Sort when k is relatively big due to memory limitation, as we need to store frequencies of those k integers.

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Assumption: If the items to be sorted are Integers with large range but of few digits, we can combine Counting Sort idea with Radix Sort to achieve the linear time complexity.

In Radix Sort, we treat each item to be sorted as a string of w digits (we pad Integers that have less than w digits with leading zeroes if necessary).

For the least significant (rightmost) digit to the most significant digit (leftmost), we pass through the N items and put them according to the active digit into 10 Queues (one for each digit [0..9]), which is like a modified Counting Sort as this one preserves stability. Then we re-concatenate the groups again for subsequent iteration.

Try Radix Sort on the example array above for clearer explanation.

Notice that we only perform O(w × (N+k)) iterations. In this example, w = 4 and k = 10.

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Now, having discussed about Radix Sort, should we use it for every sorting situation?

For example, it should be theoretically faster to sort many (N is very large) 32-bit signed integers as w ≤ 10 digits and k = 10 if we interpret those 32-bit signed integers in Decimal. O(10 × (N+10)) = O(N).

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The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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There are a few other properties that can be used to differentiate sorting algorithms on top of whether they are comparison or non-comparison, recursive or iterative.

In this section, we will talk about in-place versus not in-place, stable versus not stable, and caching performance of sorting algorithms.

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A sorting algorithm is said to be an in-place sorting algorithm if it requires only a constant amount (i.e. O(1)) of extra space during the sorting process. That's it, a few, constant number of extra variables is OK but we are not allowed to have variables that has variable length depending on the input size N.

Merge Sort (the classic version), due to its merge sub-routine that requires additional temporary array of size N, is not in-place.

Discussion: How about Bubble Sort, Selection Sort, Insertion Sort, Quick Sort (randomized or not), Counting Sort, and Radix Sort. Which ones are in-place?

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A sorting algorithm is called stable if the relative order of elements with the same key value is preserved by the algorithm after sorting is performed.

Example application of stable sort: Assume that we have student names that have been sorted in alphabetical order. Now, if this list is sorted again by tutorial group number (recall that one tutorial group usually has many students), a stable sort algorithm would ensure that all students in the same tutorial group still appear in alphabetical order of their names.

Discussion: Which of the sorting algorithms discussed in this e-Lecture are stable?
Try sorting array A = {3, 4a, 2, 4b, 1}, i.e. there are two copies of 4 (4a first, then 4b).

🕑

The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

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We are nearing the end of this e-Lecture.

Time for a few simple questions.

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Quiz: Which of these algorithms run in O(N log N) on any input array of size N?

Quick Sort (Deterministic)
Merge Sort
Insertion Sort
Bubble Sort
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Quiz: Which of these algorithms has worst case time complexity of Θ(N^2) for sorting N integers?

Bubble Sort
Selection Sort
Merge Sort
Insertion Sort

Θ is a tight time complexity analysis where the best case Ω and the worst case big-O analysis match.

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We have reached the end of sorting e-Lecture.

However, there are two other sorting algorithms in VisuAlgo that are embedded in other data structures: Heap Sort and Balanced BST Sort. We will discuss them when you go through the e-Lecture of those two data structures.

🕑

The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

🕑

The content of this interesting slide (the answer of the usually intriguing discussion point from the earlier slide) is hidden and only available for legitimate CS lecturer worldwide. This mechanism is used in the various flipped classrooms in NUS.

If you are really a CS lecturer (or an IT teacher) (outside of NUS) and are interested to know the answers, please drop an email to stevenhalim at gmail dot com (show your University staff profile/relevant proof to Steven) for Steven to manually activate this CS lecturer-only feature for you.

FAQ: This feature will NOT be given to anyone else who is not a CS lecturer.

🕑

Actually, the C++ source code for many of these basic sorting algorithms are already scattered throughout these e-Lecture slides. For other programming languages, you can translate the given C++ source code to the other programming language.

Usually, sorting is just a small part in problem solving process and nowadays, most of programming languages have their own sorting functions so we don't really have to re-code them unless absolutely necessary.

In C++, you can use std::sort (most likely a hybrid sorting algorithm: Introsort), std::stable_sort (most likely Merge Sort), or std::partial_sort (most likely Binary Heap) in STL algorithm.
In Python, you can use sort (most likely a hybrid sorting algorithm: Timsort).
In Java, you can use Collections.sort.
In OCaml, you can use List.sort compare list_name.

If the comparison function is problem-specific, we may need to supply additional comparison function to those built-in sorting routines.

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Now it is time for you to see if you have understand the basics of various sorting algorithms discussed so far.

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Now that you have reached the end of this e-Lecture, do you think sorting problem is just as simple as calling built-in sort routine?

Try these online judge problems to find out more:
Kattis - mjehuric
Kattis - sortofsorting, or
Kattis - sidewayssorting

This is not the end of the topic of sorting. When you explore other topics in VisuAlgo, you will realise that sorting is a pre-processing step for many other advanced algorithms for harder problems, e.g. as the pre-processing step for Kruskal's algorithm, creatively used in Suffix Array data structure, etc.

You have reached the last slide. Return to 'Exploration Mode' to start exploring!

Note that if you notice any bug in this visualization or if you want to request for a new visualization feature, do not hesitate to drop an email to the project leader: Dr Steven Halim via his email address: stevenhalim at gmail dot com.

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Create(A)

Sort

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Random

Sorted

Nearly sorted

Many Duplicates

A =

Go

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VisuAlgo was conceptualised in 2011 by Dr Steven Halim as a tool to help his students better understand data structures and algorithms, by allowing them to learn the basics on their own and at their own pace.

VisuAlgo contains many advanced algorithms that are discussed in Dr Steven Halim's book ('Competitive Programming', co-authored with his brother Dr Felix Halim and his friend Dr Suhendry Effendy) and beyond. Today, a few of these advanced algorithms visualization/animation can only be found in VisuAlgo.

Though specifically designed for National University of Singapore (NUS) students taking various data structure and algorithm classes (e.g., CS1010/equivalent, CS2040/equivalent, CS3230, CS3233, and CS4234), as advocators of online learning, we hope that curious minds around the world will find these visualizations useful too.

VisuAlgo is not designed to work well on small touch screens (e.g., smartphones) from the outset due to the need to cater for many complex algorithm visualizations that require lots of pixels and click-and-drag gestures for interaction. The minimum screen resolution for a respectable user experience is 1024x768 and only the landing page is relatively mobile-friendly. However, we are currently experimenting with a mobile (lite) version of VisuAlgo to be ready by April 2022.

VisuAlgo is an ongoing project and more complex visualizations are still being developed.

The most exciting development is the automated question generator and verifier (the online quiz system) that allows students to test their knowledge of basic data structures and algorithms. The questions are randomly generated via some rules and students' answers are instantly and automatically graded upon submission to our grading server. This online quiz system, when it is adopted by more CS instructors worldwide, should technically eliminate manual basic data structure and algorithm questions from typical Computer Science examinations in many Universities. By setting a small (but non-zero) weightage on passing the online quiz, a CS instructor can (significantly) increase his/her students mastery on these basic questions as the students have virtually infinite number of training questions that can be verified instantly before they take the online quiz. The training mode currently contains questions for 12 visualization modules. We will soon add the remaining 12 visualization modules so that every visualization module in VisuAlgo have online quiz component.

We have translated VisuAlgo pages into three main languages: English, Chinese, and Indonesian. Currently, we have also written public notes about VisuAlgo in various languages:

id, kr, vn, th.

#### Team

Dr Steven Halim, Senior Lecturer, School of Computing (SoC), National University of Singapore (NUS)
Dr Felix Halim, Senior Software Engineer, Google (Mountain View)

Undergraduate Student Researchers 1 (Jul 2011-Apr 2012)
Koh Zi Chun, Victor Loh Bo Huai

Final Year Project/UROP students 1 (Jul 2012-Dec 2013)
Phan Thi Quynh Trang, Peter Phandi, Albert Millardo Tjindradinata, Nguyen Hoang Duy

Final Year Project/UROP students 2 (Jun 2013-Apr 2014)
Rose Marie Tan Zhao Yun, Ivan Reinaldo

Undergraduate Student Researchers 2 (May 2014-Jul 2014)
Jonathan Irvin Gunawan, Nathan Azaria, Ian Leow Tze Wei, Nguyen Viet Dung, Nguyen Khac Tung, Steven Kester Yuwono, Cao Shengze, Mohan Jishnu

Final Year Project/UROP students 3 (Jun 2014-Apr 2015)
Erin Teo Yi Ling, Wang Zi

Final Year Project/UROP students 4 (Jun 2016-Dec 2017)
Truong Ngoc Khanh, John Kevin Tjahjadi, Gabriella Michelle, Muhammad Rais Fathin Mudzakir

Final Year Project/UROP students 5 (Aug 2021-Dec 2022)
Liu Guangyuan, Manas Vegi, Sha Long, Vuong Hoang Long

List of translators who have contributed ≥100 translations can be found at statistics page.

Acknowledgements
This project is made possible by the generous Teaching Enhancement Grant from NUS Centre for Development of Teaching and Learning (CDTL).

VisuAlgo is free of charge for Computer Science community on earth. If you like VisuAlgo, the only "payment" that we ask of you is for you to tell the existence of VisuAlgo to other Computer Science students/instructors that you know =) via Facebook/Twitter/Instagram/TikTok posts, course webpages, blog reviews, emails, etc.

If you are a data structure and algorithm student/instructor, you are allowed to use this website directly for your classes. If you take screen shots (videos) from this website, you can use the screen shots (videos) elsewhere as long as you cite the URL of this website (https://visualgo.net) and/or list of publications below as reference. However, you are NOT allowed to download VisuAlgo (client-side) files and host it on your own website as it is plagiarism. As of now, we do NOT allow other people to fork this project and create variants of VisuAlgo. Using the offline copy of (client-side) VisuAlgo for your personal usage is fine.

Note that VisuAlgo's online quiz component is by nature has heavy server-side component and there is no easy way to save the server-side scripts and databases locally. Currently, the general public can only use the 'training mode' to access these online quiz system. Currently the 'test mode' is a more controlled environment for using these randomly generated questions and automatic verification for real examinations in NUS.

List of Publications

This work has been presented briefly at the CLI Workshop at the ICPC World Finals 2012 (Poland, Warsaw) and at the IOI Conference at IOI 2012 (Sirmione-Montichiari, Italy). You can click this link to read our 2012 paper about this system (it was not yet called VisuAlgo back in 2012) and this link for the short update in 2015 (to link VisuAlgo name with the previous project).

This work is done mostly by my past students.

Bug Reports or Request for New Features

VisuAlgo is not a finished project. Dr Steven Halim is still actively improving VisuAlgo. If you are using VisuAlgo and spot a bug in any of our visualization page/online quiz tool or if you want to request for new features, please contact Dr Steven Halim. His contact is the concatenation of his name and add gmail dot com.

Version 1.1 (Updated Fri, 14 Jan 2022).

Disclosure to all visitors: We currently use Google Analytics to get an overview understanding of our site visitors. We now give option for user to Accept or Reject this tracker.

Since Wed, 22 Dec 2021, only National University of Singapore (NUS) staffs/students and approved CS lecturers outside of NUS who have written a request to Steven can login to VisuAlgo, anyone else in the world will have to use VisuAlgo as an anonymous user that is not really trackable other than what are tracked by Google Analytics.