**Difficulty:** Medium, **Asked-in:** Google, Microsoft, Adobe, Goldman Sachs, Qualcomm.

Quick sort algorithm was developed by British computer scientist Tony Hoare in 1959. It quickly became popular due to its efficiency and simplicity, and it remains one of the most widely used sorting algorithms today.

Quick sort is one of the fastest sorting algorithms based on the idea of divide-and-conquer approach. Here are some excellent reasons to learn quick sort:

- Quick sort is often the best choice for sorting because it works efficiently in O(n log n) average time.
- Quick sort is one of the best algorithms to learn recursion. The recursive structure, flow of recursion, and base case of quick sort are intuitive.
- Quick sort is an excellent algorithm to learn worst, best, and average-case analysis.
- The partition algorithm of quick sort is a good idea to learn problem-solving using the two-pointers approach. We can use a similar approach to solve several coding questions.
- Quick sort will be cache-friendly due to its in-place sorting property and fewer memory accesses compared to merge sort.
- Quick sort is an excellent algorithm to learn the idea of randomized algorithms. Randomization is a popular tool to improve the efficiency of real-time systems.

Let’s start by exploring the pattern of sorted array.

From the above observation, we can think in a reverse way: Suppose we partition an unsorted array around some input value (**pivot**) into two parts:

- All values in the left part are less than pivot.
- All values in the right part are greater than pivot.

After the partition, pivot will get placed at the correct position in the sorted order. Now we have a situation similar to a sorted array with one big difference: the left and right halves are still unsorted! If we observe further, both unsorted halves are smaller subproblems of the sorting problem, i.e., a sorting problem for smaller input sizes.

If we sort both halves recursively using the same function, the whole array will get sorted! The idea is simple: All values on the left side are less than pivot, pivot is at the correct position, and all values on the right side are greater than pivot.

The above idea of quick sort is a divide-and-conquer approach of problem-solving: We divide the sorting problem into two smaller subproblems using the partition algorithm and solve each subproblem recursively to get the final sorted array. We repeat the partition process for smaller subarrays until we reach the base case.

**Divide part:** We divide the problem into two smaller subproblems by partitioning the array X[l...r]. The partition process will return the **pivotIndex**, i.e. the index of the pivot in the sorted array.

**Subproblem 1:**Sort array X[] from l to pivotIndex - 1.**Subproblem 2**: Sort array X[] from pivotIndex + 1 to r.

**Conquer part:** Now, we solve both subproblems by recursively sorting smaller subarrays.

**Combine part:** This is a trivial case because after sorting both smaller arrays, the whole array will get sorted. In other words, there is no need to write code for the combine part.

Suppose we use the function **quickSort(int X[], int l, int r)** to sort the entire array, where l and r are the left and right ends. The initial function call will be **quickSort(X, 0, n - 1)**.

We define the **partition(X, l, r)** function to divide the array around the pivot and return the pivotIndex. We will select the pivot value inside the partition function. The critical question is: Why are we choosing the pivot value inside the partition function? Think and explore!

`pivotIndex = partition(X, l, r)`

- We recursively sort the left subarray by calling the same function with
**l**and**pivotIndex**as the left and right ends, i.e.,**quickSort(X, l, pivotIndex - 1)**. - We recursively sort the right subarray by calling the same function with
**pivotIndex + 1**and**r**as the left and right ends, i.e.,**quickSort(X, pivotIndex + 1, r)**.

In recursive algorithms, the base case is the smallest version of the problem that will be directly solved without further recursion. In the context of quick sort, the base case will occur when the sub-array being sorted has either zero elements (empty) or just one element remaining.

In other words, **l >= r** is the situation of base case in the quick sort. At this point, no further partitioning is needed, sub-array is considered sorted, and the recursion will terminate.

```
void quickSort(int X[], int l, int r)
{
if (l < r)
{
int pivotIndex = partition(X, l, r)
quickSort(X, l, pivotIndex - 1)
quickSort(X, pivotIndex + 1, r)
}
}
```

How can we divide an array around the pivot so that the values in the left part are less than the pivot and values in the right part are greater than the pivot? Can we do it in place? Let’s think!

Suppose we pick the rightmost value as the pivot, i.e., **pivot = X[r]**, and scan the array from left to right. Whenever an element is less than the pivot, we will place it at the starting part of the array and move forward. Otherwise, we will ignore that element and move forward. How can we implement it? Let’s think!

- We need two pointers to track two different parts of the array: 1) Pointer
**i**to track the starting subarray from**0 to i**, which will include values less than the pivot 2) Pointer**j**to traverse the array. - Now we scan the array from
**j = l to r - 1**. Whenever**X[j] < pivot**, we increment pointer i,**swap X[j] with X[i]**, and move to the next iteration. Otherwise, we ignore the value at index j and move to the next iteration. - By the end of the loop: all values in the subarray
**X[0…i]**will be less than the pivot, and all values in the subarray**X[i + 1 … r - 1]**will be greater than the pivot. - Now we need to place the pivot at its correct position, i.e.,
**i + 1**. For this, we**swap X[r] with X[i + 1]**and complete the partition process. Think! - Finally, we return the position of the pivot as an output, i.e., return
**i + 1**.

```
int partition (int X[], int l, int r)
{
int pivot = X[r]
int i = l - 1
for (int j = l; j < r; j = j + 1)
{
if (X[j] < pivot)
{
i = i + 1
swap (X[i], X[j])
}
}
swap (X[i + 1], X[r])
return i + 1
}
```

We are running a single loop and doing constant operations at each iteration. In the worst or best case, we are doing one comparison at each iteration. So time complexity = O(n). We are using constant extra space, so space complexity = O(1). Note: Here swapping operation will depend on comparison if (X[j] < pivot).

Suppose **T(n)** is the worst-case time complexity of quick sort for n integers. Let's analyze it by breaking down the time complexities of each process:

**Divide part:** Time complexity of the divide part is equal to the time complexity of the partition algorithm, which is **O(n)**.

**Conquer part:** We are recursively solving two subproblems, where size of subproblems will depend on the choice of pivot in the partition process. Suppose after the partition, **i** elements are in the left subarray and **n - i - 1** elements are in the right subarray. So time complexity of conquer part = Time complexity of recursively sorting the left subarray + Time complexity of recursively sorting the right subarray = **T(i) + T(n - i - 1)**.

**Combine part:** As mentioned above, this is a trivial part. Time complexity of combine part = **O(1)**.

For calculating time complexity T(n), we need to add time complexities of divide, conquer, and combine parts.

```
T(n)
= O(n) + T(i) + T(n — i — 1) + O(1)
= T(i) + T(n — i — 1) + O(n)
= T(i) + T(n — i — 1) + cn
Recurrence relation of the quick sort
T(n) = c, if n = 1
T(n) = T(i) + T(n — i — 1) + cn, if n > 1
```

The worst-case scenario will occur when the partition process always chooses the largest or smallest element as the pivot. In this case, the partition will be highly unbalanced, with one subarray containing n - 1 elements and the other subarray containing 0 elements.

When we always choose the rightmost element as the pivot, the worst-case scenario will arise when the array is already sorted in either increasing or decreasing order. In this case, each recursive call will create an unbalanced partition.

For calculating the time complexity in the worst case, we put **i = n - 1** in the above formula of T(n).

```
T(n) = T(n - 1) + T(0) + cn
T(n) = T(n - 1) + cn
```

**Worst-case analysis of quick sort using Substitution Method**

We simply expand the recurrence relation by substituting all intermediate value of T(i) (from i = n - 1 to 1). By end of this process, we ended up in a sum of arithmetic series.

```
T(n)
= T(n - 1) + cn
= T(n - 2) + c(n - 1) + cn
= T(n - 3) + c(n - 2) + c(n - 1) + cn
... and so on
= T(1) + 2c + 3c + ... + c(n - 2) + c(n - 1) + cn
= c + 2c + 3c + ... + c(n - 2) + c(n - 1) + cn
= c(1 + 2 + 3 + ... + n - 2 + n - 1 + n)
This is a simple sum of arithmetic series.
T(n) = c(n(n + 1)/2) = O(n^2)
Worst case time complexity of quick sort = O(n^2)
```

**Worst-case analysis of quick sort using Recursion Tree Method**

The recursion tree method is one of the popular techniques for recursion time complexity analysis. Here we draw a tree structure of recursive calls and highlight the extra cost associated with each recursive call. To get the overall time complexity, we add the cost level by level.

This is the recursion tree diagram of quick sort in the worst case.

We sum the total partitioning cost for each level => cn + c(n−1) + c(n−2) +⋯+ 2c + c = c (n + n−1 + n−2 + ⋯+ 2 + 1) = c[n(n+1)/2] = O(n^2).

The best-case scenario of quick sort will occur when partition process will always pick the median element as the pivot. In other words, this is a case of balanced partition, where both sub-arrays are approx. n/2 size each.

Suppose, the scenario of balanced partitioning will arise in each recursive call. Now, for calculating the time complexity in the best case, we put i = n/2 in the above formula of T(n).

```
T(n) = T(n/2) + T(n - 1 - n/2) + cn
= T(n/2) + T(n/2 - 1) + cn
~ 2 T(n/2)+ cn
T(n) = 2 T(n/2)+ cn
```

This recurrence is similar to the recurrence for merge sort, for which the solution is O(n log n). So the best-case time complexity of quicksort = O(n log n). **Note:** We can solve this using the recursion tree method or the master method.

Suppose all permutations of input are equally likely. Now, when we run the quicksort algorithm on random input, partitioning is highly unlikely to happen in the same way at each level of recursion. The idea is simple: The behaviour of quicksort will depend on the relative order of values in the input.

Here some of the splits will be reasonably well balanced and some of the splits will be pretty unbalanced. So the partition process will generate a mix of good (balanced partition) and bad (unbalanced partition) splits in the average case. These good and bad splits will be distributed randomly throughout the recursion tree.

For a better understanding of the analysis, Suppose good and bad splits appear at alternate levels of the tree.

Suppose, at the root, the partition generates a **good split**, and at the next level, the partition generates a **bad split**. The cost of the partition process will be O(n) at both levels. Therefore, the combined partitioning cost of the bad split followed by the good split is O(n).

Overall, this is equivalent to a single level of partitioning, which resembles the scenario of a balanced partition. As a result, the height of the recursion tree will be O(log n), the combined cost of each level will be O(n), and the average-case running time of quicksort will be O(n log n). **Note:** Average case time complexity is the same as the best case, but the value of the constant factor in O(n log n) will be slightly higher in the average case.

For better visualization, let’s assume that the partition algorithm always produces a partially unbalanced split in the ratio of 9 to 1. The recurrence relation for this would be: T(n) = T(9n/10) + T(n/10) + cn. Image source: CLRS Book.

We can notice following things from above diagram:

- The left subtree is decreasing fast with a factor of 1/10. So the depth of left subtree is equal to log10(n).
- The right subtree is decreasing slowly with a factor of 9/10. So the depth of right subtree is equal to log10/9(n).
**Note:**log10/9(n) = O(logn). - At each level of recursion, the cost of partition is at most cn. After doing the sum of cost at each level of recursion tree, quick sort cost is O(nlogn).
- With 9-to-1 proportional split at each level of recursion, which intuitively seems unbalanced, quicksort runs in O(nlogn) time. Even a 99-to-1 split can produce O(nlogn) time complexity!
- In general, any split of constant proportionality produces a recursion tree of depth O(logn), where cost at each level is O(n). So time complexity is O(nlogn) whenever the split has constant proportionality.

Quicksort is an in-place sorting algorithm because it does not use extra space in the code. However, every recursive program uses a call stack in the background to execute recursion. Therefore, the space complexity of quicksort will depend on the size of the recursion call stack, which will be equal to the height of the recursion tree. As mentioned in the above analysis, the height of the recursion tree depends on the partition process.

In the worst-case scenario, the partition will always be unbalanced, resulting in only one recursive call at each level of recursion. In other words, the generated recursion tree will be skewed in nature, and the height of the recursion tree will be O(n). As a result, recursion will require a call stack of size O(n) and the worst-case space complexity of quicksort will be O(n).

In the best-case scenario, the partition will always be balanced, resulting in two recursive calls at each level of recursion. In such a scenario, the generated recursion tree will be balanced in nature. So, the height of the recursion tree will be O(log n), and recursion will require a call stack of size O(log n). Therefore, the best-case space complexity of quicksort is O(log n).

In the above algorithm, we assume that all input values are distinct. How do we modify quick sort algorithm when input values are repeated? Let’s think!

We can modify the partition algorithm and separate input values into three groups: Values less than pivot, values equal to pivot and values greater than pivot. Values equal to pivot are already sorted, so only less-than and greater-than partitions need to be recursively sorted. Because of this, we will return two index **startIndex** and **endIndex i**n the modified partition algorithm.

- All elements of X[] from
**l**to**startIndex - 1**are less than pivot. - All elements of X[] from
**startIndex**to**endIndex**are equal to pivot. - All elements of X[] from
**endIndex + 1**to**r**are greater than pivot.

```
void quickSort(X[], l, r)
{
if (l < r)
{
[leftIndex, rightIndex] = partition(X, l, r)
quickSort(X, l, leftIndex - 1)
quickSort(X, rightIndex + 1, r)
}
}
```

As we know from the analysis, the efficiency of quicksort depends on the smart selection of the pivot element. So there can be many ways to choose the pivot in quicksort:

- Always choosing the last element as the pivot.
- Always choosing the first element as the pivot.
- Selecting a random element as the pivot.
- Selecting the median as the pivot.

In the above implementation, we have chosen the rightmost element as the pivot. This can result in a worst-case situation when the input array is sorted. The best idea would be to choose a random pivot that minimizes the chances of a worst-case at each level of recursion. Selecting the median element as the pivot can also be acceptable in the majority of cases.

```
mid = l + (r - l)/ 2
if X[mid] < X[l]
swap (X[l], X[mid])
if X[hi] < X[lo]
swap (X[l], X[r])
if X[mid] < X[r]
swap (X[mid], X[r])
pivot = X[r]
```

- When input is almost sorted, we can improve the running time of quicksort by taking advantage of the fast running time of insertion sort. How can we implement this? What would be the time complexity of this algorithm?
- Quick sort is an unstable sorting algorithm. This means that the relative order of equal elements might not be preserved after sorting. How can we modify the quick sort algorithm to make it stable?
- Quicksort is an in-place sorting algorithm where we use extra space only for recursion call stack but not for manipulating input. What would be the average case space complexity of quicksort? How can we optimize space complexity to O(logn) in the worst case?
- How can we implement quicksort iteratively using a stack?
- When do we prefer quicksort over merge sort for sorting arrays?
- How do we sort a linked list using the quicksort algorithm? Why do we prefer merge sort over quicksort for sorting linked lists?
- The number of comparisons in quicksort is equal to the number of comparisons in constructing a BST with a sequence of insertion operations. In other words, quicksort is also a similar version of BST sort, where both algorithms make the exact number of comparisons but in a different order.
- What value will the partition function return for the pivotIndex when all elements have the same value? What is the time complexity of quicksort when all elements are the same?
- We can use quick sort partitioning process to efficiently find the kth smallest element or largest element in an unsorted array. This problem is called Quick-Select, which provides a solution in linear time complexity on average.
- QuickSort is widely used in various programming languages and libraries due to its efficiency. Many programming languages incorporate QuickSort as their default sorting algorithm or provide it as a standard library function.

- The idea of Hoare partition
- Quicksort optimization using tail recursion approach

- Algorithms by CLRS
- The Algorithm Design Manual by Steven Skiena

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