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Can QuickSort be implemented in O(nLogn) worst case time complexity?

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  • Difficulty Level : Hard
  • Last Updated : 09 Jan, 2023
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The worst case time complexity of a typical implementation of QuickSort is O(n2). The worst case occurs when the picked pivot is always an extreme (smallest or largest) element. This happens when input array is sorted or reverse sorted and either first or last element is picked as pivot.

Although randomized QuickSort works well even when the array is sorted, there is still possible that the randomly picked element is always extreme. Can the worst case be reduced to O(nLogn)?

The answer is yes, we can achieve O(nLogn) worst case. The idea is based on the fact that the median element of an unsorted array can be found in linear time. So we find the median first, then partition the array around the median element. 
Following is C++ implementation based on above idea. Most of the functions in below program are copied from K’th Smallest/Largest Element in Unsorted Array | Set 3 (Worst Case Linear Time)


/* A worst case O(nLogn) implementation of quicksort */
using namespace std;
// Following functions are taken from
int partition(int arr[], int l, int r, int k);
int kthSmallest(int arr[], int l, int r, int k);
/* A O(nLogn) time complexity function for sorting arr[l..h] */
void quickSort(int arr[], int l, int h)
    if (l < h)
        // Find size of current subarray
        int n = h-l+1;
        // Find median of arr[].
        int med = kthSmallest(arr, l, h, n/2);
        // Partition the array around median
        int p = partition(arr, l, h, med);
        // Recur for left and right of partition
        quickSort(arr, l, p - 1);
        quickSort(arr, p + 1, h);
// A simple function to find median of arr[].  This is called
// only for an array of size 5 in this program.
int findMedian(int arr[], int n)
    sort(arr, arr+n);  // Sort the array
    return arr[n/2];   // Return middle element
// Returns k'th smallest element in arr[l..r] in worst case
int kthSmallest(int arr[], int l, int r, int k)
    // If k is smaller than number of elements in array
    if (k > 0 && k <= r - l + 1)
        int n = r-l+1; // Number of elements in arr[l..r]
        // Divide arr[] in groups of size 5, calculate median
        // of every group and store it in median[] array.
        int i, median[(n+4)/5]; // There will be floor((n+4)/5) groups;
        for (i=0; i<n/5; i++)
            median[i] = findMedian(arr+l+i*5, 5);
        if (i*5 < n) //For last group with less than 5 elements
            median[i] = findMedian(arr+l+i*5, n%5);
        // Find median of all medians using recursive call.
        // If median[] has only one element, then no need
        // of recursive call
        int medOfMed = (i == 1)? median[i-1]:
                                 kthSmallest(median, 0, i-1, i/2);
        // Partition the array around a random element and
        // get position of pivot element in sorted array
        int pos = partition(arr, l, r, medOfMed);
        // If position is same as k
        if (pos-l == k-1)
            return arr[pos];
        if (pos-l > k-1)  // If position is more, recur for left
            return kthSmallest(arr, l, pos-1, k);
        // Else recur for right subarray
        return kthSmallest(arr, pos+1, r, k-pos+l-1);
    // If k is more than number of elements in array
    return INT_MAX;
void swap(int *a, int *b)
    int temp = *a;
    *a = *b;
    *b = temp;
// It searches for x in arr[l..r], and partitions the array
// around x.
int partition(int arr[], int l, int r, int x)
    // Search for x in arr[l..r] and move it to end
    int i;
    for (i=l; i<r; i++)
        if (arr[i] == x)
    swap(&arr[i], &arr[r]);
    // Standard partition algorithm
    i = l;
    for (int j = l; j <= r - 1; j++)
        if (arr[j] <= x)
            swap(&arr[i], &arr[j]);
    swap(&arr[i], &arr[r]);
    return i;
/* Function to print an array */
void printArray(int arr[], int size)
    int i;
    for (i=0; i < size; i++)
        cout << arr[i] << " ";
    cout << endl;
// Driver program to test above functions
int main()
    int arr[] = {1000, 10, 7, 8, 9, 30, 900, 1, 5, 6, 20};
    int n = sizeof(arr)/sizeof(arr[0]);
    quickSort(arr, 0, n-1);
    cout << "Sorted array is\n";
    printArray(arr, n);
    return 0;


Sorted array is
1 5 6 7 8 9 10 20 30 900 1000 

Auxiliary Space: O(n*logn)  where n is size of array.

How is QuickSort implemented in practice – is above approach used? 
Although worst case time complexity of the above approach is O(nLogn), it is never used in practical implementations. The hidden constants in this approach are high compared to normal Quicksort. Following are some techniques used in practical implementations of QuickSort. 
1) Randomly picking up to make worst case less likely to occur (Randomized QuickSort) 
2) Calling insertion sort for small sized arrays to reduce recursive calls. 
3) QuickSort is tail recursive, so tail call optimizations is done.

So the approach discussed above is more of a theoretical approach with O(nLogn) worst case time complexity.

This article is compiled by Shivam. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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