3-Way QuickSort

In simple QuickSort algorithm, we select an element as pivot, partition the array around pivot and recur for subarrays on left and right of pivot. 
Consider an array which has many redundant elements. For example, {1, 4, 2, 4, 2, 4, 1, 2, 4, 1, 2, 2, 2, 2, 4, 1, 4, 4, 4}. If 4 is picked as pivot in Simple QuickSort, we fix only one 4 and recursively process remaining occurrences.

The idea of 3 way QuickSort is to process all occurrences of pivot.

In 3 Way QuickSort, an array arr[l..r] is divided in 3 parts:
a) arr[l..i] elements less than pivot.
b) arr[i+1..j-1] elements equal to pivot.
c) arr[j..r] elements greater than pivot. 

Below is C++ implementation of above algorithm.

// C++ program for 3-way quick sort #include <bits/stdc++.h> using namespace std;   /* This function partitions a[] in three parts    a) a[l..i] contains all elements smaller than pivot    b) a[i+1..j-1] contains all occurrences of pivot    c) a[j..r] contains all elements greater than pivot */ void partition(int a[], int l, int r, int &i, int &j) {     i = l-1, j = r;     int p = l-1, q = r;     int v = a[r];       while (true)     {         // From left, find the first element greater than         // or equal to v. This loop will definitely terminate         // as v is last element         while (a[++i] < v);           // From right, find the first element smaller than or         // equal to v         while (v < a[--j])             if (j == l)                 break;           // If i and j cross, then we are don         if (i >= j) break;           // Swap, so that smaller goes on left greater goes on right         swap(a[i], a[j]);           // Move all same left occurrence of pivot to beginning of         // array and keep count using p         if (a[i] == v)         {             p++;             swap(a[p], a[i]);         }           // Move all same right occurrence of pivot to end of array         // and keep count using q         if (a[j] == v)         {             q--;             swap(a[j], a[q]);         }     }       // Move pivot element to its correct index     swap(a[i], a[r]);       // Move all left same occurrences from beginning     // to adjacent to arr[i]     j = i-1;     for (int k = l; k < p; k++, j--)         swap(a[k], a[j]);       // Move all right same occurrences from end     // to adjacent to arr[i]     i = i+1;     for (int k = r-1; k > q; k--, i++)         swap(a[i], a[k]); }   // 3-way partition based quick sort void quicksort(int a[], int l, int r) {     if (r <= l) return;       int i, j;       // Note that i and j are passed as reference     partition(a, l, r, i, j);       // Recur     quicksort(a, l, j);     quicksort(a, i, r); }   // A utility function to print an array void printarr(int a[], int n) {     for (int i = 0; i < n; ++i)         printf("%d  ", a[i]);     printf("\n"); }   // Driver program int main() {     int a[] = {4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4};     int size = sizeof(a) / sizeof(int);     printarr(a, size);     quicksort(a, 0, size - 1);     printarr(a, size);     return 0; }

Output:

4  9  4  4  1  9  4  4  9  4  4  1  4
1  1  4  4  4  4  4  4  4  4  9  9  9

Thanks to Utkarsh for suggesting above implementation.