3Way QuickSort (Dutch National Flag) (original) (raw)

Last Updated : 23 Jul, 2025

In simple QuickSort algorithm, we select an element as pivot, partition the array around a pivot and recur for subarrays on the left and right of the 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 a pivot in Simple Quick Sort, we fix only one 4 and recursively process remaining occurrences.
The idea of 3 way Quick Sort is to process all occurrences of the pivot and is based on Dutch National Flag algorithm.

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 the implementation of the above algorithm.

C++ `

// 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 done
    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 code int main() { int a[] = { 4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4 }; int size = sizeof(a) / sizeof(int);

  // Function Call
printarr(a, size);
quicksort(a, 0, size - 1);
printarr(a, size);
return 0;

}

Java

// Java program for 3-way quick sort import java.util.*; class GFG {

static int i, 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 done
    if (i >= j)
        break;

    // Swap, so that smaller goes on left greater goes
    // on right
    int temp = a[i];
      a[i] = a[j];
      a[j] = temp;

    // Move all same left occurrence of pivot to
    // beginning of array and keep count using p
    if (a[i] == v) {
        p++;
        temp = a[i];
        a[i] = a[p];
        a[p] = temp;

    }

    // Move all same right occurrence of pivot to end of
    // array and keep count using q
    if (a[j] == v) {
        q--;
        temp = a[q];
        a[q] = a[j];
        a[j] = temp;
    }
}

// Move pivot element to its correct index
int temp = a[i];
  a[i] = a[r];
  a[r] = temp;

// Move all left same occurrences from beginning
// to adjacent to arr[i]
j = i - 1;
for (int k = l; k < p; k++, j--)
{
    temp = a[k];
      a[k] = a[j];
      a[j] = temp;
}

// Move all right same occurrences from end
// to adjacent to arr[i]
i = i + 1;
for (int k = r - 1; k > q; k--, i++)
{
    temp = a[i];
      a[i] = a[k];
      a[k] = temp;
}

}

// 3-way partition based quick sort static void quicksort(int a[], int l, int r) { if (r <= l) return;

i = 0; j = 0;

// Note that i and j are passed as reference
partition(a, l, r);

// Recur
quicksort(a, l, j);
quicksort(a, i, r);

}

// A utility function to print an array static void printarr(int a[], int n) { for (int i = 0; i < n; ++i) System.out.printf("%d ", a[i]); System.out.printf("\n"); }

// Driver code public static void main(String[] args) { int a[] = { 4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4 }; int size = a.length;

  // Function Call
printarr(a, size);
quicksort(a, 0, size - 1);
printarr(a, size);

} }

// This code is contributed by Rajput-Ji

Python3

''' This function partitions a[] in three parts a) a[first..start] contains all elements smaller than pivot b) a[start+1..mid-1] contains all occurrences of pivot c) a[mid..last] contains all elements greater than pivot

''' def partition(arr, first, last, start, mid):

pivot = arr[last]
end = last

# Iterate while mid is not greater than end.
while (mid[0] <= end):
    
    # Inter Change position of element at the starting if it's value is less than pivot.
    if (arr[mid[0]] < pivot):
        
        arr[mid[0]], arr[start[0]] = arr[start[0]], arr[mid[0]]
        
        mid[0] = mid[0] + 1
        start[0] = start[0] + 1
        
    # Inter Change position of element at the end if it's value is greater than pivot.
    elif (arr[mid[0]] > pivot):
        
        arr[mid[0]], arr[end] = arr[end], arr[mid[0]]
        
        end = end - 1
        
    else:
        mid[0] = mid[0] + 1

Function to sort the array elements in 3 cases

def quicksort(arr,first,last): # First case when an array contain only 1 element if (first >= last): return

# Second case when an array contain only 2 elements
if (last == first + 1):
    
    if (arr[first] > arr[last]):
        
        arr[first], arr[last] = arr[last], arr[first]
        
        return

# Third case when an array contain more than 2 elements
start = [first]
mid = [first]

# Function to partition the array.
partition(arr, first, last, start, mid)

# Recursively sort sublist containing elements that are less than the pivot.
quicksort(arr, first, start[0] - 1)

# Recursively sort sublist containing elements that are more than the pivot
quicksort(arr, mid[0], last)

Code Start from here

arr = [4,9,4,4,1,9,4,4,9,4,4,1,4]

Call the quicksort function.

quicksort(arr,0,len(arr) - 1)

print arr after sorting the elements

print(arr)

C#

// C# program for 3-way quick sort using System;

class GFG { // A function which is used to swap values static void swap(ref T lhs, ref T rhs) { T temp; temp = lhs; lhs = rhs; rhs = temp; } /* 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 */ public static void partition(int[] a, int l, int r, ref int i, ref 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 done
        if (i >= j)
            break;

        // Swap, so that smaller goes on left greater
        // goes on right
        swap(ref a[i], ref a[j]);

        // Move all same left occurrence of pivot to
        // beginning of array and keep count using p
        if (a[i] == v) {
            p++;
            swap(ref a[p], ref a[i]);
        }

        // Move all same right occurrence of pivot to
        // end of array and keep count using q
        if (a[j] == v) {
            q--;
            swap(ref a[j], ref a[q]);
        }
    }

    // Move pivot element to its correct index
    swap(ref a[i], ref 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(ref a[k], ref 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(ref a[i], ref a[k]);
}

// 3-way partition based quick sort
public static void quicksort(int[] a, int l, int r)
{
    if (r <= l)
        return;

    int i = 0, j = 0;

    // Note that i and j are passed as reference
    partition(a, l, r, ref i, ref j);

    // Recur
    quicksort(a, l, j);
    quicksort(a, i, r);
}

// A utility function to print an array
public static void printarr(int[] a, int n)
{
    for (int i = 0; i < n; ++i)
        Console.Write(a[i] + " ");
    Console.Write("\n");
}

// Driver code
static void Main()
{
    int[] a = { 4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4 };
    int size = a.Length;
    
      // Function Call
      printarr(a, size);
    quicksort(a, 0, size - 1);
    printarr(a, size);
}
// This code is contributed by DrRoot_

}

JavaScript

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

Time Complexity: O(N * log(N))

Where 'N' is the number of elements in the given array/list

The average number of recursive calls made to the quicksort function is log N, and every time the function is called we are traversing the given array/list which requires O(N) time. Thus, the total time complexity is O(N * log (N)).

Space Complexity: O(log N)

where ‘N’ is the number of elements in the given array/list.

Thanks to Utkarsh for suggesting above implementation.

Another Implementation using Dutch National Flag Algorithm

C++ `

// C++ program for 3-way quick sort #include <bits/stdc++.h> using namespace std;

void swap(int* a, int* b) { int temp = *a; *a = *b; *b = temp; }

// 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"); }

// It uses Dutch National Flag Algorithm void partition(int a[], int low, int high, int& i, int& j) { // To handle 2 elements if (high - low <= 1) { if (a[high] < a[low]) swap(&a[high], &a[low]); i = low; j = high; return; }

int mid = low;
int pivot = a[high];
while (mid <= high) {
    if (a[mid] < pivot)
        swap(&a[low++], &a[mid++]);
    else if (a[mid] == pivot)
        mid++;
    else if (a[mid] > pivot)
        swap(&a[mid], &a[high--]);
}

// update i and j
i = low - 1;
j = mid; // or high+1

}

// 3-way partition based quick sort void quicksort(int a[], int low, int high) { if (low >= high) // 1 or 0 elements return;

int i, j;

// Note that i and j are passed as reference
partition(a, low, high, i, j);

// Recur two halves
quicksort(a, low, i);
quicksort(a, j, high);

}

// Driver Code int main() { int a[] = { 4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4 }; // int a[] = {4, 39, 54, 14, 31, 89, 44, 34, 59, 64, 64, // 11, 41}; int a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; // int a[] = {91, 82, 73, 64, 55, 46, 37, 28, 19, 10}; // int a[] = {4, 9, 4, 4, 9, 1, 1, 1}; int size = sizeof(a) / sizeof(int);

// Function Call
printarr(a, size);
quicksort(a, 0, size - 1);
printarr(a, size);
return 0;

}

Java

// Java program for 3-way quick sort import java.util.*; class GFG {

static void swap(int[] arr, int i, int j) { int temp = arr[i]; arr[i] = arr[j]; arr[j] = temp; }

// A utility function to print an array

static void printarr(int a[], int n) { for (int i = 0; i < n; ++i) System.out.printf("%d ", a[i]); System.out.printf("\n"); }

// It uses Dutch National Flag Algorithm static void partition(int a[], int low, int high, int i, int j) { // To handle 2 elements if (high - low <= 1) { if (a[high] < a[low]) swap(a, high, low); i = low; j = high; return; }

int mid = low;
int pivot = a[high];
while (mid <= high) {
    if (a[mid] < pivot)
        swap(a, low++, mid++);
    else if (a[mid] == pivot)
        mid++;
    else if (a[mid] > pivot)
        swap(a, mid, high--);
}

// update i and j
i = low - 1;
j = mid; // or high+1

}

// 3-way partition based quick sort static void quicksort(int a[], int low, int high) { if (low >= high) // 1 or 0 elements return;

int i=low, j=high;

// Note that i and j are passed
partition(a, low, high, i, j);

// Recur two halves
quicksort(a, low, i);
quicksort(a, j, high);

}

// Driver Code public static void main(String[] args) { int a[] = { 4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4 }; // int a[] = {4, 39, 54, 14, 31, 89, 44, 34, 59, 64, 64, // 11, 41}; int a[] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}; // int a[] = {91, 82, 73, 64, 55, 46, 37, 28, 19, 10}; // int a[] = {4, 9, 4, 4, 9, 1, 1, 1}; int size = a.length;

// Function Call
printarr(a, size);
quicksort(a, 0, size - 1);
printarr(a, size);

} }

// This code is contributed by Pushpesh Raj.

Python3

python3 program for 3-way quick sort

Function to find lexicographically minimum

def swap(a,i,j) : temp = a[i] a[i] = a[j] a[j] = temp

A utility function to print an array

def printarr(a, n) :

for i in range (n) :
    print(a[i],end=' ')

print("\n")

''' 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 '''

It uses Dutch National Flag Algorithm

def partition(a, low, high, i, j) : # To handle 2 elements if high - low <= 1 : if a[high] < a[low] : swap(a,high, low) i = low j = high return

mid = low; pivot = a[high];
while mid <= high :
    if a[mid] < pivot :
        swap(a,low,mid)
        low+=1
        mid+=1
    elif a[mid] == pivot :
        mid+=1
    elif a[mid] > pivot :
        swap(a,mid,high)
        high-=1

# update i and j
i = low - 1
j = mid # or high+1

3-way partition based quick sort

def quickSort(a,low,high) :

if low >= high : # 1 or 0 elements
    return

i = low; j = high;

# Note that i and j are passed as reference
partition(a,low,high,i,j)

# Recur two halves
quickSort(a,low,i)
quickSort(a,j,high)

Driver code

if name == "main" :

a = [4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4]

size = len(a)

printarr(a,size)

quickSort(a,0,size-1)

printarr(a,size)

#this code is contributed by aditya942003patil

C#

// C# program for 3-way quick sort using System;

class GFG { // A function which is used to swap values static void swap(ref T lhs, ref T rhs) { T temp; temp = lhs; lhs = rhs; rhs = temp; }

// A utility function to print an array
public static void printarr(int[] a, int n)
{
    for (int i = 0; i < n; ++i)
        Console.Write(a[i] + " ");
    Console.Write("\n");
}

/* 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 */

// It uses Dutch National Flag Algorithm
public static void partition(int[] a, int low, int high,
                             ref int i, ref int j)
{
    // To handle 2 elements
    if (high - low <= 1) {
        if (a[high] < a[low])
            swap(ref a[high], ref a[low]);
        i = low;
        j = high;
        return;
    }

    int mid = low;
    int pivot = a[high];
    while (mid <= high) {
        if (a[mid] < pivot)
            swap(ref a[low++], ref a[mid++]);
        else if (a[mid] == pivot)
            mid++;
        else if (a[mid] > pivot)
            swap(ref a[mid], ref a[high--]);
    }

    // update i and j
    i = low - 1;
    j = mid; // or high+1
}

// 3-way partition based quick sort
public static void quicksort(int[] a, int low, int high)
{
    if (low >= high) // 1 or 0 elements
        return;

    int i = 0, j = 0;

    // Note that i and j are passed as reference
    partition(a, low, high, ref i, ref j);

    // Recur two halves
    quicksort(a, low, i);
    quicksort(a, j, high);
}

// Driver code
static void Main()
{
    int[] a = { 4, 9, 4, 4, 1, 9, 4, 4, 9, 4, 4, 1, 4 };
    // int[] a = {4, 39, 54, 14, 31, 89, 44, 34, 59, 64,
    // 64, 11, 41}; int[] a = {1, 2, 3, 4, 5, 6, 7, 8, 9,
    // 10}; int[] a = {91, 82, 73, 64, 55, 46, 37, 28,
    // 19, 10}; int[] a = {4, 9, 4, 4, 9, 1, 1, 1};
    int size = a.Length;
    
      // Function Call
    printarr(a, size);
    quicksort(a, 0, size - 1);
    printarr(a, size);
}
// This code is contributed by DrRoot_

}

JavaScript

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

Time Complexity: O(N2) The time complexity for this code is O(N*log(N)) in the average and best-case scenarios, and O(N^2) in the worst-case scenario.

Space Complexity: O(log N)
Thanks Aditya Goel for this implementation.
Reference:
https://algs4.cs.princeton.edu/lectures/23DemoPartitioning.pdf
http://www.sorting-algorithms.com/quick-sort-3-way