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Count quadruplets with sum K from given array
  • Last Updated : 04 Dec, 2020

Given an array arr[] of size N and an integer S, the task is to find the count of quadruplets present in the given array having sum S

Examples:

Input: arr[] = {1, 5, 3, 1, 2, 10}, S = 20
Output: 1
Explanation: Only quadruplet satisfying the conditions is arr[1] + arr[2] + arr[4] + arr[5] = 5 + 3 + 2 + 10 = 20.

Input: N = 6, S = 13, arr[] = {4, 5, 3, 1, 2, 4}
Output: 3
Explanation: Three quadruplets with sum 13 are: 

  1. arr[0] + arr[2] + arr[4] + arr[5] = 4 + 3 + 2 + 4 = 13
  2. arr[0] + arr[1] + arr[2] + arr[3] = 4 + 5 + 3 + 1 = 13
  3. arr[1] + arr[2] + arr[3] + arr[5] = 5 + 3 + 1 + 4 = 13
 

Naive Approach: The idea is to generate all possible combinations of length 4 from the given array. For each quadruplet, if the sum equals S, then increment the counter by 1. After checking all the quadruplets, print the counter as the total number of quadruplets having sum S.



Below is the implementation of the above approach:

C++

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// C++ program for the above approach
 
#include <iostream>
using namespace std;
 
// Function to return the number of
// quadruplets with the given sum
int countSum(int a[], int n, int sum)
{
    // Initialize variables
    int i, j, k, l;
 
    // Initialize answer
    int count = 0;
 
    // All possible first elements
    for (i = 0; i < n - 3; i++) {
 
        // All possible second elements
        for (j = i + 1; j < n - 2; j++) {
 
            // All possible third elements
            for (k = j + 1; k < n - 1; k++) {
 
                // All possible fourth elements
                for (l = k + 1; l < n; l++) {
 
                    // Increment counter by 1
                    // if quadruplet sum is S
                    if (a[i] + a[j]
                            + a[k] + a[l]
                        == sum)
                        count++;
                }
            }
        }
    }
 
    // Return the final count
    return count;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << countSum(arr, N, S);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
   
// Function to return the number of
// quadruplets with the given sum
static int countSum(int a[], int n, int sum)
{
     
    // Initialize variables
    int i, j, k, l;
 
    // Initialize answer
    int count = 0;
 
    // All possible first elements
    for(i = 0; i < n - 3; i++)
    {
         
        // All possible second elements
        for(j = i + 1; j < n - 2; j++)
        {
             
            // All possible third elements
            for(k = j + 1; k < n - 1; k++)
            {
                 
                // All possible fourth elements
                for(l = k + 1; l < n; l++)
                {
                     
                    // Increment counter by 1
                    // if quadruplet sum is S
                    if (a[i] + a[j] +
                        a[k] + a[l] == sum)
                        count++;
                }
            }
        }
    }
 
    // Return the final count
    return count;
}
 
// Driver Code
public static void main(String args[])
{
     
    // Given array arr[]
    int arr[] = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = arr.length;
 
    // Function Call
    System.out.print(countSum(arr, N, S));
}
}
 
// This code is contributed by bgangwar59

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Python3

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# Python3 program for the above approach
 
# Function to return the number of
# quadruplets with the given sum
def countSum(a, n, sum):
     
    # Initialize variables
    # i, j, k, l
 
    # Initialize answer
    count = 0
 
    # All possible first elements
    for i in range(n - 3):
         
        # All possible second elements
        for j in range(i + 1, n - 2):
             
            # All possible third elements
            for k in range(j + 1, n - 1):
                 
                # All possible fourth elements
                for l in range(k + 1, n):
                     
                    # Increment counter by 1
                    # if quadruplet sum is S
                    if (a[i] + a[j] + a[k] + a[l]== sum):
                        count += 1
                         
    # Return the final count
    return count
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr = [ 4, 5, 3, 1, 2, 4 ]
 
    # Given sum S
    S = 13
 
    N = len(arr)
     
    # Function Call
    print(countSum(arr, N, S))
     
# This code is contributed by mohit kumar 29

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C#

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// C# program for the above approach
class GFG{
   
// Function to return the number of
// quadruplets with the given sum
static int countSum(int []a, int n, int sum)
{
     
    // Initialize variables
    int i, j, k, l;
 
    // Initialize answer
    int count = 0;
 
    // All possible first elements
    for(i = 0; i < n - 3; i++)
    {
         
        // All possible second elements
        for(j = i + 1; j < n - 2; j++)
        {
             
            // All possible third elements
            for(k = j + 1; k < n - 1; k++)
            {
                 
                // All possible fourth elements
                for(l = k + 1; l < n; l++)
                {
                     
                    // Increment counter by 1
                    // if quadruplet sum is S
                    if (a[i] + a[j] +
                        a[k] + a[l] == sum)
                        count++;
                }
            }
        }
    }
 
    // Return the final count
    return count;
}
 
// Driver Code
public static void Main()
{
     
    // Given array arr[]
    int []arr = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = arr.Length;
 
    // Function Call
    System.Console.Write(countSum(arr, N, S));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

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Output:

3

Time Complexity: O(N4)
Auxiliary Space: O(N)

Better Approach: To optimize the above approach, the idea is to use a Map data structure. Follow the steps below to solve the problem:

  • Initialize the counter count with 0 to store the number of quadruplets.
  • Traverse the given array over the range [0, N – 3)using the variable i. For each element arr[i], traverse the array again over the range [i + 1, N – 2) using the variable j and do the following:
    • Find the value of the required sum(say req) as (S – arr[i] – arr[j]).
    • Initialize the count_twice with 0 that will store the count of ordered pairs in the above subarray with sum (S – arr[i] – arr[j]).
    • After finding the count_twice, update the count by count_twice / 2.
  • After the above steps, print the value of count as the result.

Below is the implementation of the above idea:

C++

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// C++ program for the above approach
 
#include <iostream>
#include <unordered_map>
using namespace std;
 
// Function to return the number of
// quadruplets having given sum
int countSum(int a[], int n, int sum)
{
 
    // Initialize variables
    int i, j, k, l;
 
    // Initialize answer
    int count = 0;
 
    // All possible first elements
    for (i = 0; i < n - 3; i++) {
 
        // All possible second element
        for (j = i + 1; j < n - 2; j++) {
            int req = sum - a[i] - a[j];
 
            // Use map to find the
            // fourth element
            unordered_map<int, int> m;
 
            // All possible third elements
            for (k = j + 1; k < n; k++)
                m[a[k]]++;
 
            int twice_count = 0;
 
            // Calculate number of valid
            // 4th elements
            for (k = j + 1; k < n; k++) {
 
                // Update the twice_count
                twice_count += m[req - a[k]];
 
                if (req - a[k] == a[k])
                    twice_count--;
            }
 
            // Unordered pairs
            count += twice_count / 2;
        }
    }
 
    // Return answer
    return count;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << countSum(arr, N, S);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to return the number of
// quadruplets having given sum
static int countSum(int a[], int n, int sum)
{
     
    // Initialize variables
    int i, j, k, l;
 
    // Initialize answer
    int count = 0;
 
    // All possible first elements
    for(i = 0; i < n - 3; i++)
    {
         
        // All possible second element
        for(j = i + 1; j < n - 2; j++)
        {
            int req = sum - a[i] - a[j];
 
            // Use map to find the
            // fourth element
            HashMap<Integer, Integer> m = new HashMap<>();
 
            // All possible third elements
            for(k = j + 1; k < n; k++)
                if (m.containsKey(a[k]))
                {
                    m.put(a[k], m.get(a[k]) + 1);
                }
                else
                {
                    m.put(a[k], 1);
                }
                 
            int twice_count = 0;
 
            // Calculate number of valid
            // 4th elements
            for(k = j + 1; k < n; k++)
            {
                 
                // Update the twice_count
                if (m.containsKey(req - a[k]))
                    twice_count += m.get(req - a[k]);
 
                if (req - a[k] == a[k])
                    twice_count--;
            }
 
            // Unordered pairs
            count += twice_count / 2;
        }
    }
 
    // Return answer
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = arr.length;
 
    // Function Call
    System.out.print(countSum(arr, N, S));
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program for the above approach
 
# Function to return the number of
# quadruplets having given sum
def countSum(a, n, sum):
     
    # Initialize variables
    # Initialize answer
    count = 0
 
    # All possible first elements
    for i in range(n - 3):
         
        # All possible second element
        for j in range(i + 1, n - 2, 1):
            req = sum - a[i] - a[j]
 
            # Use map to find the
            # fourth element
            m = {}
 
            # All possible third elements
            for k in range(j + 1, n, 1):
                m[a[k]] = m.get(a[k], 0) + 1
 
            twice_count = 0
 
            # Calculate number of valid
            # 4th elements
            for k in range(j + 1, n, 1):
                 
                # Update the twice_count
                twice_count += m.get(req - a[k], 0)
 
                if (req - a[k] == a[k]):
                    twice_count -= 1
 
            # Unordered pairs
            count += twice_count // 2
 
    # Return answer
    return count
 
# Driver Code
if __name__ == '__main__':
     
    # Given array arr[]
    arr =  [ 4, 5, 3, 1, 2, 4 ]
 
    # Given sum S
    S = 13
 
    N =  len(arr)
 
    # Function Call
    print(countSum(arr, N, S))
 
# This code is contributed by ipg2016107

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the number of
// quadruplets having given sum
static int countSum(int []a, int n,
                    int sum)
{
     
    // Initialize variables
    int i, j, k;
    //int l;
     
    // Initialize answer
    int count = 0;
 
    // All possible first elements
    for(i = 0; i < n - 3; i++)
    {
         
        // All possible second element
        for(j = i + 1; j < n - 2; j++)
        {
            int req = sum - a[i] - a[j];
 
            // Use map to find the
            // fourth element
            Dictionary<int,
                       int> m = new Dictionary<int,
                                               int>();
 
            // All possible third elements
            for(k = j + 1; k < n; k++)
                if (m.ContainsKey(a[k]))
                {
                    m[a[k]]++;
                }
                else
                {
                    m.Add(a[k], 1);
                }
                 
            int twice_count = 0;
 
            // Calculate number of valid
            // 4th elements
            for(k = j + 1; k < n; k++)
            {
                 
                // Update the twice_count
                if (m.ContainsKey(req - a[k]))
                    twice_count += m[req - a[k]];
 
                if (req - a[k] == a[k])
                    twice_count--;
            }
 
            // Unordered pairs
            count += twice_count / 2;
        }
    }
 
    // Return answer
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = arr.Length;
 
    // Function Call
    Console.Write(countSum(arr, N, S));
}
}
 
// This code is contributed by Princi Singh

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Output:

3

Time complexity: O(N3) where N is the size of the given array, 
Auxiliary Space: O(N)

Efficient Approach: The idea is similar to the above approach using a map. In this approach, fix the 3rd element, then find and store the frequency of sums of all possible first two elements of any quadruplet of the given array. Follow the below steps to solve the problem:

  1. Initialize the counter count to store the total quadruplets with the given sum S and a map to store all possible sums for the first two elements of each possible quadruplet.
  2. Traverse the given array over the range [0, N – 1] using the variable i where arr[i] is the fixed 3rd element.
  3. Then for each element arr[i] in the above step, traverse the given array over the range [i + 1, N – 1] using the variable j and increment the counter count by map[arr[i] + arr[j]].
  4. After traversing from the above loop, for each element arr[i], traverse the array arr[] from j = 0 to i – 1 and increment the frequency of any sum arr[i] + arr[j] of the first two elements of any possible quadruplet by 1 i.e., increment map[arr[i] + arr[j]] by 1.
  5. Repeat the above steps for each element arr[i] and then print the counter count as the total number of quadruplets with the given sum S.

Below is the implementation of the above idea:

C++

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// C++ program for the above approach
 
#include <iostream>
#include <unordered_map>
using namespace std;
 
// Function to return the number of
// quadruplets having the given sum
int countSum(int a[], int n, int sum)
{
 
    // Initialize variables
    int i, j, k;
 
    // Initialize answer
    int count = 0;
 
    // Store the frequency of sum
    // of first two elements
    unordered_map<int, int> m;
 
    // Traverse from 0 to N-1, where
    // arr[i] is the 3rd element
    for (i = 0; i < n - 1; i++) {
 
        // All possible 4th elements
        for (j = i + 1; j < n; j++) {
 
            // Sum of last two element
            int temp = a[i] + a[j];
 
            // Frequency of sum of first
            // two elements
            if (temp < sum)
                count += m[sum - temp];
        }
        for (j = 0; j < i; j++) {
 
            // Store frequency of all possible
            // sums of first two elements
            int temp = a[i] + a[j];
 
            if (temp < sum)
                m[temp]++;
        }
    }
 
    // Return the answer
    return count;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function Call
    cout << countSum(arr, N, S);
 
    return 0;
}

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Java

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// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to return the number of
// quadruplets having the given sum
static int countSum(int a[], int n, int sum)
{
     
    // Initialize variables
    int i, j, k;
 
    // Initialize answer
    int count = 0;
 
    // Store the frequency of sum
    // of first two elements
    HashMap<Integer, Integer> m = new HashMap<>();
 
    // Traverse from 0 to N-1, where
    // arr[i] is the 3rd element
    for(i = 0; i < n - 1; i++)
    {
         
        // All possible 4th elements
        for(j = i + 1; j < n; j++)
        {
             
            // Sum of last two element
            int temp = a[i] + a[j];
 
            // Frequency of sum of first
            // two elements
            if (temp < sum && m.containsKey(sum - temp))
                count += m.get(sum - temp);
        }
        for(j = 0; j < i; j++)
        {
             
            // Store frequency of all possible
            // sums of first two elements
            int temp = a[i] + a[j];
 
            if (temp < sum)
                if (m.containsKey(temp))
                    m.put(temp, m.get(temp) + 1);
                else
                    m.put(temp, 1);
        }
    }
     
    // Return the answer
    return count;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = arr.length;
 
    // Function Call
    System.out.print(countSum(arr, N, S));
}
}
 
// This code is contributed by Princi Singh

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Python3

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# Python3 program for the above approach
from collections import defaultdict
 
# Function to return the number of
# quadruplets having the given sum
def countSum(a, n, sum):
     
    # Initialize answer
    count = 0
 
    # Store the frequency of sum
    # of first two elements
    m = defaultdict(int)
 
    # Traverse from 0 to N-1, where
    # arr[i] is the 3rd element
    for i in range(n - 1):
 
        # All possible 4th elements
        for j in range(i + 1, n):
 
            # Sum of last two element
            temp = a[i] + a[j]
 
            # Frequency of sum of first
            # two elements
            if (temp < sum):
                count += m[sum - temp]
 
        for j in range(i):
 
            # Store frequency of all possible
            # sums of first two elements
            temp = a[i] + a[j]
 
            if (temp < sum):
                m[temp] += 1
 
    # Return the answer
    return count
 
# Driver Code
if __name__ == "__main__":
 
    # Given array arr[]
    arr = [ 4, 5, 3, 1, 2, 4 ]
 
    # Given sum S
    S = 13
 
    N = len(arr)
 
    # Function Call
    print(countSum(arr, N, S))
 
# This code is contributed by chitranayal

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C#

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// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the number of
// quadruplets having the given sum
static int countSum(int []a, int n, int sum)
{
     
    // Initialize variables
    int i, j;
 
    // Initialize answer
    int count = 0;
 
    // Store the frequency of sum
    // of first two elements
    Dictionary<int,
               int> m = new Dictionary<int,
                                       int>();
 
    // Traverse from 0 to N-1, where
    // arr[i] is the 3rd element
    for(i = 0; i < n - 1; i++)
    {
         
        // All possible 4th elements
        for(j = i + 1; j < n; j++)
        {
             
            // Sum of last two element
            int temp = a[i] + a[j];
             
            // Frequency of sum of first
            // two elements
            if (temp < sum && m.ContainsKey(sum - temp))
                count += m[sum - temp];
        }
         
        for(j = 0; j < i; j++)
        {
             
            // Store frequency of all possible
            // sums of first two elements
            int temp = a[i] + a[j];
 
            if (temp < sum)
                if (m.ContainsKey(temp))
                    m[temp]++;
                else
                    m.Add(temp, 1);
        }
    }
     
    // Return the answer
    return count;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 4, 5, 3, 1, 2, 4 };
 
    // Given sum S
    int S = 13;
 
    int N = arr.Length;
     
    // Function Call
    Console.Write(countSum(arr, N, S));
}
}
 
// This code is contributed by shikhasingrajput

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Output:

3

Time Complexity: O(N2)
Auxiliary Space: O(N)

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