Find four elements that sum to a given value | Set 1 (n^3 solution)

Given an array of integers, find all combination of four elements in the array whose sum is equal to a given value X.
For example, if the given array is {10, 2, 3, 4, 5, 9, 7, 8} and X = 23, then your function should print “3 5 7 8” (3 + 5 + 7 + 8 = 23).

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A Naive Solution is to generate all possible quadruples and compare the sum of every quadruple with X. The following code implements this simple method using four nested loops

C++

// C++ program for naive solution to 
// print all combination of 4 elements 
// in A[] with sum equal to X  
#include <bits/stdc++.h> 
using namespace std; 
  
/* A naive solution to print all combination  
of 4 elements in A[]with sum equal to X */
void findFourElements(int A[], int n, int X) 
{ 
      
// Fix the first element and find other three 
for (int i = 0; i < n - 3; i++) 
{ 
    // Fix the second element and find other two 
    for (int j = i + 1; j < n - 2; j++) 
    { 
          
        // Fix the third element and find the fourth 
        for (int k = j + 1; k < n - 1; k++) 
        { 
            // find the fourth 
            for (int l = k + 1; l < n; l++) 
            if (A[i] + A[j] + A[k] + A[l] == X) 
                cout << A[i] <<", " << A[j]  
                     << ", " << A[k] << ", " << A[l]; 
        }  
    } 
} 
} 
  
// Driver Code 
int main() 
{ 
    int A[] = {10, 20, 30, 40, 1, 2}; 
    int n = sizeof(A) / sizeof(A[0]); 
    int X = 91; 
    findFourElements (A, n, X); 
    return 0; 
} 
  
// This code is contributed 
// by Akanksha Rai 

C

#include <stdio.h> 
  
/* A naive solution to print all combination of 4 elements in A[] 
  with sum equal to X */
void findFourElements(int A[], int n, int X) 
{ 
  // Fix the first element and find other three 
  for (int i = 0; i < n-3; i++) 
  { 
    // Fix the second element and find other two 
    for (int j = i+1; j < n-2; j++) 
    { 
      // Fix the third element and find the fourth 
      for (int k = j+1; k < n-1; k++) 
      { 
        // find the fourth 
        for (int l = k+1; l < n; l++) 
           if (A[i] + A[j] + A[k] + A[l] == X) 
              printf("%d, %d, %d, %d", A[i], A[j], A[k], A[l]); 
      } 
    } 
  } 
} 
  
// Driver program to test above function 
int main() 
{ 
    int A[] = {10, 20, 30, 40, 1, 2}; 
    int n = sizeof(A) / sizeof(A[0]); 
    int X = 91; 
    findFourElements (A, n, X); 
    return 0; 
} 

Java

class FindFourElements  
{ 
  
    /* A naive solution to print all combination of 4 elements in A[] 
       with sum equal to X */
    void findFourElements(int A[], int n, int X)  
    { 
        // Fix the first element and find other three 
        for (int i = 0; i < n - 3; i++)  
        { 
            // Fix the second element and find other two 
            for (int j = i + 1; j < n - 2; j++)  
            { 
                // Fix the third element and find the fourth 
                for (int k = j + 1; k < n - 1; k++)  
                { 
                    // find the fourth 
                    for (int l = k + 1; l < n; l++)  
                    { 
                        if (A[i] + A[j] + A[k] + A[l] == X)  
                            System.out.print(A[i]+" "+A[j]+" "+A[k]  
                                                                 +" "+A[l]); 
                    } 
                } 
            } 
        } 
    } 
  
    // Driver program to test above functions 
    public static void main(String[] args)  
    { 
        FindFourElements findfour = new FindFourElements(); 
        int A[] = {10, 20, 30, 40, 1, 2}; 
        int n = A.length; 
        int X = 91; 
        findfour.findFourElements(A, n, X); 
    } 
} 

Python3

# A naive solution to print all combination 
# of 4 elements in A[] with sum equal to X  
def findFourElements(A, n, X): 
      
    # Fix the first element and find  
    # other three 
    for i in range(0,n-3): 
          
        # Fix the second element and  
        # find other two 
        for j in range(i+1,n-2): 
              
            # Fix the third element  
            # and find the fourth 
            for k in range(j+1,n-1): 
                  
                # find the fourth 
                for l in range(k+1,n): 
                      
                    if A[i] + A[j] + A[k] + A[l] == X: 
                        print ("%d, %d, %d, %d"
                        %( A[i], A[j], A[k], A[l])) 
  
# Driver program to test above function 
A = [10, 2, 3, 4, 5, 9, 7, 8] 
n = len(A) 
X = 23
findFourElements (A, n, X) 
  
# This code is contributed by shreyanshi_arun 

C#

// C# program for naive solution to 
// print all combination of 4 elements 
// in A[] with sum equal to X  
using System; 
  
class FindFourElements  
{ 
    void findFourElements(int []A, int n, int X)  
    { 
        // Fix the first element and find other three 
        for (int i = 0; i < n - 3; i++)  
        { 
            // Fix the second element and find other two 
            for (int j = i + 1; j < n - 2; j++)  
            { 
                // Fix the third element and find the fourth 
                for (int k = j + 1; k < n - 1; k++)  
                { 
                    // find the fourth 
                    for (int l = k + 1; l < n; l++)  
                    { 
                        if (A[i] + A[j] + A[k] + A[l] == X)  
                        Console.Write(A[i] + " " + A[j] + 
                                " " + A[k] + " " + A[l]); 
                    } 
                } 
            } 
        } 
    } 
  
    // Driver program to test above functions 
    public static void Main()  
    { 
        FindFourElements findfour = new FindFourElements(); 
        int []A = {10, 20, 30, 40, 1, 2}; 
        int n = A.Length; 
        int X = 91; 
        findfour.findFourElements(A, n, X); 
    } 
} 
  
// This code is contributed by nitin mittal 

PHP

<?php 
/* A naive solution to print all combination  
of 4 elements in A[] with sum equal to X */
  
function findFourElements($A, $n, $X) 
{ 
    // Fix the first element and find other 
    // three 
    for ($i = 0; $i < $n-3; $i++) 
    { 
        // Fix the second element and find 
        // other two 
        for ($j = $i+1; $j < $n-2; $j++) 
        { 
            // Fix the third element and  
            // find the fourth 
            for ($k = $j+1; $k < $n-1; $k++) 
            { 
                // find the fourth 
                for ($l = $k+1; $l < $n; $l++) 
                    if ($A[$i] + $A[$j] + 
                        $A[$k] + $A[$l] == $X) 
                          
                        echo $A[$i], ", " , $A[$j], 
                        ", ", $A[$k], ", ", $A[$l]; 
            } 
        } 
    } 
} 
  
// Driver program to test above function 
    $A = array (10, 20, 30, 40, 1, 2); 
    $n = sizeof($A) ; 
    $X = 91; 
    findFourElements ($A, $n, $X); 
      
// This code is contributed by m_kit 
?> 

Output:

20, 30, 40, 1

Time Complexity: O(n^4)

The time complexity can be improved to O(n^3) with the use of sorting as a preprocessing step, and then using method 1 of this post to reduce a loop.

Following are the detailed steps.
1) Sort the input array.
2) Fix the first element as A[i] where i is from 0 to n–3. After fixing the first element of quadruple, fix the second element as A[j] where j varies from i+1 to n-2. Find remaining two elements in O(n) time, using the method 1 of this post

Following is the implementation of O(n^3) solution.

C++

// C++ program for to  print all combination  
// of 4 elements in A[] with sum equal to X  
#include<bits/stdc++.h>  
using namespace std;  
  
/* Following function is needed  
for library function qsort(). */
int compare (const void *a, const void * b)  
{  
    return ( *(int *)a - *(int *)b );  
}  
  
/* A sorting based solution to print  
all combination of 4 elements in A[]  
with sum equal to X */
void find4Numbers(int A[], int n, int X)  
{  
    int l, r;  
  
    // Sort the array in increasing 
    // order, using library function  
    // for quick sort  
    qsort (A, n, sizeof(A[0]), compare);  
  
    /* Now fix the first 2 elements  
    one by one and find  
    the other two elements */
    for (int i = 0; i < n - 3; i++)  
    {  
        for (int j = i+1; j < n - 2; j++)  
        {  
            // Initialize two variables as  
            // indexes of the first and last  
            // elements in the remaining elements  
            l = j + 1;  
            r = n-1;  
  
            // To find the remaining two  
            // elements, move the index  
            // variables (l & r) toward each other.  
            while (l < r)  
            {  
                if( A[i] + A[j] + A[l] + A[r] == X)  
                {  
                    cout << A[i]<<", " << A[j] <<  
                        ", " << A[l] << ", " << A[r];  
                    l++; r--;  
                }  
                else if (A[i] + A[j] + A[l] + A[r] < X)  
                    l++;  
                else // A[i] + A[j] + A[l] + A[r] > X  
                    r--;  
            } // end of while  
        } // end of inner for loop  
    } // end of outer for loop  
}  
  
/* Driver code */
int main()  
{  
    int A[] = {1, 4, 45, 6, 10, 12};  
    int X = 21;  
    int n = sizeof(A) / sizeof(A[0]);  
    find4Numbers(A, n, X);  
    return 0;  
}  
  
// This code is contributed by rathbhupendra 

C

# include <stdio.h> 
# include <stdlib.h> 
  
/* Following function is needed for library function qsort(). Refer 
   http://www.cplusplus.com/reference/clibrary/cstdlib/qsort/ */
int compare (const void *a, const void * b) 
{  return ( *(int *)a - *(int *)b ); } 
  
/* A sorting based solution to print all combination of 4 elements in A[] 
   with sum equal to X */
void find4Numbers(int A[], int n, int X) 
{ 
    int l, r; 
  
    // Sort the array in increasing order, using library 
    // function for quick sort 
    qsort (A, n, sizeof(A[0]), compare); 
  
    /* Now fix the first 2 elements one by one and find 
       the other two elements */
    for (int i = 0; i < n - 3; i++) 
    { 
        for (int j = i+1; j < n - 2; j++) 
        { 
            // Initialize two variables as indexes of the first and last  
            // elements in the remaining elements 
            l = j + 1; 
            r = n-1; 
  
            // To find the remaining two elements, move the index  
            // variables (l & r) toward each other. 
            while (l < r) 
            { 
                if( A[i] + A[j] + A[l] + A[r] == X) 
                { 
                   printf("%d, %d, %d, %d", A[i], A[j], 
                                           A[l], A[r]); 
                   l++; r--; 
                } 
                else if (A[i] + A[j] + A[l] + A[r] < X) 
                    l++; 
                else // A[i] + A[j] + A[l] + A[r] > X 
                    r--; 
            } // end of while 
        } // end of inner for loop 
    } // end of outer for loop 
} 
  
/* Driver program to test above function */
int main() 
{ 
    int A[] = {1, 4, 45, 6, 10, 12}; 
    int X = 21; 
    int n = sizeof(A)/sizeof(A[0]); 
    find4Numbers(A, n, X); 
    return 0; 
} 

Java

import java.util.Arrays; 
  
  
class FindFourElements 
{ 
    /* A sorting based solution to print all combination of 4 elements in A[] 
       with sum equal to X */
    void find4Numbers(int A[], int n, int X)  
    { 
        int l, r; 
  
        // Sort the array in increasing order, using library 
        // function for quick sort 
        Arrays.sort(A); 
  
        /* Now fix the first 2 elements one by one and find 
           the other two elements */
        for (int i = 0; i < n - 3; i++)  
        { 
            for (int j = i + 1; j < n - 2; j++)  
            { 
                // Initialize two variables as indexes of the first and last  
                // elements in the remaining elements 
                l = j + 1; 
                r = n - 1; 
  
                // To find the remaining two elements, move the index  
                // variables (l & r) toward each other. 
                while (l < r)  
                { 
                    if (A[i] + A[j] + A[l] + A[r] == X)  
                    { 
                        System.out.println(A[i]+" "+A[j]+" "+A[l]+" "+A[r]); 
                        l++; 
                        r--; 
                    }  
                    else if (A[i] + A[j] + A[l] + A[r] < X) 
                        l++; 
                    else // A[i] + A[j] + A[l] + A[r] > X 
                        r--; 
                } // end of while 
            } // end of inner for loop 
        } // end of outer for loop 
    } 
  
    // Driver program to test above functions 
    public static void main(String[] args)  
    { 
        FindFourElements findfour = new FindFourElements(); 
        int A[] = {1, 4, 45, 6, 10, 12}; 
        int n = A.length; 
        int X = 21; 
        findfour.find4Numbers(A, n, X); 
    } 
} 
  
// This code has been contributed by Mayank Jaiswal 

Python 3

# A sorting based solution to print all combination  
# of 4 elements in A[] with sum equal to X  
def find4Numbers(A, n, X): 
  
    # Sort the array in increasing order,  
    # using library function for quick sort 
    A.sort() 
  
    # Now fix the first 2 elements one by  
    # one and find the other two elements  
    for i in range(n - 3): 
        for j in range(i + 1, n - 2): 
              
            # Initialize two variables as indexes  
            # of the first and last elements in  
            # the remaining elements 
            l = j + 1
            r = n - 1
  
            # To find the remaining two elements,  
            # move the index variables (l & r) 
            # toward each other. 
            while (l < r): 
                if(A[i] + A[j] + A[l] + A[r] == X): 
                    print(A[i], ",", A[j], ",",  
                          A[l], ",", A[r]) 
                    l += 1
                    r -= 1
                  
                elif (A[i] + A[j] + A[l] + A[r] < X): 
                    l += 1
                else: # A[i] + A[j] + A[l] + A[r] > X 
                    r -= 1
  
# Driver Code 
if __name__ == "__main__": 
      
    A = [1, 4, 45, 6, 10, 12] 
    X = 21
    n = len(A) 
    find4Numbers(A, n, X) 
  
# This code is contributed by ita_c 

C#

// C# program for to  print all combination  
// of 4 elements in A[] with sum equal to X  
using System;  
  
class FindFourElements 
{ 
    // A sorting based solution to print all 
    // combination of 4 elements in A[] with 
    // sum equal to X  
    void find4Numbers(int []A, int n, int X)  
    { 
        int l, r; 
  
        // Sort the array in increasing order,  
        // using library function for quick sort 
        Array.Sort(A); 
  
        /* Now fix the first 2 elements one by one  
           and find the other two elements */
        for (int i = 0; i < n - 3; i++)  
        { 
            for (int j = i + 1; j < n - 2; j++)  
            { 
                // Initialize two variables as indexes of  
                // the first and last elements in the  
                // remaining elements 
                l = j + 1; 
                r = n - 1; 
  
                // To find the remaining two elements, move the  
                // index variables (l & r) toward each other. 
                while (l < r)  
                { 
                    if (A[i] + A[j] + A[l] + A[r] == X)  
                    { 
                        Console.Write(A[i] + " " + A[j] + 
                                " " + A[l] + " " + A[r]); 
                        l++; 
                        r--; 
                    }  
                    else if (A[i] + A[j] + A[l] + A[r] < X) 
                        l++; 
                          
                    else // A[i] + A[j] + A[l] + A[r] > X 
                        r--; 
                          
                } // end of while 
                  
            } // end of inner for loop 
              
        } // end of outer for loop 
    } 
  
    // Driver program to test above functions 
    public static void Main()  
    { 
        FindFourElements findfour = new FindFourElements(); 
        int []A = {1, 4, 45, 6, 10, 12}; 
        int n = A.Length; 
        int X = 21; 
        findfour.find4Numbers(A, n, X); 
    } 
} 
  
// This code has been contributed by nitin mittal 

PHP

<?php 
// PHP program for to print all  
// combination of 4 elements in  
// A[] with sum equal to X  
  
// A sorting based solution to 
// print all combination of 4  
// elements in A[] with sum 
// equal to X  
function find4Numbers($A, $n, $X)  
{ 
    $l; $r; 
  
    // Sort the array in increasing  
    // order, using library function 
    // for quick sort 
    sort($A); 
  
    // Now fix the first 2 elements  
    // one by one and find the other  
    // two elements 
    for ($i = 0; $i < $n - 3; $i++)  
    { 
        for ($j = $i + 1; $j < $n - 2; $j++)  
        { 
            // Initialize two variables  
            // as indexes of the first  
            // and last elements in the  
            // remaining elements 
            $l = $j + 1; 
            $r = $n - 1; 
  
            // To find the remaining two  
            // elements, move the index  
            // variables (l & r) towards 
            // each other. 
            while ($l < $r)  
            { 
                if ($A[$i] + $A[$j] +  
                    $A[$l] + $A[$r] == $X)  
                { 
                    echo $A[$i] , " " , $A[$j] ,  
                                  " " , $A[$l] ,  
                                  " " , $A[$r]; 
                    $l++; 
                    $r--; 
                }  
                else if ($A[$i] + $A[$j] +  
                         $A[$l] + $A[$r] < $X) 
                    $l++; 
                      
                // A[i] + A[j] + A[l] + A[r] > X 
                else 
                    $r--; 
                      
            }  
        }  
    }  
} 
  
// Driver Code 
$A = array(1, 4, 45, 6, 10, 12); 
$n = count($A); 
$X = 21; 
find4Numbers($A, $n, $X); 
  
// This code is contributed 
// by nitin mittal 
?> 

Output :

1, 4, 6, 10

Time Complexity : O(n^3)

This problem can also be solved in O(n^2Logn) complexity. Please refer below post for details

Find four elements that sum to a given value | Set 2 ( O(n^2Logn) Solution)

Please write comments if you find any of the above codes/algorithms incorrect, or find other ways to solve the same problem.

My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

C++

C

Java

Python 3

C#

PHP

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