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).

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++

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// 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

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C

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#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;
}

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Java

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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);
    }
}

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Python3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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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 nā€“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++

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// 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

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C

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# include <stdio.h>
# include <stdlib.h>
  
/* Following function is needed for library function qsort(). Refer
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;
}

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Java

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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

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Python 3

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# 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

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C#

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// 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

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PHP

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<?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
?>

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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.

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