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

2.7

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).

Source: Amazon Interview Question

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

#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 funtion
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);
    }
}


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

# 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

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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