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

Sources: Find Specific Sum and Amazon Interview Question

We have discussed a O(n^3) algorithm in the previous post on this topic. The problem can be solved in O(n^2Logn) time with the help of auxiliary space.

Thanks to itsnimish for suggesting this method. Following is the detailed process.

Let the input array be A[].

1) Create an auxiliary array aux[] and store sum of all possible pairs in aux[]. The size of aux[] will be n*(n-1)/2 where n is the size of A[].

2) Sort the auxiliary array aux[].

3) Now the problem reduces to find two elements in aux[] with sum equal to X. We can use method 1 of this post to find the two elements efficiently. There is following important point to note though. An element of aux[] represents a pair from A[]. While picking two elements from aux[], we must check whether the two elements have an element of A[] in common. For example, if first element sum of A[1] and A[2], and second element is sum of A[2] and A[4], then these two elements of aux[] don’t represent four distinct elements of input array A[].

Following is C implementation of this method.

#include <stdio.h>
#include <stdlib.h>

// The following structure is needed to store pair sums in aux[]
struct pairSum
    int first; // index (int A[]) of first element in pair
    int sec; // index of second element in pair
    int sum;  // sum of the pair

// Following function is needed for library function qsort()
int compare (const void *a, const void * b)
    return ( (*(pairSum *)a).sum - (*(pairSum*)b).sum );

// Function to check if two given pairs have any common element or not
bool noCommon(struct pairSum a, struct pairSum b)
    if (a.first == b.first || a.first == b.sec ||
            a.sec == b.first || a.sec == b.sec)
        return false;
    return true;

// The function finds four elements with given sum X
void findFourElements (int arr[], int n, int X)
    int i, j;

    // Create an auxiliary array to store all pair sums
    int size = (n*(n-1))/2;
    struct pairSum aux[size];

    /* Generate all possible pairs from A[] and store sums
       of all possible pairs in aux[] */
    int k = 0;
    for (i = 0; i < n-1; i++)
        for (j = i+1; j < n; j++)
            aux[k].sum = arr[i] + arr[j];
            aux[k].first = i;
            aux[k].sec = j;

    // Sort the aux[] array using library function for sorting
    qsort (aux, size, sizeof(aux[0]), compare);

    // Now start two index variables from two corners of array
    // and move them toward each other.
    i = 0;
    j = size-1;
    while (i < size && j >=0 )
        if ((aux[i].sum + aux[j].sum == X) && noCommon(aux[i], aux[j]))
            printf ("%d, %d, %d, %d\n", arr[aux[i].first], arr[aux[i].sec],
                                     arr[aux[j].first], arr[aux[j].sec]);
        else if (aux[i].sum + aux[j].sum < X)

// Driver program to test above function
int main()
    int arr[] = {10, 20, 30, 40, 1, 2};
    int n = sizeof(arr) / sizeof(arr[0]);
    int X = 91;
    findFourElements (arr, n, X);
    return 0;


20, 1, 30, 40

Please note that the above code prints only one quadruple. If we remove the return statement and add statements “i++; j–;”, then it prints same quadruple five times. The code can modified to print all quadruples only once. It has been kept this way to keep it simple.

Time complexity: The step 1 takes O(n^2) time. The second step is sorting an array of size O(n^2). Sorting can be done in O(n^2Logn) time using merge sort or heap sort or any other O(nLogn) algorithm. The third step takes O(n^2) time. So overall complexity is O(n^2Logn).

Auxiliary Space: O(n^2). The big size of auxiliary array can be a concern in this method.

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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