# 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).

Source: Amazon Interview Question

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

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;
k++;
}
}

// 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]);
return;
}
else if (aux[i].sum + aux[j].sum < X)
i++;
else
j--;
}
}

// 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;
}
```

Output:

`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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