# Find four elements that sum to a given value | Set 2

Given an array of integers, find any one combination of four elements in the array whose sum is equal to a given value X.

For example,

```Input: array = {10, 2, 3, 4, 5, 9, 7, 8}
X = 23
Output: 3 5 7 8
Sum of output is equal to 23,
i.e. 3 + 5 + 7 + 8 = 23.

Input: array = {1, 2, 3, 4, 5, 9, 7, 8}
X = 16
Output: 1 3 5 7
Sum of output is equal to 16,
i.e. 1 + 3 + 5 + 7 = 16.
```

## 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.

Method 1: Two Pointers Algorithm.
Approach: 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 and A, and second element is sum of A and A, then these two elements of aux[] don’t represent four distinct elements of input array A[].

Below is the implementation of the above approach:

## C++

 `#include ` `using` `namespace` `std; ` ` `  `// The following structure is needed ` `// to store pair sums in aux[] ` `class` `pairSum { ` `public``: ` `    ``// index (int A[]) of first element in pair ` `    ``int` `first; ` ` `  `    ``// index of second element in pair ` `    ``int` `sec; ` ` `  `    ``// sum of the pair ` `    ``int` `sum; ` `}; ` ` `  `// 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(pairSum a, 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; ` `    ``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), 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])) { ` `            ``cout << arr[aux[i].first] ` `                 ``<< ``", "` `<< arr[aux[i].sec] ` `                 ``<< ``", "` `<< arr[aux[j].first] ` `                 ``<< ``", "` `<< arr[aux[j].sec] ` `                 ``<< endl; ` `            ``return``; ` `        ``} ` `        ``else` `if` `(aux[i].sum + aux[j].sum < X) ` `            ``i++; ` `        ``else` `            ``j--; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 20, 30, 40, 1, 2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `X = 91; ` `    ``findFourElements(arr, n, X); ` `    ``return` `0; ` `} ` ` `  `// This is code is contributed by rathbhupendra `

## C

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

Complexity Analysis:

• Time complexity: O(n^2Logn).
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 size of the auxiliary array is O(n^2). The big size of the auxiliary array can be a concern in this method.

Method 2: Hashing Based Solution[O(n2)]
Approach:

1. Store sums of all pairs in a hash table
2. Traverse through all pairs again and search for X – (current pair sum) in the hash table.
3. If a pair is found with the required sum, then make sure that all elements are distinct array elements and an element is not considered more than once.

Below image is a dry run of the above approach: Below is the implementation of the above approach:

## C++

 `// A hashing based  CPP program ` `// to find if there are ` `// four elements with given sum. ` `#include ` `using` `namespace` `std; ` ` `  `// The function finds four ` `// elements with given sum X ` `void` `findFourElements( ` `    ``int` `arr[], ``int` `n, ``int` `X) ` `{ ` `    ``// Store sums of all pairs ` `    ``// in a hash table ` `    ``unordered_map<``int``, pair<``int``, ``int``> > mp; ` `    ``for` `(``int` `i = 0; i < n - 1; i++) ` `        ``for` `(``int` `j = i + 1; j < n; j++) ` `            ``mp[arr[i] + arr[j]] = { i, j }; ` ` `  `    ``// Traverse through all pairs and search ` `    ``// for X - (current pair sum). ` `    ``for` `(``int` `i = 0; i < n - 1; i++) { ` `        ``for` `(``int` `j = i + 1; j < n; j++) { ` `            ``int` `sum = arr[i] + arr[j]; ` ` `  `            ``// If X - sum is present in hash table, ` `            ``if` `(mp.find(X - sum) != mp.end()) { ` ` `  `                ``// Making sure that all elements are ` `                ``// distinct array elements and an element ` `                ``// is not considered more than once. ` `                ``pair<``int``, ``int``> p = mp[X - sum]; ` `                ``if` `(p.first != i && p.first != j ` `                    ``&& p.second != i && p.second != j) { ` `                    ``cout << arr[i] << ``", "` `                         ``<< arr[j] << ``", "` `                         ``<< arr[p.first] << ``", "` `                         ``<< arr[p.second]; ` `                    ``return``; ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` `    ``int` `arr[] = { 10, 20, 30, 40, 1, 2 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` `    ``int` `X = 91; ` `    ``findFourElements(arr, n, X); ` `    ``return` `0; ` `} `

## Java

 `// A hashing based Java program to find ` `// if there are four elements with given sum. ` `import` `java.util.HashMap; ` `class` `GFG { ` `    ``static` `class` `pair { ` `        ``int` `first, second; ` `        ``public` `pair(``int` `first, ``int` `second) ` `        ``{ ` `            ``this``.first = first; ` `            ``this``.second = second; ` `        ``} ` `    ``} ` ` `  `    ``// The function finds four elements ` `    ``// with given sum X ` `    ``static` `void` `findFourElements( ` `        ``int` `arr[], ` `        ``int` `n, ``int` `X) ` `    ``{ ` `        ``// Store sums of all pairs in a hash table ` `        ``HashMap ` `            ``mp = ``new` `HashMap(); ` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) ` `                ``mp.put(arr[i] + arr[j], ` `                       ``new` `pair(i, j)); ` ` `  `        ``// Traverse through all pairs and search ` `        ``// for X - (current pair sum). ` `        ``for` `(``int` `i = ``0``; i < n - ``1``; i++) { ` `            ``for` `(``int` `j = i + ``1``; j < n; j++) { ` `                ``int` `sum = arr[i] + arr[j]; ` ` `  `                ``// If X - sum is present in hash table, ` `                ``if` `(mp.containsKey(X - sum)) { ` ` `  `                    ``// Making sure that all elements are ` `                    ``// distinct array elements and an element ` `                    ``// is not considered more than once. ` `                    ``pair p = mp.get(X - sum); ` `                    ``if` `(p.first != i && p.first != j ` `                        ``&& p.second != i && p.second != j) { ` `                        ``System.out.print( ` `                            ``arr[i] + ``", "` `+ arr[j] ` `                            ``+ ``", "` `+ arr[p.first] ` `                            ``+ ``", "` `+ arr[p.second]); ` `                        ``return``; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` `        ``int` `arr[] = { ``10``, ``20``, ``30``, ``40``, ``1``, ``2` `}; ` `        ``int` `n = arr.length; ` `        ``int` `X = ``91``; ` `        ``findFourElements(arr, n, X); ` `    ``} ` `} ` ` `  `// This code is contributed by Princi Singh `

## Python3

 `# A hashing based Python program to find if there are  ` `# four elements with given summ. ` ` `  `# The function finds four elements with given summ X ` `def` `findFourElements (arr, n, X): ` `     `  `    ``# Store summs of all pairs in a hash table ` `    ``mp ``=` `{} ` `    ``for` `i ``in` `range``(n ``-` `1``): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``mp[arr[i] ``+` `arr[j]] ``=` `[i, j] ` `             `  `    ``# Traverse through all pairs and search ` `    ``# for X - (current pair summ).  ` `    ``for` `i ``in` `range``(n ``-` `1``): ` `        ``for` `j ``in` `range``(i ``+` `1``, n): ` `            ``summ ``=` `arr[i] ``+` `arr[j] ` `             `  `            ``# If X - summ is present in hash table,      ` `            ``if` `(X ``-` `summ) ``in` `mp: ` `                 `  `                ``# Making sure that all elements are ` `                ``# distinct array elements and an element ` `                ``# is not considered more than once. ` `                ``p ``=` `mp[X ``-` `summ] ` `                ``if` `(p[``0``] !``=` `i ``and` `p[``0``] !``=` `j ``and` `p[``1``] !``=` `i ``and` `p[``1``] !``=` `j): ` `                    ``print``(arr[i], ``", "``, arr[j], ``", "``, arr[p[``0``]], ``", "``, arr[p[``1``]], sep ``=``"") ` `                    ``return` ` `  `# Driver code ` `arr ``=` `[``10``, ``20``, ``30``, ``40``, ``1``, ``2``] ` `n ``=` `len``(arr) ` `X ``=` `91` `findFourElements(arr, n, X) ` ` `  `# This is code is contributed by shubhamsingh10 `

## C#

 `// A hashing based C# program to find ` `// if there are four elements with given sum. ` `using` `System; ` `using` `System.Collections.Generic; ` ` `  `class` `GFG { ` `    ``public` `class` `pair { ` `        ``public` `int` `first, second; ` `        ``public` `pair(``int` `first, ``int` `second) ` `        ``{ ` `            ``this``.first = first; ` `            ``this``.second = second; ` `        ``} ` `    ``} ` ` `  `    ``// The function finds four elements ` `    ``// with given sum X ` `    ``static` `void` `findFourElements(``int``[] arr, ` `                                 ``int` `n, ``int` `X) ` `    ``{ ` `        ``// Store sums of all pairs in a hash table ` `        ``Dictionary<``int``, ` `                   ``pair> ` `            ``mp = ``new` `Dictionary<``int``, ` `                                ``pair>(); ` `        ``for` `(``int` `i = 0; i < n - 1; i++) ` `            ``for` `(``int` `j = i + 1; j < n; j++) ` `                ``if` `(mp.ContainsKey(arr[i] + arr[j])) ` `                    ``mp[arr[i] + arr[j]] = ``new` `pair(i, j); ` `                ``else` `                    ``mp.Add(arr[i] + arr[j], ``new` `pair(i, j)); ` ` `  `        ``// Traverse through all pairs and search ` `        ``// for X - (current pair sum). ` `        ``for` `(``int` `i = 0; i < n - 1; i++) { ` `            ``for` `(``int` `j = i + 1; j < n; j++) { ` `                ``int` `sum = arr[i] + arr[j]; ` ` `  `                ``// If X - sum is present in hash table, ` `                ``if` `(mp.ContainsKey(X - sum)) { ` ` `  `                    ``// Making sure that all elements are ` `                    ``// distinct array elements and an element ` `                    ``// is not considered more than once. ` `                    ``pair p = mp[X - sum]; ` `                    ``if` `(p.first != i && p.first != j && p.second != i && p.second != j) { ` `                        ``Console.Write(arr[i] + ``", "` `+ arr[j] + ``", "` `+ arr[p.first] + ``", "` `+ arr[p.second]); ` `                        ``return``; ` `                    ``} ` `                ``} ` `            ``} ` `        ``} ` `    ``} ` ` `  `    ``// Driver Code ` `    ``public` `static` `void` `Main(String[] args) ` `    ``{ ` `        ``int``[] arr = { 10, 20, 30, 40, 1, 2 }; ` `        ``int` `n = arr.Length; ` `        ``int` `X = 91; ` `        ``findFourElements(arr, n, X); ` `    ``} ` `} ` ` `  `// This code is contributed by 29AjayKumar `

Output:

`20, 30, 40, 1`

Complexity Analysis:

• Time complexity: O(n^2).
Nested traversal is needed to store all pairs in the hash Map.
• Auxiliary Space: O(n^2).
All n*(n-1) pairs are stored in hash Map so the space required is O(n^2)

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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