Find four elements that sum to a given value | Set 3 (Hashmap)

Given an array of integers, Check if there exist four elements at different indexes in the array whose sum is equal to a given value k. For example, if the given array is {1 5 1 0 6 0} and k = 7, then your function should print “YES” as (1+5+1+0=7).

Examples:

Input  : arr[] = {1 5 1 0 6 0} 
             k = 7
Output : YES

Input :  arr[] = {38 7 44 42 28 16 10 37 
                  33 2 38 29 26 8 25} 
            k = 22
Output : NO

We have discussed different solutions in below two sets. Find four elements that sum to a given value | Set 1 (n^3 solution) Find four elements that sum to a given value | Set 2 ( O(n^2Logn) Solution) In this post, an optimized solution is discussed that works in O(n²) on average. The idea is to create a hashmap to store pair sums.

Loop i = 0 to n-1 :
 Loop j = i + 1 to n-1  
   calculate sum = arr[i] + arr[j]
     If (k-sum) exist in hash 
      a) Check in hash table for all
         pairs of indexes which form
         (k-sum).
      b) If there is any pair with no 
         common indexes.
           return true 
    Else update hash table
    EndLoop;
EndLoop;

Implementation:

CPP

// C++ program to find if there exist 4 elements
// with given sum
#include <bits/stdc++.h>
using namespace std;

// function to check if there exist four
// elements whose sum is equal to k
bool findfour(int arr[], int n, int k)
{
    // map to store sum and indexes for
    // a pair sum
    unordered_map<int, vector<pair<int, int> > > hash;

    for (int i = 0; i < n; i++) {
        for (int j = i + 1; j < n; j++) {

            // calculate the sum of each pair
            int sum = arr[i] + arr[j];

            // if k-sum exist in map
            if (hash.find(k - sum) != hash.end()) {
                auto num = hash.find(k - sum);
                vector<pair<int, int> > v = num->second;

                // check for index coincidence as if
                // there is a common that means all
                // the four numbers are not from
                // different indexes and one of the
                // index is repeated
                for (int k = 0; k < num->second.size();
                     k++) {

                    pair<int, int> it = v[k];

                    // if all indexes are different then
                    // it means four number exist
                    // set the flag and break the loop
                    if (it.first != i && it.first != j
                        && it.second != i && it.second != j)
                        return true;
                }
            }

            // store the sum and index pair in hashmap
            hash[sum].push_back(make_pair(i, j));
        }
    }
    hash.clear();
    return false;
}

// Driver code
int main()
{
    int k = 7;
    int arr[] = { 1, 5, 1, 0, 6, 0 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (findfour(arr, n, k))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

Output

YES

Time complexity: O(n^2)

Auxiliary space: O(n) for hashmap

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