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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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
no common indexes.
return true
Else  update hash table
EndLoop;
EndLoop;

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

 // C++ program to find if there exist 4 elements // with given sum #include 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 > > 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 > 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 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);     if (findfour(arr, n, k))         cout             << "YES" << endl;     else         cout << "NO" << endl;     return 0; }

Output:

YES

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