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