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Here we will see a problem, assuming an array is given. There are n elements. Another value S is also given. We must find an element K in the array such that if all elements greater than K are equal to K, then the sum of all elements of the final array equals S. If not possible, then -1 is returned.
Assume the elements are {12, 6, 3, 7, 8} and the sum value is 15, then the output is 3. The final array is {3, 3, 3, 3, 3}, and the sum of the array elements is S = 15
getVal(arr, n, S) −
Begin sort arr as increasing order sum := 0 for i in range 0 to n-1, do if sum + (arr[i] * (n - i)) is same as S, then return arr[i] end if sum := sum + arr[i] done return -1 End
#include <iostream> #include <algorithm> using namespace std; int getVal(int arr[], int n, int S) { sort(arr, arr + n); int sum = 0; for (int i = 0; i < n; i++) { if (sum + (arr[i] * (n - i)) == S) //if current value is satisfying, then return arr[i] return arr[i]; sum += arr[i]; } return -1; } int main() { int S = 15; int arr[] = { 12, 3, 6, 7, 8 }; int n = sizeof(arr) / sizeof(arr[0]); cout << getVal(arr, n, S); }
3
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