Suppose we have an array A containing n elements. There is another hidden array B of size n. These elements can be negative or positive. For each index i in the range 1 to n, the following will be done -
Initially set A[i] to 0
Then add B[i] to A[i], subtract B[i 1], then add B[i 2] and so on
So if the input is something like A = [6, -4, 8, -2, 3] then the output will be [2, 4, 6, 1, 3]
To solve this problem we will follow the following steps-
for initialize i := 0, when i < size of A, update (increase i by 1), do: print (A[i] + A[i + 1])
Let us see the following implementation for better understanding-
#include <bits/stdc++.h> using namespace std; void solve(vector<int> A){ for (int i = 0; i < A.size(); i++) cout << A[i] + A[i + 1] << ", "; } int main(){ vector<int> A = { 6, -4, 8, -2, 3 }; solve(A); }
{ 6, -4, 8, -2, 3 }
2, 4, 6, 1, 3,
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