C++ program to count at least how many operations needed for given conditions

Suppose we have an array A with N elements. In each operation, we pick an element and increase or decrease it by 1. We have to find at least how many operations are needed to satisfy following conditions −

For every i in range 1 to n, the sum of the terms from 1st through ith term is not 0
For every i in range 1 to n - 1, the sign of the terms from the 1st through ith term is different from sign of the sum of the terms from the 1st through (i+1)th term.

So, if the input is like A = [1, -3, 1, 0], then the output will be 4, because we can transform the sequence like 1, -2, 2, -2 by four operations. The sums of the first one, two, three and four terms are 1, -1, 1 and -1.

Steps

To solve this, we will follow these steps −

n := size of A
ret := 0
sum := 0
for each ai in A, do
   nsum := sum + ai
   if s > 0, then:
      if nsum <= 0, then:
         ret := ret + |nsum|
         ai := ai + |nsum|
      Otherwise
         if nsum >= 0, then:
            ret := ret + nsum + 1
            ai := ai - (nsum + 1)
      sum := sum + ai
      s := s * -1
return ret

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;

int util(vector<int> A, int s){
   int n = A.size();
   int ret = 0;
   int sum = 0;
   for (int ai : A){
      int nsum = sum + ai;
      if (s > 0){
         if (nsum <= 0){
            ret += abs(nsum) + 1;
            ai = ai + abs(nsum) + 1;
         }
      } else{
         if (nsum >= 0){
            ret += nsum + 1;
            ai = ai - (nsum + 1);
         }
      }
      sum += ai;
      s *= -1;
   }
   return ret;
}
int solve(vector<int> A){
   int res = min(util(A, 1), util(A, -1));
   return res;
}
int main(){
   vector<int> A = { 1, -3, 1, 0 };
   cout << solve(A) << endl;
}