Count Operations Needed for Given Conditions in C++



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 −

Open Compiler
#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; }

Input

{ 1, -3, 1, 0 }

Output

4
Updated on: 2022-03-03T06:00:04+05:30

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