C++ code to count days to complete reading book

Suppose we have an array A with n elements and have another value t. On ith day Amal spends A[i] seconds in work. In the free time he reads a book. The entire book will take t seconds to complete. We have to find how many days he will need to read the complete book.

So, if the input is like A = [86400, 86398]; t = 2, then the output will be 2, because one day has 86400 seconds, and first day is totally blocked. On second day he will get 2 seconds to complete the book.

Steps

To solve this, we will follow these steps −

cnt := 1
n := size of A
for initialize i := 0, when i < n, update (increase i by 1), do:
   x := A[i]
   t := t - 86400 - x
   if t <= 0, then:
      return cnt
   (increase cnt by 1)

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(vector<int> A, int t){
   int cnt = 1;
   int n = A.size();
   for (int i = 0; i < n; i++){
      int x = A[i];
      t -= 86400 - x;
      if (t <= 0){
         return cnt;
      }
      ++cnt;
   }
}
int main(){
   vector<int> A = { 86400, 86398 };
   int t = 2;
   cout << solve(A, t) << endl;
}

Input

{ 86400, 86398 }, 2

Output

2

Updated on: 29-Mar-2022

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