# 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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