C++ program to count final number of characters are there after typing n characters in a crazy writer

Suppose we have an array A with n elements, and another value c is there. There is a crazy word processor present in our system where we can type characters but if we do not type for consecutive c seconds, all of the written letters will be removed. A[i] represents the time when we have typed the ith character. We have to find the final number of characters that will remain on the screen after typing all n characters.

So, if the input is like A = [1, 3, 8, 14, 19, 20]; c = 5, then the output will be 3, because at the second 8 there will be 3 words on the screen. Then, everything disappears at the second 13. At the seconds 14 and 19 another two characters are typed, and finally, at the second 20, one more character is typed, and a total of 3 letters remain on the screen.

Steps

To solve this, we will follow these steps −

s := 1
n := size of A
for initialize i := 1, when i < n, update (increase i by 1), do:
   if (A[i] - A[i - 1]) <= c, then:
      (increase s by 1)
   Otherwise
      s := 1
return s

Example

Let us see the following implementation to get better understanding −

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

int solve(vector<int> A, int c) {
   int s = 1;
   int n = A.size();
   for (int i = 1; i < n; i++) {
      if ((A[i] - A[i - 1]) <= c) {
         s++;
      } else {
         s = 1;
      }
   }
   return s;
}
int main() {
   vector<int> A = { 1, 3, 8, 14, 19, 20 };
   int c = 5;
   cout << solve(A, c) << endl;
}