Minimum Jumps to Reach Home by Frog in C++

Suppose we have one binary string S with n bits and another number d. On a number line, a frog wants to reach point n, starting from the point 1. The frog can jump to the right at a distance not more than d. For each point from 1 to n if there is a lily flower it is marked as 1, and 0 if not. The frog can jump only in points with a lilies. We have to find the minimal number of jumps that the frog needs to reach n. If not possible, return -1.

So, if the input is like S = "10010101"; d = 4, then the output will be 2, because from position 1, it jumps to 4, then at the index 8(n).

Steps

To solve this, we will follow these steps −

n := size of s
x := 0
y := 0
while (x < n - 1 and y <= n), do:
   if s[x] is same as '1', then:
      x := x + d
      increase y by 1
   Otherwise
      (decrease x by 1)
if y >= n, then:
   return -1
Otherwise
   return y

Example

Let us see the following implementation to get better understanding −

#include <bits/stdc++.h>
using namespace std;
int solve(string s, int d){
   int n = s.size();
   int x = 0, y = 0;
   while (x < n - 1 && y <= n){
      if (s[x] == '1')
         x += d, ++y;
      else
         --x;
   }
   if (y >= n)
      return -1;
   else
      return y;
}
int main(){
   string S = "10010101";
   int d = 4;
   cout << solve(S, d) << endl;
}

Input

"10010101", 4

Output

2