C++ code to find minimum jump to reach home by frog

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