Check if bike racer A can reach racer B with K jumps

Find if bike racer A defends bike racer B in race when he can jump on his path on seeing a ‘.’ and if he sees an obstacle ‘@’ in his path he cannot path through it.
Note : Race direction is decided by direction of A-B 

Examples :

Input : 5 2 @A@B@
Output : A can reach B

Input : 6 1 B....A
Output : A can reach B

Input : 7 3 B..@..A
Output : It is not possible for A to reach B

Input : 6 2 ..AB..
Output : It is not possible for A to reach B

Approach: 

This can be solve using following algorithm
1. find the first ocuurance of A and B in string and calculate the absoulute
  difference between them
2. if diff%k !=0 means we can not reach B  in the given jumps
3.  mainatain a check flag and initialise it with true initially
4. run a loop from the minimum index among A and B till the maximum index among
   between A and B , update the loop with the given number of jumps (k) and do the 
  following
  a) if any where the '@' occurs , set check as false means we can not move as it is
    an obstacle , given in the problem above
5. check the value of check_ varible if it is true
   means we can reach , print message A can reach B
   if it is false print message
   A can not reach B

Below is the implementation code: 

//C++ program to find if A defend B in bike race
#include<bits/stdc++.h>
using namespace std;

//Driver function
int main(){ 
	int n = 6;
	//cin>>n;
	int k = 1;
	//cin>>k;
	
	string s = "B....A";
	// cin>>s;
	// find the first indexes of A and B
	int index1 = s.find_first_of('A');
	int index2 = s.find_first_of('B');
	
	// find aabsoulute difference
	int diff = abs(index1 - index2);

	bool check = true;
	// if(diff == 0){
	//	 check = true;
	// }
	if(diff%k != 0){ // we can not reach
		check = false;
	} 

	for(int i=min(index1, index2); i<max(index1, index2); i+=k){
		if(s[i] == '@'){
			check = false; // it is an obstacle , we can not move
		}
	}

	if(check){ // true
		cout<<"A can reach B";
	}else{ // false
		cout<<"It is not possible for A to reach B";
	}

	return 0;
}

Output :  

A can reach B