Find 0 with Farthest 1s in a Binary Array

 

Given a string (seats) of 1s and 0s, where 1 represents a filled seat and 0 represents an empty seat in a row. Find an empty seat with maximum distance from an occupied seat. Return the maximum distance.

Examples:

Input: Seats = "1000101"
Output: 2
Explanation: Geek can take 3rd place and have a distance of 2 in left and 2 in right.

Input: Seats = "1000"
Output: 3
Explanation: Geek can take the rightmost seat to have a distance of 3.

 

class gfg{

    public int maxDis(String seats[]){

        int n = seats.length();

        int res = -1;

        int emptySeats = 0;

        for(int i = 0 ; i<n;i++){

            if(seats.charAt(i) == '0'){

                emptySeats++;

            else if(seats.charAt(i) == '1' && res == -1){

                    res = emptySeats;

                    emptySeats = 0;

            else{

                    res = Math.max(res,(int)Math.ceil(emptySeats/2.0));

                    emptySeat = 0;

            }

    }

    res = Math.max(res,emptySeats);

    return res;

}

}