Maximize Social Distancing - Google Top Interview Questions


Problem Statement :


You are given a list of integers seats containing 1s and 0s. 
Each element seats[i] represents a seat and is either occupied if seats[i] = 1 or empty if seats[i] = 0.

Given that there’s at least one empty seat and at least one occupied seat, return the maximum distance from an empty seat to the closest occupied seat.

Constraints

n ≤ 100,000 where n is the length of seats

Example 1

Input

seats = [1, 0, 1, 0, 0, 0, 1]

Output

2

Explanation

We can sit at seats[4].

Example 2

Input

seats = [1, 0, 0, 0]

Output

3

Explanation

We can sit at seats[3].


Solution :



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                        Solution in C++ :

int solve(vector<int>& seats) {
    int ans = 1, last = -1;
    for (int i = 0; i < seats.size(); ++i) {
        if (seats[i] == 1) {
            if (last == -1)
                ans = max(ans, i);
            else
                ans = max(ans, (i - last) / 2);
            last = i;
        }
    }
    ans = max(ans, (int)seats.size() - last - 1);
    return ans;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[] seats) {
        int pre = 0;
        boolean seen = false;

        int maxConsecutives = 0;
        int currConsecutives = 0;

        for (int i = 0; i < seats.length; i++) {
            if (seats[i] == 0) {
                currConsecutives++;
                maxConsecutives = Integer.max(maxConsecutives, currConsecutives);

                if (!seen) {
                    pre++;
                }
            } else if (seats[i] == 1) {
                seen = true;
                currConsecutives = 0;
            }
        }

        return Integer.max(
            Integer.max(pre, currConsecutives), (int) Math.ceil(maxConsecutives / 2.0));
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, seats):
        indices = [i for i, s in enumerate(seats) if s]
        d = max(((i2 - i1) // 2) for i1, i2 in zip(indices, indices[1:])) if len(indices) > 1 else 0
        d = max(indices[0], d, len(seats) - indices[-1] - 1)
        return d


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