**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 :

` ````
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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