# Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

## Solution & Complexity

1. Time : O(nlogn). Space : O(1); Sort the unsorted array, and find the maximum difference.
2. Time : O(n). Space : O(n). Drawer Theory. If we put n numbers into (n+1) drawers, then there must be at least one empty drawer. So we can find the maximum difference between two succesive non-empty drawers.

Suppose there are N elements and they range from A to B.

Then the maximum gap will be no smaller than ceiling[(B - A) / (N - 1)]

Let the length of a bucket to be len = ceiling[(B - A) / (N - 1)], then we will have at most num = (B - A) / len + 1 of bucket

for any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K - A) / len and therefore maintain the maximum and minimum elements in each bucket.

Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken from two elements in the same buckets.

For each non-empty buckets p, find the next non-empty buckets q, then q.min - p.max could be the potential answer to the question. Return the maximum of all those values.

## Code

``````public class Solution {
public int maximumGap_1(int[] num) {
Arrays.sort(num);
int res = 0;
for (int i = 1; i < num.length; ++i) {
res = Math.max(res, num[i] - num[i - 1]);
}
return res;
}
class node {
public int low;
public int high;
public node() {
low = -1;
high = -1;
}
}

public int maximumGap_2(int[] num) {
int n = num.length;
if (n < 2) return 0;
int minVal = num[0], maxVal = num[0];
for (int i = 1; i < n; ++i) {
minVal = Math.min(minVal, num[i]);
maxVal = Math.max(maxVal, num[i]);
}
//delta = (maxVal + 1 - minVal) / (n + 1)
//idx = (val - minVal) / delta = (val - minVal) * (n + 1) / (maxVal + 1 - minVal)
node[] pool = new node[n+2];
for (int i = 0; i < n+2; ++i) pool[i] = new node();
for (int i = 0; i < n; ++i) {
int idx =(int)(Long.valueOf(num[i] - minVal)* Long.valueOf(n + 1) / Long.valueOf(maxVal + 1 - minVal));
if (pool[idx].low == -1) {
pool[idx].low = pool[idx].high = num[i];
} else {
pool[idx].low = Math.min(pool[idx].low, num[i]);
pool[idx].high = Math.max(pool[idx].high, num[i]);
}
}
int pre = pool[0].high;
int res = 0;
for (int i = 1; i < n + 2; ++i) {
if (pool[i].low != -1) {
res = Math.max(res, pool[i].low - pre);
pre = pool[i].high;
}
}
return res;
}
}
``````