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# Maximum Gap LeetCode Solution

## Problem – Maximum Gap

Given an integer array `nums`, return the maximum difference between two successive elements in its sorted form. If the array contains less than two elements, return `0`.

You must write an algorithm that runs in linear time and uses linear extra space.

Example 1:

``````Input: nums = [3,6,9,1]
Output: 3
Explanation: The sorted form of the array is [1,3,6,9], either (3,6) or (6,9) has the maximum difference 3.``````

Example 2:

``````Input: nums = 
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.``````

Constraints:

• `1 <= nums.length <= 105`
• `0 <= nums[i] <= 109`

### Maximum Gap LeetCode Solution in Java

``````public class Solution {
public int maximumGap(int[] num) {
if (num == null || num.length < 2)
return 0;
// get the max and min value of the array
int min = num;
int max = num;
for (int i:num) {
min = Math.min(min, i);
max = Math.max(max, i);
}
// the minimum possibale gap, ceiling of the integer division
int gap = (int)Math.ceil((double)(max - min)/(num.length - 1));
int[] bucketsMIN = new int[num.length - 1]; // store the min value in that bucket
int[] bucketsMAX = new int[num.length - 1]; // store the max value in that bucket
Arrays.fill(bucketsMIN, Integer.MAX_VALUE);
Arrays.fill(bucketsMAX, Integer.MIN_VALUE);
// put numbers into buckets
for (int i:num) {
if (i == min || i == max)
continue;
int idx = (i - min) / gap; // index of the right position in the buckets
bucketsMIN[idx] = Math.min(i, bucketsMIN[idx]);
bucketsMAX[idx] = Math.max(i, bucketsMAX[idx]);
}
// scan the buckets for the max gap
int maxGap = Integer.MIN_VALUE;
int previous = min;
for (int i = 0; i < num.length - 1; i++) {
if (bucketsMIN[i] == Integer.MAX_VALUE && bucketsMAX[i] == Integer.MIN_VALUE)
// empty bucket
continue;
// min value minus the previous value is the current gap
maxGap = Math.max(maxGap, bucketsMIN[i] - previous);
// update previous bucket value
previous = bucketsMAX[i];
}
maxGap = Math.max(maxGap, max - previous); // updata the final max value gap
return maxGap;
}
``````

### Maximum Gap LeetCode Solution in Python

``````class Solution:
def maximumGap(self, nums):
lo, hi, n = min(nums), max(nums), len(nums)
if n <= 2 or hi == lo: return hi - lo
B = defaultdict(list)
for num in nums:
ind = n-2 if num == hi else (num - lo)*(n-1)//(hi-lo)
B[ind].append(num)

cands = [[min(B[i]), max(B[i])] for i in range(n-1) if B[i]]
return max(y-x for x,y in zip(cands, cands[1:]))
``````

### Maximum Gap LeetCode Solution in C++

``````class Solution {
public:
int maximumGap(vector<int>& nums) {
int n = nums.size();
if (n < 2) return 0;
auto lu = minmax_element(nums.begin(), nums.end());
int l = *lu.first, u = *lu.second;
int gap = max((u - l) / (n - 1), 1);
int m = (u - l) / gap + 1;
vector<int> bucketsMin(m, INT_MAX);
vector<int> bucketsMax(m, INT_MIN);
for (int num : nums) {
int k = (num - l) / gap;
if (num < bucketsMin[k]) bucketsMin[k] = num;
if (num > bucketsMax[k]) bucketsMax[k] = num;
}
int i = 0, j;
gap = bucketsMax - bucketsMin;
while (i < m) {
j = i + 1;
while (j < m && bucketsMin[j] == INT_MAX && bucketsMax[j] == INT_MIN)
j++;
if (j == m) break;
gap = max(gap, bucketsMin[j] - bucketsMax[i]);
i = j;
}
return gap;
}
};
``````
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