Maximum Gap LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Maximum Gap in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemMaximum Gap– LeetCode Problem

Maximum Gap– LeetCode Problem


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 = [10]
Output: 0
Explanation: The array contains less than 2 elements, therefore return 0.


  • 1 <= nums.length <= 105
  • 0 <= nums[i] <= 109
Maximum Gap– LeetCode Solutions
Maximum Gap Solution in C++:
struct Bucket {
  int min;
  int max;

class Solution {
  int maximumGap(vector<int>& nums) {
    if (nums.size() < 2)
      return 0;

    const int mini = *min_element(begin(nums), end(nums));
    const int maxi = *max_element(begin(nums), end(nums));
    if (mini == maxi)
      return 0;

    const int gap = ceil((maxi - mini) / (double)(nums.size() - 1));
    const int bucketSize = (maxi - mini) / gap + 1;
    vector<Bucket> buckets(bucketSize, {INT_MAX, INT_MIN});

    for (const int num : nums) {
      const int i = (num - mini) / gap;
      buckets[i].min = min(buckets[i].min, num);
      buckets[i].max = max(buckets[i].max, num);

    int ans = 0;
    int prevMax = mini;

    for (const Bucket& bucket : buckets) {
      if (bucket.min == INT_MAX)
        continue;  // empty bucket
      ans = max(ans, bucket.min - prevMax);
      prevMax = bucket.max;

    return ans;
Maximum Gap Solution in Java:
class Bucket {
  public int min;
  public int max;

  public Bucket(int min, int max) {
    this.min = min;
    this.max = max;

class Solution {
  public int maximumGap(int[] nums) {
    if (nums.length < 2)
      return 0;

    final int min =;
    final int max =;
    if (min == max)
      return 0;

    final int gap = (int) Math.ceil((double) (max - min) / (nums.length - 1));
    final int bucketsLength = (max - min) / gap + 1;
    Bucket[] buckets = new Bucket[bucketsLength];

    for (int i = 0; i < buckets.length; ++i)
      buckets[i] = new Bucket(Integer.MAX_VALUE, Integer.MIN_VALUE);

    for (final int num : nums) {
      final int i = (num - min) / gap;
      buckets[i].min = Math.min(buckets[i].min, num);
      buckets[i].max = Math.max(buckets[i].max, num);

    int ans = 0;
    int prevMax = min;

    for (final Bucket bucket : buckets) {
      if (bucket.min == Integer.MAX_VALUE) // empty bucket
      ans = Math.max(ans, bucket.min - prevMax);
      prevMax = bucket.max;

    return ans;
Maximum Gap Solution in Python:
class Bucket:
  def __init__(self, mini: int, maxi: int): = mini
    self.maxi = maxi

class Solution:
  def maximumGap(self, nums: List[int]) -> int:
    if len(nums) < 2:
      return 0

    mini = min(nums)
    maxi = max(nums)
    if mini == maxi:
      return 0

    gap = ceil((maxi - mini) / (len(nums) - 1))
    bucketSize = (maxi - mini) // gap + 1
    buckets = [Bucket(inf, -inf) for _ in range(bucketSize)]

    for num in nums:
      i = (num - mini) // gap
      buckets[i].mini = min(buckets[i].mini, num)
      buckets[i].maxi = max(buckets[i].maxi, num)

    ans = 0
    prevMax = mini

    for bucket in buckets:
      if == inf:
        continue  # empty bucket
      ans = max(ans, - prevMax)
      prevMax = bucket.maxi

    return ans

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