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Home Code Solutions

Maximum Gap – LeetCode Solution

Maximum Gap - LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

BhautikBhalala by BhautikBhalala
September 5, 2022
Reading Time: 2 mins read
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Table of Contents

  • Maximum Gap – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • C++ Maximum Gap LeetCode Solution
  • Java Maximum Gap LeetCode Solution
  • Python 3 Maximum Gap LeetCode Solution
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Maximum Gap – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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.

 

Constraints:

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

C++ Maximum Gap LeetCode Solution


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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[0] - bucketsMin[0];
        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;
    }
};

Java Maximum Gap LeetCode Solution


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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[0];
    int max = num[0];
    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;
}

 

}

 



Python 3 Maximum Gap LeetCode Solution


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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[0]-x[1] for x,y in zip(cands, cands[1:]))



 

Array-1180
String-562
Hash Table-412
Dynamic Programming-390
Math-368
Sorting-264
Greedy-257
Depth-First Search-256
Database-215
Breadth-First Search-200
Tree-195
Binary Search-191
Matrix-176
Binary Tree-160
Two Pointers-151
Bit Manipulation-140
Stack-133
Heap (Priority Queue)-117
Design-116
Graph-108
Simulation-103
Prefix Sum-96
Backtracking-92
Counting-86
Sliding Window-73
Linked List-69
Union Find-66
Ordered Set-48
Monotonic Stack-47
Recursion-43
Trie-41
Binary Search Tree-40
Divide and Conquer-40
Enumeration-39
Bitmask-37
Queue-33
Memoization-32
Topological Sort-31
Geometry-30
Segment Tree-27
Game Theory-24
Hash Function-24
Binary Indexed Tree-21
Interactive-18
Data Stream-17
String Matching-17
Rolling Hash-17
Shortest Path-16
Number Theory-16
Combinatorics-15
Randomized-12
Monotonic Queue-9
Iterator-9
Merge Sort-9
Concurrency-9
Doubly-Linked List-8
Brainteaser-8
Probability and Statistics-7
Quickselect-7
Bucket Sort-6
Suffix Array-6
Minimum Spanning Tree-5
Counting Sort-5
Shell-4
Line Sweep-4
Reservoir Sampling-4
Eulerian Circuit-3
Radix Sort-3
Strongly Connected Componen-t2
Rejection Sampling-2
Biconnected Component-1

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Leetcode All Problems SolutionsMaximum Subarray – LeetCode Solution Leetcode All Problems SolutionsMaximum Product Subarray – LeetCode Solution Leetcode All Problems SolutionsRemove Duplicates from Sorted Array – LeetCode Solution Leetcode All Problems SolutionsN-Queens – LeetCode Solution Leetcode All Problems SolutionsMinimum Path Sum – LeetCode Solution Leetcode All Problems SolutionsSingle Number II – LeetCode Solution
Tags: Cc++14full solutionGojavajava 15java 7java 8java8javascriptkotlinLeetCodeLeetCodeSolutionMaximum Gappypy 3Python 2python 3rubyrustSolution
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