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 = [10]
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[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;
}

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[0]-x[1] 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[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;
    }
};
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