164. Maximum Gap

Description

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

Solutions

Solution 1

Python Code
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class Solution:
    def maximumGap(self, nums: List[int]) -> int:
        n = len(nums)
        if n < 2:
            return 0
        mi, mx = min(nums), max(nums)
        bucket_size = max(1, (mx - mi) // (n - 1))
        bucket_count = (mx - mi) // bucket_size + 1
        buckets = [[inf, -inf] for _ in range(bucket_count)]
        for v in nums:
            i = (v - mi) // bucket_size
            buckets[i][0] = min(buckets[i][0], v)
            buckets[i][1] = max(buckets[i][1], v)
        ans = 0
        prev = inf
        for curmin, curmax in buckets:
            if curmin > curmax:
                continue
            ans = max(ans, curmin - prev)
            prev = curmax
        return ans

Java Code
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class Solution {
    public int maximumGap(int[] nums) {
        int n = nums.length;
        if (n < 2) {
            return 0;
        }
        int inf = 0x3f3f3f3f;
        int mi = inf, mx = -inf;
        for (int v : nums) {
            mi = Math.min(mi, v);
            mx = Math.max(mx, v);
        }
        int bucketSize = Math.max(1, (mx - mi) / (n - 1));
        int bucketCount = (mx - mi) / bucketSize + 1;
        int[][] buckets = new int[bucketCount][2];
        for (var bucket : buckets) {
            bucket[0] = inf;
            bucket[1] = -inf;
        }
        for (int v : nums) {
            int i = (v - mi) / bucketSize;
            buckets[i][0] = Math.min(buckets[i][0], v);
            buckets[i][1] = Math.max(buckets[i][1], v);
        }
        int prev = inf;
        int ans = 0;
        for (var bucket : buckets) {
            if (bucket[0] > bucket[1]) {
                continue;
            }
            ans = Math.max(ans, bucket[0] - prev);
            prev = bucket[1];
        }
        return ans;
    }
}

C++ Code
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using pii = pair<int, int>;

class Solution {
public:
    const int inf = 0x3f3f3f3f;
    int maximumGap(vector<int>& nums) {
        int n = nums.size();
        if (n < 2) return 0;
        int mi = inf, mx = -inf;
        for (int v : nums) {
            mi = min(mi, v);
            mx = max(mx, v);
        }
        int bucketSize = max(1, (mx - mi) / (n - 1));
        int bucketCount = (mx - mi) / bucketSize + 1;
        vector<pii> buckets(bucketCount, {inf, -inf});
        for (int v : nums) {
            int i = (v - mi) / bucketSize;
            buckets[i].first = min(buckets[i].first, v);
            buckets[i].second = max(buckets[i].second, v);
        }
        int ans = 0;
        int prev = inf;
        for (auto [curmin, curmax] : buckets) {
            if (curmin > curmax) continue;
            ans = max(ans, curmin - prev);
            prev = curmax;
        }
        return ans;
    }
};

Go Code
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func maximumGap(nums []int) int {
	n := len(nums)
	if n < 2 {
		return 0
	}
	inf := 0x3f3f3f3f
	mi, mx := inf, -inf
	for _, v := range nums {
		mi = min(mi, v)
		mx = max(mx, v)
	}
	bucketSize := max(1, (mx-mi)/(n-1))
	bucketCount := (mx-mi)/bucketSize + 1
	buckets := make([][]int, bucketCount)
	for i := range buckets {
		buckets[i] = []int{inf, -inf}
	}
	for _, v := range nums {
		i := (v - mi) / bucketSize
		buckets[i][0] = min(buckets[i][0], v)
		buckets[i][1] = max(buckets[i][1], v)
	}
	ans := 0
	prev := inf
	for _, bucket := range buckets {
		if bucket[0] > bucket[1] {
			continue
		}
		ans = max(ans, bucket[0]-prev)
		prev = bucket[1]
	}
	return ans
}

C# Code
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using System;
using System.Linq;

public class Solution {
    public int MaximumGap(int[] nums) {
        if (nums.Length < 2) return 0;
        var max = nums.Max();
        var min = nums.Min();
        var bucketSize = Math.Max(1, (max - min) / (nums.Length - 1));
        var buckets = new Tuple<int, int>[(max - min) / bucketSize + 1];
        foreach (var num in nums)
        {
            var index = (num - min) / bucketSize;
            if (buckets[index] == null)
            {
                buckets[index] = Tuple.Create(num, num);
            }
            else
            {
                buckets[index] = Tuple.Create(Math.Min(buckets[index].Item1, num), Math.Max(buckets[index].Item2, num));
            }
        }

        var result = 0;
        Tuple<int, int> lastBucket = null;
        for (var i = 0; i < buckets.Length; ++i)
        {
            if (buckets[i] != null)
            {
                if (lastBucket != null)
                {
                    result = Math.Max(result, buckets[i].Item1 - lastBucket.Item2);
                }
                lastBucket = buckets[i];
            }
        }
        return result;
    }
}