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: Discuss Different Cases

Let $m$ represent the length of string $s$, and $n$ represent the length of string $t$. We can assume that $m$ is always greater than or equal to $n$.

If $m-n > 1$, return false directly;

Otherwise, iterate through $s$ and $t$, if $s[i]$ is not equal to $t[i]$:

If $m \neq n$, compare $s[i+1:]$ with $t[i:]$, return true if they are equal, otherwise return false;
If $m = n$, compare $s[i:]$ with $t[i:]$, return true if they are equal, otherwise return false.

If the iteration ends, it means that all the characters of $s$ and $t$ that have been iterated are equal, at this time it needs to satisfy $m=n+1$.

The time complexity is $O(m)$, where $m$ is the length of string $s$. The space complexity is $O(1)$.

Python3JavaC++GoC#

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

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;
    }
}

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;
    }
};

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
}

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;
    }
}

164. Maximum Gap

Description

Solutions

Solution 1: Discuss Different Cases

Comments