164. Maximum Gap

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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:

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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:

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

Constraints:

  • 1 <= nums.length <= 10^5
  • 0 <= nums[i] <= 10^9

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    int maximumGap(vector<int>& nums) {
        if (nums.size() == 1) {
            return 0;
        }
        int mx = ranges::max(nums), mi = ranges::min(nums), diff = mx - mi, n = nums.size();
        int range = ceil((double)diff / (n - 1));
        if (!range) {
            return 0;
        }
        vector<pair<int, int>> buckets(n, {INT_MAX, INT_MIN});
        for (auto& num : nums) {
            // first let's figure out which bucket to put this item
            int idx = (num - mi) / range;
            buckets[idx].first = min(buckets[idx].first, num);
            buckets[idx].second = max(buckets[idx].second, num);
        }
        int idx = 0, res = range;
        while (idx < n) {
            if (buckets[idx].first == INT_MAX) {
                idx++;
                continue;
            }
            int cur_mx = buckets[idx].second;
            idx++;
            while (idx < n && buckets[idx].first == INT_MAX) {
                idx++;
            }
            if (idx == n) {
                break;
            }
            int cur_mi = buckets[idx].first;
            res = max(res, cur_mi - cur_mx);
        }
        return res;
    }
};