164

164. Maximum Gap (Hard)

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 <= 104
• 0 <= nums[i] <= 109

Companies:
Amazon

Related Topics:
Sort

Solution 1.

// OJ: https://leetcode.com/problems/maximum-gap/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
// Ref: https://discuss.leetcode.com/topic/5999/bucket-sort-java-solution-with-explanation-o-n-time-and-space
class Solution {
public:
    int maximumGap(vector<int>& A) {
        if (A.size() == 1) return 0;
        int minVal = *min_element(begin(A), end(A));
        int maxVal = *max_element(begin(A), end(A));
        int N = A.size(), gap = (maxVal - minVal + N - 2) / (N - 1);
        vector<int> mn(N - 1, INT_MAX), mx(N - 1, INT_MIN);
        for (int n : A) {
            if (n == minVal || n == maxVal) continue;
            int i = (n - minVal) / gap;
            mn[i] = min(mn[i], n);
            mx[i] = max(mx[i], n);
        }
        int ans = gap, prev = minVal;
        for (int i = 0; i < N - 1; ++i) {
            if (mn[i] == INT_MAX) continue;
            ans = max(ans, mn[i] - prev);
            prev = mx[i];
        }
        return max(ans, maxVal - prev);
    }
};