164. Maximum Gap

Given an unsorted array, find the maximum difference between the successive elements in its sorted form.

Return 0 if the array contains less than 2 elements.

Example 1:

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

Note:

• You may assume all elements in the array are non-negative integers and fit in the 32-bit signed integer range.

• Try to solve it in linear time/space.

// Bucket Sort
int maximumGap(vector<int>& nums) { // time: O(n); space: O(n)
    int n = nums.size();
    if (n < 2) return 0; // invalid input
    int mx = nums[0], mn = nums[0];
    // find the max/min in the input
    for (int num : nums) {
        mx = max(mx, num);
        mn = min(mn, num);
    }
    // the minimum possible gap
    int gap = max(1, (mx - mn) / (n - 1)); // make sure the value is at least 1
    int m = (mx - mn) / gap + 1;
    vector<vector<int> > buckets(m);
    // put numbers into buckets
    for (int num : nums) {
        int idx = (num - mn) / gap;
        if (buckets[idx].empty()) {
            buckets[idx] = vector<int>(2, num);
        } else {
            buckets[idx][0] = min(buckets[idx][0], num);
            buckets[idx][1] = max(buckets[idx][1], num);
        }
    }
    int prev = 0;
    gap = 0;
    // scan and find the max gap
    for (int i = 0; i < m; ++i) {
        if (buckets[i].empty()) continue;
        if (buckets[i][0] - buckets[prev][1] > gap)
            gap = buckets[i][0] - buckets[prev][1];
        prev = i;
    }
    return gap;
}