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.

 

AC code:

class Solution {
public:
    int maximumGap(vector<int>& nums) {
        int len = nums.size();
        if (len < 2) return 0;
        int minn = nums[0];
        int maxx = nums[0]; 
        for (int i = 1; i < len; ++i) {
            minn = min(minn, nums[i]);
            maxx = max(maxx, nums[i]);
        }
        int gaps = (int)ceil((double)(maxx - minn) / (len-1));  // the numbers of burket
        vector<int> gap_min(len-1, INT_MAX);
        vector<int> gap_max(len-1, INT_MIN);
        for (int i = 0; i < len; ++i) {
            if (nums[i] == minn || nums[i] == maxx)
                continue;
            int idx = (nums[i] - minn) / gaps;
            gap_min[idx] = min(gap_min[idx], nums[i]);
            gap_max[idx] = max(gap_max[idx], nums[i]);
        }
        int maxdis = INT_MIN;
        int prev = minn;
        for (int i = 0; i < len-1; ++i) {
            if (gap_min[i] == INT_MAX && gap_max[i] == INT_MIN)
                continue;
            maxdis = max(maxdis, gap_min[i] - prev);
            prev = gap_max[i];
        }
        maxdis = max(maxx-prev, maxdis);
        return maxdis;
    }
};

Runtime: 8 ms, faster than 56.67% of C++ online submissions for Maximum Gap.

 

Analylis:

step1:

calculating numbers between the min number and the max number in the array. 

step2:

using gaps buckets to equallt distributed the numbers.

step3:

find the max and min numbers in each buckets.

step4:

the maximum gap is max(gaps, buckets[i]'s min_number - buckets[i-1]'s max_number);