164. Maximum Gap

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

Try to solve it in linear time/space.

Return 0 if the array contains less than 2 elements.

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

public class Solution {
    public int maximumGap(int[] nums) {
        // corner case:if the input is null or the length of the input array is zero or one
        if(nums==null||nums.length<2) return 0;
        int min = Integer.MAX_VALUE;
        int max = Integer.MIN_VALUE;
        //find the min and max of the array
        for(int i=0;i<nums.length;i++){
            min = Math.min(min,nums[i]);
            max = Math.max(max,nums[i]);
        }
        // Math.ceil is the min Integer that >= i;Math.floor is the max integer that <=i;
        // the gap represents the average gap value between two numbers
        int gap = (int)Math.ceil((double)(max-min)/(nums.length-1));
        int[] bucketsMax = new int[nums.length-1];
        int[] bucketsMin = new int[nums.length-1];
        Arrays.fill(bucketsMax,Integer.MIN_VALUE);
        Arrays.fill(bucketsMin,Integer.MAX_VALUE);
        // put numbers into buckets
        for(int i:nums){
            if(i==min||i==max) continue;
            int idx = (i-min)/gap;
            bucketsMax[idx] = Math.max(i,bucketsMax[idx]);
            bucketsMin[idx] = Math.min(i,bucketsMin[idx]);
        }
        int maxGap = Integer.MIN_VALUE;
        int previous = min;
        // find the max neighboring buckets
        for(int i=0;i<nums.length-1;i++){
            if(bucketsMin[i]==Integer.MAX_VALUE||bucketsMax[i] ==Integer.MIN_VALUE){
                continue;
            }
            maxGap = Math.max(maxGap,bucketsMin[i]-previous);
            previous = bucketsMax[i];
        }
        maxGap = Math.max(maxGap,max - previous);
        return maxGap;
    }
}
// the total run time could be O(n) time,and the space complexity could be O(n)