164. Maximum Gap

 1 public class Solution {
 2     public int maximumGap(int[] nums) {
 3         // corner case:if the input is null or the length of the input array is zero or one
 4         if(nums==null||nums.length<2) return 0;
 5         int min = Integer.MAX_VALUE;
 6         int max = Integer.MIN_VALUE;
 7         //find the min and max of the array
 8         for(int i=0;i<nums.length;i++){
 9             min = Math.min(min,nums[i]);
10             max = Math.max(max,nums[i]);
11         }
12         // Math.ceil is the min Integer that >= i;Math.floor is the max integer that <=i;
13         // the gap represents the average gap value between two numbers
14         int gap = (int)Math.ceil((double)(max-min)/(nums.length-1));
15         int[] bucketsMax = new int[nums.length-1];
16         int[] bucketsMin = new int[nums.length-1];
17         Arrays.fill(bucketsMax,Integer.MIN_VALUE);
18         Arrays.fill(bucketsMin,Integer.MAX_VALUE);
19         // put numbers into buckets
20         for(int i:nums){
21             if(i==min||i==max) continue;
22             int idx = (i-min)/gap;
23             bucketsMax[idx] = Math.max(i,bucketsMax[idx]);
24             bucketsMin[idx] = Math.min(i,bucketsMin[idx]);
25         }
26         int maxGap = Integer.MIN_VALUE;
27         int previous = min;
28         // find the max neighboring buckets
29         for(int i=0;i<nums.length-1;i++){
30             if(bucketsMin[i]==Integer.MAX_VALUE||bucketsMax[i] ==Integer.MIN_VALUE){
31                 continue;
32             }
33             maxGap = Math.max(maxGap,bucketsMin[i]-previous);
34             previous = bucketsMax[i];
35         }
36         maxGap = Math.max(maxGap,max - previous);
37         return maxGap;
38     }
39 }
40 // the total run time could be O(n) time,and the space complexity could be O(n)