This one is to use bucket sort.
Review bucket sort process first:
Suppose there are N elements and they range from A to B, then, the maximum gap will be no smaller then ceiling((B - A) / (n-1))
Let the length of a bucket to be len = ceiling((B-A) / N - 1), then, we will have at most num = (B - A) / len + 1 numbers of buckets. (might have empty bucket)
For any number K in the array, we can easily find out which bucket it belongs by calculating loc = (K -A) / len and therefore maintain
the maximum and minmum elements in each bucket.
Since the maximum difference between elements in the same buckets will be at most len - 1, so the final answer will not be taken
from two elements in the same buckets. For each non-empty buckets p, find the next non-empty buckets q, then q.min-p.max could be
the potential answer to the question. return the maximum of all those values.
An example will make the explain clear.
Suppose, we have array, [2, 3, 7, 9, 100] (it actually doesn't matter whether the given array is sorted or not.)
So, maximum = 100, min = 2 --> bucket len = (max - min) / (size-1) -- > 98/4 --> 25 (ceiling value)
Then, bucket number is : (100 - 2)/ len + 1 --> 98 / 25 + 1 --> 5. We thus have 5 buckets.
Location of input number is calculated by (num[i] - min) / len
2 ----> loc = 0
3 ---> loc = 0
7 ----> loc = 0
9---> loc = 0
100--> loc = 4. In this case, bucket 1, 2, 3 are empty (we didn't use them)
Following code is copied from others.
int maximumGap(vector<int> &num) { if (num.size() < 2) return 0; int maxNum = num[0]; int minNum = num[0]; for (int i : num) { maxNum=max(maxNum,i); minNum=min(minNum,i); } // bucket len int len = (maxNum - minNum) / num.size() + 1; // bucket number vector<vector<int>> buckets((maxNum - minNum) / len + 1); for (int x : num) { int i = (x - minNum) / len; if (buckets[i].empty()) { buckets[i].reserve(2); buckets[i].push_back(x); buckets[i].push_back(x); } else { if (x < buckets[i][0]) buckets[i][0] = x; if (x > buckets[i][1]) buckets[i][1] = x; } } int gap = 0; int prev = 0; for (int i = 1; i < buckets.size(); i++) { if (buckets[i].empty()) continue; gap = max(gap, buckets[i][0] - buckets[prev][1]); prev = i; } return gap; }