Maximum Gap

Description:

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. so that objects of the same color are adjacent, with the colors in the order 1, 2, ... k.

Example

Given [1, 9, 2, 5], the sorted form of it is [1, 2, 5, 9], the maximum gap is between 5 and 9 = 4.

Note

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

Challenge

Sort is easy but will cost O(nlogn) time. Try to solve it in linear time and space.

Solution:

class Solution {
public:
	/**
	 * @param nums: a vector of integers
	 * @return: the maximum difference
	 */
	int maximumGap(vector<int> nums) {
		auto sz = (int)nums.size();
		if (sz <= 1) return 0;
		int maxn = nums[0];
		int minn = nums[0];
		for (int i = 1; i < sz; ++i) {
			maxn = max(maxn, nums[i]);
			minn = min(minn, nums[i]);
		}
		int bucketSize = (maxn-minn) / sz + 1;
		// int bucketSize = max((maxn-minn)/(sz-1), 1)
		int bucketNum = (maxn-minn) / bucketSize + 1;
		vector<int> maxBuckets(bucketNum, -1);
		vector<int> minBuckets(bucketNum, maxn);
		for (auto& e : nums) {
			auto ind = (e-minn)/bucketSize;
			maxBuckets[ind] = max(maxBuckets[ind], e);
			minBuckets[ind] = min(minBuckets[ind], e);
		}
		int rc = 0;
		int i = 0;
		while (i < bucketNum && maxBuckets[i] == -1) ++i;
		int pre = maxBuckets[i];
		while (i < bucketNum) {
			while (i < bucketNum && maxBuckets[i] == -1) ++i;
			if (i < bucketNum) {
				rc = max(rc, minBuckets[i]-pre);
				pre = maxBuckets[i];
				++i;
			}
		}
		return rc;
	}
};