910. Smallest Range II

https://leetcode.com/problems/smallest-range-ii/

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 3
Explanation: B = [4,6,3]

Note:

1. 1 <= A.length <= 10000

2. 0 <= A[i] <= 10000

3. 0 <= K <= 10000

Thoughts

对给定数组每个元素可选+K或-K，问选择后argmin(最大值-最小值）。最大和最小=>sorting：从小到大排序，右边那头往下压，左边的往上提，从而使得最大和最小之间的差最小。假设A[i]是左边往上提的最右边的元素，最大值可能是A[i] + K或A[-1] - K，那最小值可能是A[i + 1] - K或A[0] + K。遍历所有的i看全局最优。

Code

class Solution:
    def smallestRangeII(self, A: List[int], K: int) -> int:
        A.sort()
        res = A[-1] - A[0]
        for i in range(len(A) - 1):
            diff = max(A[i] + K, A[-1] - K) - min(A[0] + K, A[i + 1] - K)
            res = min(res, diff)
        return res
        

class Solution {
public:
    int smallestRangeII(vector<int>& A, int K) {
        sort(A.begin(), A.end());
        int n = A.size(), mi = A[0] + K, mx = A[n - 1] - K, res = A[n - 1] - A[0];
        for (int i = 0; i < n - 1; ++i) {
            mi = min(A[0] + K, A[i + 1] - K);
            mx = max(A[n - 1] - K, A[i] + K);
            res = min(res, mx - mi);
        }

        return res;
    }
};

Analysis

时间复杂度O(nlgn)