leetcode Smallest Range II

最新推荐文章于 2025-06-02 00:45:00 发布

原创最新推荐文章于 2025-06-02 00:45:00 发布 · 223 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

475 篇文章

订阅专栏

leetcode

231 篇文章

订阅专栏

本文探讨了如何通过增加或减去整数K来调整数组中每个元素，以最小化数组最大值与最小值之间的差距。提供了详细的算法解决方案及代码实现。

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 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000

Solution:
A[0]-K,A[0]+K
…
A[i]-K,A[i]+K
A[i+1]-K,A[i+1]+K
…
A[len-1]-K,A[len-1]+K

A[0]-K and A[len-1]+K won’t have effects in the final range. The final lower bound will be one of (A[0]+K, A[i]-K, A[i+1]-K) and the final upper bound will be one of (A[i]+K, A[i+1]+K, A[len-1]-K). But for different timestamp, the combination is different.
i=0, lower bound: (A[0]+K, A[1]-K) upper bound:(A[0]+K, A[len-1]-K)
i=1, lower bound: (A[0]+K, A[2]-K) upper bound:(A[1]+K, A[len-1]-K)
…
i=len-2, lower bound: (A[0]+K, A[len-1]-K) upper bound:(A[len-2]+K, A[len-1]-K)

So the code is:

class Solution:
    def smallestRangeII(self, A, K):
        la = len(A)
        if (la == 0):
            return 0
        A.sort()
        res = A[la-1]-A[0]
        for i in range(la-1):
            mi = min(A[0]+K,A[i+1]-K)
            ma = max(A[la-1]-K, A[i]+K)
            res = min(ma-mi, res)
        return res