Problem
Given an array A
of integers, for each integer A[i]
we may choose any x
with -K <= x <= K
, and add x
to A[i]
.
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: 0 Explanation: B = [3,3,3] or B = [4,4,4]
Note:
1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000
Solution 0: Brute Force (TLE)
Try all pairs
Time complexity: O(n^2)
Space complexity: O(1)
Solution 1: Math
Time complexity: O(n)
Space complexity: O(1)
Find the min/max element of the array.
min + k v.s. max – k
ans = max(0, (max – min) – 2 * k))
C++
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// Author: Huahua class Solution { public: int smallestRangeI(vector<int>& A, int K) { int a_min = *min_element(begin(A), end(A)); int a_max = *max_element(begin(A), end(A)); return max(0, (a_max - a_min) - 2 * K); } }; |
Python3
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# Author: Huahua, 72 ms class Solution: def smallestRangeI(self, A, K): return max(0, max(A) - min(A) - 2 * K); |
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