Given an array
A
of integers, for each integerA[i]
we may choose anyx
with-K <= x <= K
, and addx
toA[i]
.After this process, we have some array
B
.Return the smallest possible difference between the maximum value of
B
and the minimum value ofB
.
Example 1:
Input: A = [1], K = 0Output: 0Explanation: B = [1]Example 2:
Input: A = [0,10], K = 2Output: 6Explanation: B = [2,8]Example 3:
Input: A = [1,3,6], K = 3Output: 0Explanation: B = [3,3,3] or B = [4,4,4]
Note:
1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000
Approach #1: Math. [Java]
class Solution { public int smallestRangeI(int[] A, int K) { int min = Integer.MAX_VALUE; int max = Integer.MIN_VALUE; for (int i = 0; i < A.length; ++i) { if (min > A[i]) min = A[i]; if (max < A[i]) max = A[i]; } if (max - min < 2 * K) return 0; return max - min - 2 * K; }}
Analysis:
If you can find that the result only relate with (max - min) and 2 * K, it will become easy to solve.