910. Smallest Range II #

Problem #

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

Problem Summary #

Given an integer array A, for each integer A[i], you can choose x = -K or x = K (K is always a non-negative integer), and add x to A[i]. After this process, array B is obtained. Return the smallest possible difference between the maximum value of B and the minimum value of B.

Solution Explanation #

• Easy problem. First sort the array and find the largest difference in array A. Then scan through the array once, using two pointers, choosing x = -K or x = K. Each choice updates the smallest difference between the maximum and minimum values. After one pass, the answer that satisfies the problem can be found.

Code #

package leetcode

import "sort"

func smallestRangeII(A []int, K int) int {
	n := len(A)
	sort.Ints(A)
	res := A[n-1] - A[0]
	for i := 0; i < n-1; i++ {
		a, b := A[i], A[i+1]
		high := max(A[n-1]-K, a+K)
		low := min(A[0]+K, b-K)
		res = min(res, high-low)
	}
	return res
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}