[LeetCode 910] 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

 

The key observation to solve this problem is that if A is already in sorted order, then it makes no sense to do -K on a smaller number and +K on a bigger number since this will only generate a bigger difference.  So either we keep the relative differences among all elements the same by applying +K or -K to all, or we apply +K on some smaller numbers and -K on bigger numbers, hoping to find a better answer. 

 

1.  Sort A and set the current answer as A[n - 1] - A[0].

2. From A[0] to A[n - 2], use the current number A[i] as the split point, apply +K to A[0, i] and -K to A[i + 1, n - 1], compute the new max and min and update answer.

  

class Solution {
    public int smallestRangeII(int[] A, int K) {
        Arrays.sort(A);
        int n = A.length, minV = A[0], maxV = A[n - 1], ans = maxV - minV;
        for(int i = 0; i < n - 1; i++) {
            maxV = Math.max(A[i] + K, A[n - 1] - K);
            minV = Math.min(A[0] + K, A[i + 1] - K);
            ans = Math.min(ans, maxV - minV);
        }
        return ans;
    }
}