[NEW][Array] 910. Smallest Range II

Author: Cappuccinuo

Array/910. Smallest Range II/README.md

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+Time spent :  15min(have little thought, see hint)
+
+GIVEN: 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`.
+
+RETURNS: the smallest possible difference between the maximum value of `B` and the minimum value of `B`.
+
+EXAMPLES: 
+
+**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]
+```
+
+Algorithm:
+
+For an array like [1, 3, 4, 7]， we can somewhat split the array into two parts, one parts all add K, one parts all subtract K, like (1, 3) + K, (4, 7) - K. For every this split, we can get max - min.
+
+We sort the array first. Then we traverse the array from index 0 to N - 2, a = A[i], b = A[i + 1]. So the max is a + K, or A[N - 1] - K, same for min, is b - K or A[0] + K. Every time we compare the max - min with our ans and choose the smallest one.

Array/910. Smallest Range II/Solution.java

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