feat: solve No.2215

Author: BaffinLee

2201-2300/2215. Find the Difference of Two Arrays.md

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+# 2215. Find the Difference of Two Arrays
+
+- Difficulty: Easy.
+- Related Topics: Array, Hash Table.
+- Similar Questions: Intersection of Two Arrays, Intersection of Two Arrays II, Intersection of Multiple Arrays.
+
+## Problem
+
+Given two **0-indexed** integer arrays `nums1` and `nums2`, return **a list** `answer` **of size** `2` **where:**
+
+
+	
+- `answer[0]` **is a list of all **distinct** integers in** `nums1` **which are **not** present in** `nums2`**.**
+	
+- `answer[1]` **is a list of all **distinct** integers in** `nums2` **which are **not** present in** `nums1`.
+
+
+**Note** that the integers in the lists may be returned in **any** order.
+
+ 
+Example 1:
+
+```
+Input: nums1 = [1,2,3], nums2 = [2,4,6]
+Output: [[1,3],[4,6]]
+Explanation:
+For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].
+For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].
+```
+
+Example 2:
+
+```
+Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]
+Output: [[3],[]]
+Explanation:
+For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].
+Every integer in nums2 is present in nums1. Therefore, answer[1] = [].
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= nums1.length, nums2.length <= 1000`
+	
+- `-1000 <= nums1[i], nums2[i] <= 1000`
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {number[]} nums1
+ * @param {number[]} nums2
+ * @return {number[][]}
+ */
+var findDifference = function(nums1, nums2) {
+    var map = {};
+    for (var i = 0; i < nums1.length; i++) {
+        map[nums1[i]] = (map[nums1[i]] || 0) | 1;
+    }
+    for (var i = 0; i < nums2.length; i++) {
+        map[nums2[i]] = (map[nums2[i]] || 0) | 2;
+    }
+    var res = [new Set(), new Set()];
+    for (var i = 0; i < nums1.length; i++) {
+        if (!(map[nums1[i]] & 2)) {
+            res[0].add(nums1[i]);
+        }
+    }
+    for (var i = 0; i < nums2.length; i++) {
+        if (!(map[nums2[i]] & 1)) {
+            res[1].add(nums2[i]);
+        }
+    }
+    return res.map(item => Array.from(item));
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).