feat: solve No.1980

Author: BaffinLee

1901-2000/1980. Find Unique Binary String.md

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+# 1980. Find Unique Binary String
+
+- Difficulty: Medium.
+- Related Topics: Array, String, Backtracking.
+- Similar Questions: Missing Number, Find All Numbers Disappeared in an Array, Random Pick with Blacklist.
+
+## Problem
+
+Given an array of strings `nums` containing `n` **unique** binary strings each of length `n`, return **a binary string of length **`n`** that **does not appear** in **`nums`**. If there are multiple answers, you may return **any** of them**.
+
+ 
+Example 1:
+
+```
+Input: nums = ["01","10"]
+Output: "11"
+Explanation: "11" does not appear in nums. "00" would also be correct.
+```
+
+Example 2:
+
+```
+Input: nums = ["00","01"]
+Output: "11"
+Explanation: "11" does not appear in nums. "10" would also be correct.
+```
+
+Example 3:
+
+```
+Input: nums = ["111","011","001"]
+Output: "101"
+Explanation: "101" does not appear in nums. "000", "010", "100", and "110" would also be correct.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `n == nums.length`
+	
+- `1 <= n <= 16`
+	
+- `nums[i].length == n`
+	
+- `nums[i] `is either `'0'` or `'1'`.
+	
+- All the strings of `nums` are **unique**.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string[]} nums
+ * @return {string}
+ */
+var findDifferentBinaryString = function(nums) {
+    var str = '';
+    for (var i = 0; i <= nums.length; i++) {
+        str = i.toString(2);
+        str = '0'.repeat(nums.length - str.length) + str;
+        if (!nums.includes(str)) {
+            return str;
+        }
+    }
+};
+```
+
+**Explain:**
+
+Since array `nums` only contains `n` numbers, if we got `n + 1` numbers, there mush have at lease one number not in that array.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(1).