feat: No.1047,2125

Author: BaffinLee

Diff for: 1001-1100/1047. Remove All Adjacent Duplicates In String.md

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+# 1047. Remove All Adjacent Duplicates In String
+
+- Difficulty: Easy.
+- Related Topics: String, Stack.
+- Similar Questions: Remove All Adjacent Duplicates in String II, Removing Stars From a String, Minimize String Length.
+
+## Problem
+
+You are given a string `s` consisting of lowercase English letters. A **duplicate removal** consists of choosing two **adjacent** and **equal** letters and removing them.
+
+We repeatedly make **duplicate removals** on `s` until we no longer can.
+
+Return **the final string after all such duplicate removals have been made**. It can be proven that the answer is **unique**.
+
+ 
+Example 1:
+
+```
+Input: s = "abbaca"
+Output: "ca"
+Explanation: 
+For example, in "abbaca" we could remove "bb" since the letters are adjacent and equal, and this is the only possible move.  The result of this move is that the string is "aaca", of which only "aa" is possible, so the final string is "ca".
+```
+
+Example 2:
+
+```
+Input: s = "azxxzy"
+Output: "ay"
+```
+
+ 
+**Constraints:**
+
+
+	
+- `1 <= s.length <= 105`
+	
+- `s` consists of lowercase English letters.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string} s
+ * @return {string}
+ */
+var removeDuplicates = function(s) {
+    var stack = [];
+    for (var i = 0; i <= s.length; i++) {
+        if (stack.length >= 2 && stack[stack.length - 1] === stack[stack.length - 2]) {
+            stack.pop();
+            stack.pop();
+        }
+        i !== s.length && stack.push(s[i]);
+    }
+    return stack.join('');
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n).
+* Space complexity : O(n).

Diff for: 2101-2200/2125. Number of Laser Beams in a Bank.md

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+# 2125. Number of Laser Beams in a Bank
+
+- Difficulty: Medium.
+- Related Topics: Array, Math, String, Matrix.
+- Similar Questions: Set Matrix Zeroes.
+
+## Problem
+
+Anti-theft security devices are activated inside a bank. You are given a **0-indexed** binary string array `bank` representing the floor plan of the bank, which is an `m x n` 2D matrix. `bank[i]` represents the `ith` row, consisting of `'0'`s and `'1'`s. `'0'` means the cell is empty, while`'1'` means the cell has a security device.
+
+There is **one** laser beam between any **two** security devices **if both** conditions are met:
+
+
+	
+- The two devices are located on two **different rows**: `r1` and `r2`, where `r1 < r2`.
+	
+- For **each** row `i` where `r1 < i < r2`, there are **no security devices** in the `ith` row.
+
+
+Laser beams are independent, i.e., one beam does not interfere nor join with another.
+
+Return **the total number of laser beams in the bank**.
+
+ 
+Example 1:
+
+![](https://assets.leetcode.com/uploads/2021/12/24/laser1.jpg)
+
+```
+Input: bank = ["011001","000000","010100","001000"]
+Output: 8
+Explanation: Between each of the following device pairs, there is one beam. In total, there are 8 beams:
+ * bank[0][1] -- bank[2][1]
+ * bank[0][1] -- bank[2][3]
+ * bank[0][2] -- bank[2][1]
+ * bank[0][2] -- bank[2][3]
+ * bank[0][5] -- bank[2][1]
+ * bank[0][5] -- bank[2][3]
+ * bank[2][1] -- bank[3][2]
+ * bank[2][3] -- bank[3][2]
+Note that there is no beam between any device on the 0th row with any on the 3rd row.
+This is because the 2nd row contains security devices, which breaks the second condition.
+```
+
+Example 2:
+
+![](https://assets.leetcode.com/uploads/2021/12/24/laser2.jpg)
+
+```
+Input: bank = ["000","111","000"]
+Output: 0
+Explanation: There does not exist two devices located on two different rows.
+```
+
+ 
+**Constraints:**
+
+
+	
+- `m == bank.length`
+	
+- `n == bank[i].length`
+	
+- `1 <= m, n <= 500`
+	
+- `bank[i][j]` is either `'0'` or `'1'`.
+
+
+
+## Solution
+
+```javascript
+/**
+ * @param {string[]} bank
+ * @return {number}
+ */
+var numberOfBeams = function(bank) {
+    var res = 0;
+    var lastRowDeviceNum = 0;
+    for (var i = 0; i < bank.length; i++) {
+        var deviceNum = 0;
+        for (var j = 0; j < bank[i].length; j++) {
+            if (bank[i][j] === '1') deviceNum++;
+        }
+        if (deviceNum === 0) continue;
+        res += lastRowDeviceNum * deviceNum;
+        lastRowDeviceNum = deviceNum;
+    }
+    return res;
+};
+```
+
+**Explain:**
+
+nope.
+
+**Complexity:**
+
+* Time complexity : O(n * m).
+* Space complexity : O(1).