From c3955ad3909540854b0e51771529ee61386d5a72 Mon Sep 17 00:00:00 2001
From: MH Hung <mh.hung1994@gmail.com>
Date: Fri, 5 Sep 2025 16:52:45 +0800
Subject: [PATCH] docs(makeTheIntegerZero): add README

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+# [2749] Minimum Operations To Make The Integer Zero
+
+## 題目資訊
+- **難度**: Medium
+- **標籤**: Bit Manipulation, Brainteaser, Enumeration
+- **題目連結**: [LeetCode](https://leetcode.com/problems/minimum-operations-to-make-the-integer-zero/)
+- **練習日期**: 2025-09-05
+
+## 題目描述
+You are given two integers `num1` and `num2`.
+
+In one operation, you can choose integer i in the range `[0, 60]` and subtract `2^i + num2` from `num1`.
+
+Return the integer denoting the *minimum* number of operations needed to make `num1` equal to `0`.
+
+If it is impossible to make `num1` equal to `0`, return `-1`.
+
+## 解題思路
+
+### 初步分析
+- 這題主要考察什麼概念？
+   1. 位元操作 (Bit Manipulation)：需要理解二進制表示和位元運算
+   2. 數學建模：將實際問題轉換為數學等式
+   3. 枚舉 (Enumeration)：嘗試不同的操作次數 k
+   4. 約束條件判斷：理解多個限制條件的邏輯關係
+- 有什麼關鍵限制條件？
+   1. target ≥ 0
+   2. bitCount(target) ≤ k
+   3. k ≤ target
+- 預期時間/空間複雜度？
+   - 時間複雜度：O(60 × log(target)) ≈ O(1)
+   - 空間複雜度：O(1)
+
+### 解法概述
+1. **解法**: 
+   - 思路：
+      - 目標：讓 num1 變成 0
+      -  每次操作：num1 = num1 - (2^i + num2)
+      - k 次操作後：num1 - k*num2 - (2^i1 + 2^i2 + ... + 2^ik) = 0
+      - 重新整理：target = num1 - k*num2 = 2^i1 + 2^i2 + ... + 2^ik
+   - 時間複雜度：O(1)
+   - 空間複雜度：O(1)
+
+## 測試案例
+
+### 範例輸入輸出
+```
+Input: num1 = 3, num2 = -2
+Output: 3
+Explanation: 
+We can make 3 equal to 0 with the following operations:
+- We choose i = 2 and subtract 22 + (-2) from 3, 3 - (4 + (-2)) = 1.
+- We choose i = 2 and subtract 22 + (-2) from 1, 1 - (4 + (-2)) = -1.
+- We choose i = 0 and subtract 20 + (-2) from -1, (-1) - (1 + (-2)) = 0.
+It can be proven, that 3 is the minimum number of operations that we need to perform.
+```
+
+### 邊界情況
+ - `1 <= num1 <= 10^9`
+ - `-10^9 <= num2 <= 10^9`
+
+## 學習筆記
+
+### 今天學到什麼？
+- 二位元的操作寫法
+
+### 遇到的困難
+- 二位元的操作
+   1. 方法一: 逐位檢查法
+      - 程式碼：
+      ``` C#
+      count += (int)(n & 1);
+      n >>= 1;
+      ```
+      - 運作原理：
+         1. n & 1：檢查最右邊的位元是否為 1
+         2. count += (int)(n & 1)：如果是 1 就加到計數器
+         3. n >>= 1：把 n 右移一位（去掉已檢查的位元）
+         4. 重複直到 n 變成 0
+   2. 方法二: 移除最右邊 1
+      - 程式碼：
+      ``` C#
+      count++;
+      n &= n - 1;
+      ```
+      - 運作原理：
+         1. n - 1：讓最右邊的 1 變成 0，其右邊的 0 都變成 1
+         2. n & (n - 1)：神奇地移除了最右邊的 1
+         3. count++：每移除一個 1，計數器就加 1
+         4. 重複直到 n 變成 0
+
+### 改善方向
+- 
+
+### 相關題目
+- [#991](https://leetcode.com/problems/broken-calculator) Broken Calculator
+- [#1658](https://leetcode.com/problems/minimum-operations-to-reduce-x-to-zero/) Minimum Operations to Reduce X to Zero
+
+---
+**總結**:
+1. 學習二位元的使用技巧