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63880d4717 feat(makeTheIntegerZeio): add C# solution 2025-09-05 16:54:03 +08:00
c3955ad390 docs(makeTheIntegerZero): add README 2025-09-05 16:52:45 +08:00
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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

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// LeetCode 2749: Minimum Operations To Make The Integer Zero
// 難度: Medium
// 日期: 2025-09-05
using System;
using System.Collections.Generic;
using System.Linq;
public class Solution
{
public int MakeTheIntegerZero(int num1, int num2)
{
for (int k = 1; k <= 60; k++)
{
long target = (long)num1 - (long)k * num2;
if (target < 0) continue;
int bitCount = CountSetBits(target);
if (bitCount <= k && k <= target)
{
return k;
}
}
return -1;
}
private int CountSetBits(long n)
{
int count = 0;
while (n > 0)
{
// 方法一
// count += (int)(n & 1);
// n >>= 1;
// 方法二
count++;
n &= n - 1;
}
return count;
}
}
public class Program
{
public static void Main()
{
Solution solution = new Solution();
TestCase1(solution);
TestCase2(solution);
}
static void TestCase1(Solution solution)
{
int num1 = 3, num2 = -2;
int expected = 3;
int actual = solution.MakeTheIntegerZero(num1, num2);
Console.WriteLine("Test 1:");
Console.WriteLine($"Input: num1 = {num1}, num2 = {num2}");
Console.WriteLine($"Expected: {expected}");
Console.WriteLine($"Actual: {actual}");
Console.WriteLine($"Result: {(actual == expected ? "PASS" : "FAIL")}");
Console.WriteLine();
}
static void TestCase2(Solution solution)
{
int num1 = 5, num2 = 7;
int expected = -1;
int actual = solution.MakeTheIntegerZero(num1, num2);
Console.WriteLine("Test 2:");
Console.WriteLine($"Input: num1 = {num1}, num2 = {num2}");
Console.WriteLine($"Expected: {expected}");
Console.WriteLine($"Actual: {actual}");
Console.WriteLine($"Result: {(actual == expected ? "PASS" : "FAIL")}");
Console.WriteLine();
}
}

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<PropertyGroup>
<OutputType>Exe</OutputType>
<TargetFramework>net8.0</TargetFramework>
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</Project>