From 125b165d67601740eb79eea8ee18b2ba03b197b5 Mon Sep 17 00:00:00 2001 From: Libin YANG Date: Mon, 22 Sep 2025 21:28:46 +0800 Subject: [PATCH] feat: add solutions to lc problem: No.2429 --- .../2400-2499/2429.Minimize XOR/README.md | 136 ++++++------------ .../2400-2499/2429.Minimize XOR/README_EN.md | 136 ++++++------------ .../2400-2499/2429.Minimize XOR/Solution.cs | 19 +++ .../2400-2499/2429.Minimize XOR/Solution.rs | 27 ++++ .../2400-2499/2429.Minimize XOR/Solution2.cpp | 14 -- .../2400-2499/2429.Minimize XOR/Solution2.go | 11 -- .../2429.Minimize XOR/Solution2.java | 13 -- .../2400-2499/2429.Minimize XOR/Solution2.py | 11 -- .../2400-2499/2429.Minimize XOR/Solution2.ts | 20 --- 9 files changed, 134 insertions(+), 253 deletions(-) create mode 100644 solution/2400-2499/2429.Minimize XOR/Solution.cs create mode 100644 solution/2400-2499/2429.Minimize XOR/Solution.rs delete mode 100644 solution/2400-2499/2429.Minimize XOR/Solution2.cpp delete mode 100644 solution/2400-2499/2429.Minimize XOR/Solution2.go delete mode 100644 solution/2400-2499/2429.Minimize XOR/Solution2.java delete mode 100644 solution/2400-2499/2429.Minimize XOR/Solution2.py delete mode 100644 solution/2400-2499/2429.Minimize XOR/Solution2.ts diff --git a/solution/2400-2499/2429.Minimize XOR/README.md b/solution/2400-2499/2429.Minimize XOR/README.md index d08cad94b4057..38d620ed51462 100644 --- a/solution/2400-2499/2429.Minimize XOR/README.md +++ b/solution/2400-2499/2429.Minimize XOR/README.md @@ -70,9 +70,9 @@ num1 和 num2 的二进制表示分别是 0001 和 1100 。 ### 方法一：贪心 + 位运算 -根据题目描述，我们先求出 $num2$ 的置位数 $cnt$，然后从高位到低位枚举 $num1$ 的每一位，如果该位为 $1$，则将 $x$ 的对应位设为 $1$，并将 $cnt$ 减 $1$，直到 $cnt$ 为 $0$。如果此时 $cnt$ 仍不为 $0$，则从低位开始将 $num1$ 的每一位为 $0$ 的位置设为 $1$，并将 $cnt$ 减 $1$，直到 $cnt$ 为 $0$。 +根据题目描述，我们先求出 $\textit{nums2}$ 的置位数 $\textit{cnt}$，然后从高位到低位枚举 $\textit{num1}$ 的每一位，如果该位为 $1$，则将 $x$ 的对应位设为 $1$，并将 $\textit{cnt}$ 减 $1$，直到 $\textit{cnt}$ 为 $0$。如果此时 $\textit{cnt}$ 仍不为 $0$，则从低位开始将 $\textit{num1}$ 的每一位为 $0$ 的位置设为 $1$，并将 $\textit{cnt}$ 减 $1$，直到 $\textit{cnt}$ 为 $0$。 -时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。其中 $n$ 为 $num1$ 和 $num2$ 的最大值。 +时间复杂度 $O(\log n)$，其中 $n$ 为 $\textit{num1}$ 和 $\textit{num2}$ 的最大值。空间复杂度 $O(1)$。 @@ -191,107 +191,59 @@ function minimizeXor(num1: number, num2: number): number { } ``` - - - - - - -### 方法二 - - +#### Rust -#### Python3 - -```python -class Solution: - def minimizeXor(self, num1: int, num2: int) -> int: - cnt1 = num1.bit_count() - cnt2 = num2.bit_count() - while cnt1 > cnt2: - num1 &= num1 - 1 - cnt1 -= 1 - while cnt1 < cnt2: - num1 |= num1 + 1 - cnt1 += 1 - return num1 -``` - -#### Java - -```java -class Solution { - public int minimizeXor(int num1, int num2) { - int cnt1 = Integer.bitCount(num1); - int cnt2 = Integer.bitCount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= (num1 - 1); +```rust +impl Solution { + pub fn minimize_xor(num1: i32, mut num2: i32) -> i32 { + let mut cnt = 0; + while num2 > 0 { + num2 -= num2 & -num2; + cnt += 1; } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= (num1 + 1); + let mut x = 0; + let mut c = cnt; + for i in (0..=30).rev() { + if c > 0 && (num1 >> i) & 1 == 1 { + x |= 1 << i; + c -= 1; + } } - return num1; + for i in 0..=30 { + if c == 0 { + break; + } + if ((num1 >> i) & 1) == 0 { + x |= 1 << i; + c -= 1; + } + } + x } } ``` -#### C++ +#### C# -```cpp -class Solution { -public: - int minimizeXor(int num1, int num2) { - int cnt1 = __builtin_popcount(num1); - int cnt2 = __builtin_popcount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= (num1 - 1); +```cs +public class Solution { + public int MinimizeXor(int num1, int num2) { + int cnt = BitOperations.PopCount((uint)num2); + int x = 0; + for (int i = 30; i >= 0 && cnt > 0; --i) { + if (((num1 >> i) & 1) == 1) { + x |= 1 << i; + --cnt; + } } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= (num1 + 1); + for (int i = 0; cnt > 0; ++i) { + if (((num1 >> i) & 1) == 0) { + x |= 1 << i; + --cnt; + } } - return num1; - } -}; -``` - -#### Go - -```go -func minimizeXor(num1 int, num2 int) int { - cnt1 := bits.OnesCount(uint(num1)) - cnt2 := bits.OnesCount(uint(num2)) - for ; cnt1 > cnt2; cnt1-- { - num1 &= (num1 - 1) - } - for ; cnt1 < cnt2; cnt1++ { - num1 |= (num1 + 1) - } - return num1 -} -``` - -#### TypeScript - -```ts -function minimizeXor(num1: number, num2: number): number { - let cnt1 = bitCount(num1); - let cnt2 = bitCount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= num1 - 1; - } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= num1 + 1; + return x; } - return num1; -} - -function bitCount(i: number): number { - i = i - ((i >>> 1) & 0x55555555); - i = (i & 0x33333333) + ((i >>> 2) & 0x33333333); - i = (i + (i >>> 4)) & 0x0f0f0f0f; - i = i + (i >>> 8); - i = i + (i >>> 16); - return i & 0x3f; } ``` diff --git a/solution/2400-2499/2429.Minimize XOR/README_EN.md b/solution/2400-2499/2429.Minimize XOR/README_EN.md index 61c60717d64f9..6d909b9596539 100644 --- a/solution/2400-2499/2429.Minimize XOR/README_EN.md +++ b/solution/2400-2499/2429.Minimize XOR/README_EN.md @@ -68,9 +68,9 @@ The integer 3 has the same number of set bits as num2, and the ### Solution 1: Greedy + Bit Manipulation -According to the problem description, we first calculate the number of set bits $cnt$ in $num2$, then enumerate each bit of $num1$ from high to low. If the bit is $1$, we set the corresponding bit in $x$ to $1$ and decrement $cnt$ by $1$, until $cnt$ is $0$. If $cnt$ is still not $0$ at this point, we start from the low bit and set each bit of $num1$ that is $0$ to $1$, and decrement $cnt$ by $1$, until $cnt$ is $0$. +According to the problem description, we first calculate the number of set bits in $\textit{num2}$, denoted as $\textit{cnt}$. Then, we iterate from the highest to the lowest bit of $\textit{num1}$; if the current bit is $1$, we set the corresponding bit in $x$ to $1$ and decrement $\textit{cnt}$, until $\textit{cnt}$ becomes $0$. If $\textit{cnt}$ is still not $0$, we iterate from the lowest bit upwards, setting positions where $\textit{num1}$ has $0$ to $1$ in $x$, and decrement $\textit{cnt}$ until it reaches $0$. -The time complexity is $O(\log n)$, and the space complexity is $O(1)$. Here, $n$ is the maximum value of $num1$ and $num2$. +The time complexity is $O(\log n)$, where $n$ is the maximum value of $\textit{num1}$ and $\textit{num2}$. The space complexity is $O(1)$. @@ -189,107 +189,59 @@ function minimizeXor(num1: number, num2: number): number { } ``` - - - - - - -### Solution 2 - - +#### Rust -#### Python3 - -```python -class Solution: - def minimizeXor(self, num1: int, num2: int) -> int: - cnt1 = num1.bit_count() - cnt2 = num2.bit_count() - while cnt1 > cnt2: - num1 &= num1 - 1 - cnt1 -= 1 - while cnt1 < cnt2: - num1 |= num1 + 1 - cnt1 += 1 - return num1 -``` - -#### Java - -```java -class Solution { - public int minimizeXor(int num1, int num2) { - int cnt1 = Integer.bitCount(num1); - int cnt2 = Integer.bitCount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= (num1 - 1); +```rust +impl Solution { + pub fn minimize_xor(num1: i32, mut num2: i32) -> i32 { + let mut cnt = 0; + while num2 > 0 { + num2 -= num2 & -num2; + cnt += 1; } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= (num1 + 1); + let mut x = 0; + let mut c = cnt; + for i in (0..=30).rev() { + if c > 0 && (num1 >> i) & 1 == 1 { + x |= 1 << i; + c -= 1; + } } - return num1; + for i in 0..=30 { + if c == 0 { + break; + } + if ((num1 >> i) & 1) == 0 { + x |= 1 << i; + c -= 1; + } + } + x } } ``` -#### C++ +#### C# -```cpp -class Solution { -public: - int minimizeXor(int num1, int num2) { - int cnt1 = __builtin_popcount(num1); - int cnt2 = __builtin_popcount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= (num1 - 1); +```cs +public class Solution { + public int MinimizeXor(int num1, int num2) { + int cnt = BitOperations.PopCount((uint)num2); + int x = 0; + for (int i = 30; i >= 0 && cnt > 0; --i) { + if (((num1 >> i) & 1) == 1) { + x |= 1 << i; + --cnt; + } } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= (num1 + 1); + for (int i = 0; cnt > 0; ++i) { + if (((num1 >> i) & 1) == 0) { + x |= 1 << i; + --cnt; + } } - return num1; - } -}; -``` - -#### Go - -```go -func minimizeXor(num1 int, num2 int) int { - cnt1 := bits.OnesCount(uint(num1)) - cnt2 := bits.OnesCount(uint(num2)) - for ; cnt1 > cnt2; cnt1-- { - num1 &= (num1 - 1) - } - for ; cnt1 < cnt2; cnt1++ { - num1 |= (num1 + 1) - } - return num1 -} -``` - -#### TypeScript - -```ts -function minimizeXor(num1: number, num2: number): number { - let cnt1 = bitCount(num1); - let cnt2 = bitCount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= num1 - 1; - } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= num1 + 1; + return x; } - return num1; -} - -function bitCount(i: number): number { - i = i - ((i >>> 1) & 0x55555555); - i = (i & 0x33333333) + ((i >>> 2) & 0x33333333); - i = (i + (i >>> 4)) & 0x0f0f0f0f; - i = i + (i >>> 8); - i = i + (i >>> 16); - return i & 0x3f; } ``` diff --git a/solution/2400-2499/2429.Minimize XOR/Solution.cs b/solution/2400-2499/2429.Minimize XOR/Solution.cs new file mode 100644 index 0000000000000..ba4914569b0d6 --- /dev/null +++ b/solution/2400-2499/2429.Minimize XOR/Solution.cs @@ -0,0 +1,19 @@ +public class Solution { + public int MinimizeXor(int num1, int num2) { + int cnt = BitOperations.PopCount((uint)num2); + int x = 0; + for (int i = 30; i >= 0 && cnt > 0; --i) { + if (((num1 >> i) & 1) == 1) { + x |= 1 << i; + --cnt; + } + } + for (int i = 0; cnt > 0; ++i) { + if (((num1 >> i) & 1) == 0) { + x |= 1 << i; + --cnt; + } + } + return x; + } +} diff --git a/solution/2400-2499/2429.Minimize XOR/Solution.rs b/solution/2400-2499/2429.Minimize XOR/Solution.rs new file mode 100644 index 0000000000000..14770aa125877 --- /dev/null +++ b/solution/2400-2499/2429.Minimize XOR/Solution.rs @@ -0,0 +1,27 @@ +impl Solution { + pub fn minimize_xor(num1: i32, mut num2: i32) -> i32 { + let mut cnt = 0; + while num2 > 0 { + num2 -= num2 & -num2; + cnt += 1; + } + let mut x = 0; + let mut c = cnt; + for i in (0..=30).rev() { + if c > 0 && (num1 >> i) & 1 == 1 { + x |= 1 << i; + c -= 1; + } + } + for i in 0..=30 { + if c == 0 { + break; + } + if ((num1 >> i) & 1) == 0 { + x |= 1 << i; + c -= 1; + } + } + x + } +} diff --git a/solution/2400-2499/2429.Minimize XOR/Solution2.cpp b/solution/2400-2499/2429.Minimize XOR/Solution2.cpp deleted file mode 100644 index d0e507aa2b528..0000000000000 --- a/solution/2400-2499/2429.Minimize XOR/Solution2.cpp +++ /dev/null @@ -1,14 +0,0 @@ -class Solution { -public: - int minimizeXor(int num1, int num2) { - int cnt1 = __builtin_popcount(num1); - int cnt2 = __builtin_popcount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= (num1 - 1); - } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= (num1 + 1); - } - return num1; - } -}; \ No newline at end of file diff --git a/solution/2400-2499/2429.Minimize XOR/Solution2.go b/solution/2400-2499/2429.Minimize XOR/Solution2.go deleted file mode 100644 index 94eb188121ea3..0000000000000 --- a/solution/2400-2499/2429.Minimize XOR/Solution2.go +++ /dev/null @@ -1,11 +0,0 @@ -func minimizeXor(num1 int, num2 int) int { - cnt1 := bits.OnesCount(uint(num1)) - cnt2 := bits.OnesCount(uint(num2)) - for ; cnt1 > cnt2; cnt1-- { - num1 &= (num1 - 1) - } - for ; cnt1 < cnt2; cnt1++ { - num1 |= (num1 + 1) - } - return num1 -} \ No newline at end of file diff --git a/solution/2400-2499/2429.Minimize XOR/Solution2.java b/solution/2400-2499/2429.Minimize XOR/Solution2.java deleted file mode 100644 index 7e5636a6eaf87..0000000000000 --- a/solution/2400-2499/2429.Minimize XOR/Solution2.java +++ /dev/null @@ -1,13 +0,0 @@ -class Solution { - public int minimizeXor(int num1, int num2) { - int cnt1 = Integer.bitCount(num1); - int cnt2 = Integer.bitCount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= (num1 - 1); - } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= (num1 + 1); - } - return num1; - } -} \ No newline at end of file diff --git a/solution/2400-2499/2429.Minimize XOR/Solution2.py b/solution/2400-2499/2429.Minimize XOR/Solution2.py deleted file mode 100644 index bd07409846075..0000000000000 --- a/solution/2400-2499/2429.Minimize XOR/Solution2.py +++ /dev/null @@ -1,11 +0,0 @@ -class Solution: - def minimizeXor(self, num1: int, num2: int) -> int: - cnt1 = num1.bit_count() - cnt2 = num2.bit_count() - while cnt1 > cnt2: - num1 &= num1 - 1 - cnt1 -= 1 - while cnt1 < cnt2: - num1 |= num1 + 1 - cnt1 += 1 - return num1 diff --git a/solution/2400-2499/2429.Minimize XOR/Solution2.ts b/solution/2400-2499/2429.Minimize XOR/Solution2.ts deleted file mode 100644 index 64d948aa6871b..0000000000000 --- a/solution/2400-2499/2429.Minimize XOR/Solution2.ts +++ /dev/null @@ -1,20 +0,0 @@ -function minimizeXor(num1: number, num2: number): number { - let cnt1 = bitCount(num1); - let cnt2 = bitCount(num2); - for (; cnt1 > cnt2; --cnt1) { - num1 &= num1 - 1; - } - for (; cnt1 < cnt2; ++cnt1) { - num1 |= num1 + 1; - } - return num1; -} - -function bitCount(i: number): number { - i = i - ((i >>> 1) & 0x55555555); - i = (i & 0x33333333) + ((i >>> 2) & 0x33333333); - i = (i + (i >>> 4)) & 0x0f0f0f0f; - i = i + (i >>> 8); - i = i + (i >>> 16); - return i & 0x3f; -}