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feat: add solutions to lc problem: No.2939 (doocs#1986)
No.2939.Maximum Xor Product
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‎solution/2900-2999/2939.Maximum Xor Product/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：贪心 + 位运算**
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根据题目描述，我们可以给 $a$ 和 $b$ 在二进制下 $[0..n)$ 位上同时分配一个数字，最终使得 $a$ 和 $b$ 的乘积最大。
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因此，我们首先提取 $a$ 和 $b$ 高于 $n$ 位的部分，分别记为 $ax$ 和 $bx$。
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接下来，从大到小考虑 $[0..n)$ 位上的每一位，我们将 $a$ 和 $b$ 的当前位分别记为 $x$ 和 $y$。
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如果 $x = y$，那么我们可以将 $ax$ 和 $bx$ 的当前位同时置为 $1$，因此，我们更新 $ax = ax \mid 1 << i$ 和 $bx = bx \mid 1 << i$。否则，如果 $ax \lt bx$，要使得最终的乘积最大，我们应该让 $ax$ 的当前位置为 $1$，否则我们可以将 $bx$ 的当前位置为 $1$。
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最后，我们返回 $ax \times bx \bmod (10^9 + 7)$ 即为答案。
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时间复杂度 $O(n)$，其中 $n$ 为题目给定的整数。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def maximumXorProduct(self, a: int, b: int, n: int) -> int:
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mod = 10**9 + 7
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ax, bx = (a >> n) << n, (b >> n) << n
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for i in range(n - 1, -1, -1):
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x = a >> i & 1
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y = b >> i & 1
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if x == y:
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ax |= 1 << i
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bx |= 1 << i
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elif ax > bx:
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bx |= 1 << i
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else:
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ax |= 1 << i
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return ax * bx % mod
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int maximumXorProduct(long a, long b, int n) {
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final int mod = (int) 1e9 + 7;
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long ax = (a >> n) << n;
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long bx = (b >> n) << n;
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for (int i = n - 1; i >= 0; --i) {
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long x = a >> i & 1;
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long y = b >> i & 1;
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if (x == y) {
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ax |= 1L << i;
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bx |= 1L << i;
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} else if (ax < bx) {
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ax |= 1L << i;
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} else {
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bx |= 1L << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return (int) (ax * bx % mod);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maximumXorProduct(long long a, long long b, int n) {
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const int mod = 1e9 + 7;
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long long ax = (a >> n) << n, bx = (b >> n) << n;
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for (int i = n - 1; ~i; --i) {
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int x = a >> i & 1, y = b >> i & 1;
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if (x == y) {
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ax |= 1LL << i;
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bx |= 1LL << i;
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} else if (ax < bx) {
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ax |= 1LL << i;
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} else {
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bx |= 1LL << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return ax * bx % mod;
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}
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};
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```
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### **Go**
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```go
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func maximumXorProduct(a int64, b int64, n int) int {
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const mod int64 = 1e9 + 7
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ax := (a >> n) << n
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bx := (b >> n) << n
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for i := n - 1; i >= 0; i-- {
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x, y := (a>>i)&1, (b>>i)&1
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if x == y {
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ax |= 1 << i
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bx |= 1 << i
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} else if ax < bx {
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ax |= 1 << i
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} else {
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bx |= 1 << i
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}
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}
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ax %= mod
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bx %= mod
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return int(ax * bx % mod)
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}
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```
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### **TypeScript**
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```ts
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function maximumXorProduct(a: number, b: number, n: number): number {
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const mod = BigInt(1e9 + 7);
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let ax = (BigInt(a) >> BigInt(n)) << BigInt(n);
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let bx = (BigInt(b) >> BigInt(n)) << BigInt(n);
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for (let i = BigInt(n - 1); ~i; --i) {
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const x = (BigInt(a) >> i) & 1n;
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const y = (BigInt(b) >> i) & 1n;
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if (x === y) {
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ax |= 1n << i;
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bx |= 1n << i;
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} else if (ax < bx) {
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ax |= 1n << i;
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} else {
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bx |= 1n << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return Number((ax * bx) % mod);
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}
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```
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### **...**

‎solution/2900-2999/2939.Maximum Xor Product/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -47,30 +47,142 @@ It can be shown that 12 is the maximum value of (a XOR x) * (b XOR x) for all 0
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## Solutions
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**Solution 1: Greedy + Bitwise Operation**
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According to the problem description, we can assign a number to the $[0..n)$ bits of $a$ and $b$ in binary at the same time, so that the product of $a$ and $b$ is maximized.
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Therefore, we first extract the parts of $a$ and $b$ that are higher than the $n$ bits, denoted as $ax$ and $bx$.
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Next, we consider each bit in $[0..n)$ from high to low. We denote the current bits of $a$ and $b$ as $x$ and $y$.
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If $x = y$, then we can set the current bit of $ax$ and $bx$ to $1$ at the same time. Therefore, we update $ax = ax \mid 1 << i$ and $bx = bx \mid 1 << i$. Otherwise, if $ax < bx$, to maximize the final product, we should set the current bit of $ax$ to $1$. Otherwise, we can set the current bit of $bx$ to $1$.
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Finally, we return $ax \times bx \bmod (10^9 + 7)$ as the answer.
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The time complexity is $O(n)$, where $n$ is the integer given in the problem. The space complexity is $O(1)$.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def maximumXorProduct(self, a: int, b: int, n: int) -> int:
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mod = 10**9 + 7
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ax, bx = (a >> n) << n, (b >> n) << n
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for i in range(n - 1, -1, -1):
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x = a >> i & 1
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y = b >> i & 1
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if x == y:
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ax |= 1 << i
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bx |= 1 << i
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elif ax > bx:
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bx |= 1 << i
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else:
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ax |= 1 << i
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return ax * bx % mod
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```
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### **Java**
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```java
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class Solution {
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public int maximumXorProduct(long a, long b, int n) {
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final int mod = (int) 1e9 + 7;
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long ax = (a >> n) << n;
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long bx = (b >> n) << n;
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for (int i = n - 1; i >= 0; --i) {
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long x = a >> i & 1;
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long y = b >> i & 1;
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if (x == y) {
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ax |= 1L << i;
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bx |= 1L << i;
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} else if (ax < bx) {
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ax |= 1L << i;
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} else {
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bx |= 1L << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return (int) (ax * bx % mod);
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int maximumXorProduct(long long a, long long b, int n) {
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const int mod = 1e9 + 7;
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long long ax = (a >> n) << n, bx = (b >> n) << n;
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for (int i = n - 1; ~i; --i) {
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int x = a >> i & 1, y = b >> i & 1;
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if (x == y) {
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ax |= 1LL << i;
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bx |= 1LL << i;
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} else if (ax < bx) {
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ax |= 1LL << i;
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} else {
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bx |= 1LL << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return ax * bx % mod;
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}
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};
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```
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### **Go**
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```go
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func maximumXorProduct(a int64, b int64, n int) int {
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const mod int64 = 1e9 + 7
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ax := (a >> n) << n
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bx := (b >> n) << n
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for i := n - 1; i >= 0; i-- {
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x, y := (a>>i)&1, (b>>i)&1
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if x == y {
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ax |= 1 << i
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bx |= 1 << i
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} else if ax < bx {
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ax |= 1 << i
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} else {
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bx |= 1 << i
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}
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}
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ax %= mod
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bx %= mod
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return int(ax * bx % mod)
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}
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```
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### **TypeScript**
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```ts
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function maximumXorProduct(a: number, b: number, n: number): number {
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const mod = BigInt(1e9 + 7);
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let ax = (BigInt(a) >> BigInt(n)) << BigInt(n);
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let bx = (BigInt(b) >> BigInt(n)) << BigInt(n);
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for (let i = BigInt(n - 1); ~i; --i) {
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const x = (BigInt(a) >> i) & 1n;
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const y = (BigInt(b) >> i) & 1n;
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if (x === y) {
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ax |= 1n << i;
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bx |= 1n << i;
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} else if (ax < bx) {
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ax |= 1n << i;
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} else {
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bx |= 1n << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return Number((ax * bx) % mod);
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}
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```
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### **...**

‎solution/2900-2999/2939.Maximum Xor Product/Solution.cpp

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class Solution {
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public:
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int maximumXorProduct(long long a, long long b, int n) {
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const int mod = 1e9 + 7;
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long long ax = (a >> n) << n, bx = (b >> n) << n;
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for (int i = n - 1; ~i; --i) {
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int x = a >> i & 1, y = b >> i & 1;
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if (x == y) {
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ax |= 1LL << i;
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bx |= 1LL << i;
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} else if (ax < bx) {
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ax |= 1LL << i;
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} else {
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bx |= 1LL << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return ax * bx % mod;
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}
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};

‎solution/2900-2999/2939.Maximum Xor Product/Solution.go

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func maximumXorProduct(a int64, b int64, n int) int {
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const mod int64 = 1e9 + 7
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ax := (a >> n) << n
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bx := (b >> n) << n
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for i := n - 1; i >= 0; i-- {
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x, y := (a>>i)&1, (b>>i)&1
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if x == y {
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ax |= 1 << i
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bx |= 1 << i
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} else if ax < bx {
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ax |= 1 << i
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} else {
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bx |= 1 << i
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}
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}
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ax %= mod
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bx %= mod
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return int(ax * bx % mod)
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}

‎solution/2900-2999/2939.Maximum Xor Product/Solution.java

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class Solution {
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public int maximumXorProduct(long a, long b, int n) {
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final int mod = (int) 1e9 + 7;
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long ax = (a >> n) << n;
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long bx = (b >> n) << n;
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for (int i = n - 1; i >= 0; --i) {
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long x = a >> i & 1;
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long y = b >> i & 1;
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if (x == y) {
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ax |= 1L << i;
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bx |= 1L << i;
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} else if (ax < bx) {
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ax |= 1L << i;
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} else {
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bx |= 1L << i;
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}
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}
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ax %= mod;
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bx %= mod;
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return (int) (ax * bx % mod);
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}
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}

‎solution/2900-2999/2939.Maximum Xor Product/Solution.py

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class Solution:
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def maximumXorProduct(self, a: int, b: int, n: int) -> int:
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mod = 10**9 + 7
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ax, bx = (a >> n) << n, (b >> n) << n
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for i in range(n - 1, -1, -1):
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x = a >> i & 1
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y = b >> i & 1
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if x == y:
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ax |= 1 << i
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bx |= 1 << i
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elif ax > bx:
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bx |= 1 << i
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else:
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ax |= 1 << i
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return ax * bx % mod

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