feat: add solutions to lc problem: No.2939 (#1986)

No.2939.Maximum Xor Product

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

@@ -53,34 +53,146 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：贪心 + 位运算**
+
+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。
+
+因此，我们首先提取 $a$ 和 $b$ 高于 $n$ 位的部分，分别记为 $ax$ 和 $bx$。
+
+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 $y$。
+
+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 $1$。
+
+最后，我们返回 $ax \times bx \bmod (10^9 + 7)$ 即为答案。
+
+时间复杂度 $O(n)$，其中 $n$ 为题目给定的整数。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def maximumXorProduct(self, a: int, b: int, n: int) -> int:
+        mod = 10**9 + 7
+        ax, bx = (a >> n) << n, (b >> n) << n
+        for i in range(n - 1, -1, -1):
+            x = a >> i & 1
+            y = b >> i & 1
+            if x == y:
+                ax |= 1 << i
+                bx |= 1 << i
+            elif ax > bx:
+                bx |= 1 << i
+            else:
+                ax |= 1 << i
+        return ax * bx % mod
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public int maximumXorProduct(long a, long b, int n) {
+        final int mod = (int) 1e9 + 7;
+        long ax = (a >> n) << n;
+        long bx = (b >> n) << n;
+        for (int i = n - 1; i >= 0; --i) {
+            long x = a >> i & 1;
+            long y = b >> i & 1;
+            if (x == y) {
+                ax |= 1L << i;
+                bx |= 1L << i;
+            } else if (ax < bx) {
+                ax |= 1L << i;
+            } else {
+                bx |= 1L << i;
+            }
+        }
+        ax %= mod;
+        bx %= mod;
+        return (int) (ax * bx % mod);
+    }
+}
 ```
 
 ### **C++**
 
 ```cpp
-
+class Solution {
+public:
+    int maximumXorProduct(long long a, long long b, int n) {
+        const int mod = 1e9 + 7;
+        long long ax = (a >> n) << n, bx = (b >> n) << n;
+        for (int i = n - 1; ~i; --i) {
+            int x = a >> i & 1, y = b >> i & 1;
+            if (x == y) {
+                ax |= 1LL << i;
+                bx |= 1LL << i;
+            } else if (ax < bx) {
+                ax |= 1LL << i;
+            } else {
+                bx |= 1LL << i;
+            }
+        }
+        ax %= mod;
+        bx %= mod;
+        return ax * bx % mod;
+    }
+};
 ```
 
 ### **Go**
 
 ```go
+func maximumXorProduct(a int64, b int64, n int) int {
+	const mod int64 = 1e9 + 7
+	ax := (a >> n) << n
+	bx := (b >> n) << n
+	for i := n - 1; i >= 0; i-- {
+		x, y := (a>>i)&1, (b>>i)&1
+		if x == y {
+			ax |= 1 << i
+			bx |= 1 << i
+		} else if ax < bx {
+			ax |= 1 << i
+		} else {
+			bx |= 1 << i
+		}
+	}
+	ax %= mod
+	bx %= mod
+	return int(ax * bx % mod)
+}
+```
 
+### **TypeScript**
+
+```ts
+function maximumXorProduct(a: number, b: number, n: number): number {
+    const mod = BigInt(1e9 + 7);
+    let ax = (BigInt(a) >> BigInt(n)) << BigInt(n);
+    let bx = (BigInt(b) >> BigInt(n)) << BigInt(n);
+    for (let i = BigInt(n - 1); ~i; --i) {
+        const x = (BigInt(a) >> i) & 1n;
+        const y = (BigInt(b) >> i) & 1n;
+        if (x === y) {
+            ax |= 1n << i;
+            bx |= 1n << i;
+        } else if (ax < bx) {
+            ax |= 1n << i;
+        } else {
+            bx |= 1n << i;
+        }
+    }
+    ax %= mod;
+    bx %= mod;
+    return Number((ax * bx) % mod);
+}
 ```
 
 ### **...**

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

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

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

@@ -0,0 +1,21 @@
+class Solution {
+public:
+    int maximumXorProduct(long long a, long long b, int n) {
+        const int mod = 1e9 + 7;
+        long long ax = (a >> n) << n, bx = (b >> n) << n;
+        for (int i = n - 1; ~i; --i) {
+            int x = a >> i & 1, y = b >> i & 1;
+            if (x == y) {
+                ax |= 1LL << i;
+                bx |= 1LL << i;
+            } else if (ax < bx) {
+                ax |= 1LL << i;
+            } else {
+                bx |= 1LL << i;
+            }
+        }
+        ax %= mod;
+        bx %= mod;
+        return ax * bx % mod;
+    }
+};

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

@@ -0,0 +1,19 @@
+func maximumXorProduct(a int64, b int64, n int) int {
+	const mod int64 = 1e9 + 7
+	ax := (a >> n) << n
+	bx := (b >> n) << n
+	for i := n - 1; i >= 0; i-- {
+		x, y := (a>>i)&1, (b>>i)&1
+		if x == y {
+			ax |= 1 << i
+			bx |= 1 << i
+		} else if ax < bx {
+			ax |= 1 << i
+		} else {
+			bx |= 1 << i
+		}
+	}
+	ax %= mod
+	bx %= mod
+	return int(ax * bx % mod)
+}

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

@@ -0,0 +1,22 @@
+class Solution {
+    public int maximumXorProduct(long a, long b, int n) {
+        final int mod = (int) 1e9 + 7;
+        long ax = (a >> n) << n;
+        long bx = (b >> n) << n;
+        for (int i = n - 1; i >= 0; --i) {
+            long x = a >> i & 1;
+            long y = b >> i & 1;
+            if (x == y) {
+                ax |= 1L << i;
+                bx |= 1L << i;
+            } else if (ax < bx) {
+                ax |= 1L << i;
+            } else {
+                bx |= 1L << i;
+            }
+        }
+        ax %= mod;
+        bx %= mod;
+        return (int) (ax * bx % mod);
+    }
+}

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

@@ -0,0 +1,15 @@
+class Solution:
+    def maximumXorProduct(self, a: int, b: int, n: int) -> int:
+        mod = 10**9 + 7
+        ax, bx = (a >> n) << n, (b >> n) << n
+        for i in range(n - 1, -1, -1):
+            x = a >> i & 1
+            y = b >> i & 1
+            if x == y:
+                ax |= 1 << i
+                bx |= 1 << i
+            elif ax > bx:
+                bx |= 1 << i
+            else:
+                ax |= 1 << i
+        return ax * bx % mod