feat: add solutions to lc problem: No.3287 (#3965)

No.3287.Find the Maximum Sequence Value of Array

Author: yanglbme

solution/3200-3299/3287.Find the Maximum Sequence Value of Array/README.md

@@ -72,32 +72,295 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：动态规划 + 前后缀分解 + 枚举
+
+我们考虑将序列分成两部分，前 $k$ 个元素和后 $k$ 个元素，分别计算前后缀的所有可能的异或值。
+
+定义 $f[i][j][x]$ 表示前 $i$ 个元素中取 $j$ 个元素，是否存在一个子集的异或值为 $x$，定义 $g[i][j][y]$ 表示从下标 $i$ 开始取 $j$ 个元素，是否存在一个子集的异或值为 $y$。
+
+考虑 $f[i][j][x]$ 的转移方程，对于第 $i$ 个元素（从 $0$ 开始），我们可以选择不取，也可以选择取，因此有：
+
+$$
+f[i + 1][j][x] = f[i + 1][j][x] \lor f[i][j][x] \\
+f[i + 1][j + 1][x \lor \text{nums}[i]] = f[i + 1][j + 1][x \lor \text{nums}[i]] \lor f[i][j][x]
+$$
+
+对于 $g[i][j][y]$ 的转移方程，同样对于第 $i$ 个元素（从 $n - 1$ 开始），我们可以选择不取，也可以选择取，因此有：
+
+$$
+g[i - 1][j][y] = g[i - 1][j][y] \lor g[i][j][y] \\
+g[i - 1][j + 1][y \lor \text{nums}[i - 1]] = g[i - 1][j + 1][y \lor \text{nums}[i - 1]] \lor g[i][j][y]
+$$
+
+最后，我们在 $[k, n - k]$ 的范围内枚举 $i$，对于每一个 $i$，我们枚举 $x$ 和 $y$，其中 $0 \leq x, y < 2^7$，如果 $f[i][k][x]$ 和 $g[i][k][y]$ 均为真，那么我们更新答案 $\text{ans} = \max(\text{ans}, x \oplus y)$。
+
+时间复杂度 $O(n \times m \times k)$，空间复杂度 $O(n \times m \times k)$，其中 $n$ 为数组长度，而 $m = 2^7$。
 
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 #### Python3
 
 ```python
-
+class Solution:
+    def maxValue(self, nums: List[int], k: int) -> int:
+        m = 1 << 7
+        n = len(nums)
+        f = [[[False] * m for _ in range(k + 2)] for _ in range(n + 1)]
+        f[0][0][0] = True
+        for i in range(n):
+            for j in range(k + 1):
+                for x in range(m):
+                    f[i + 1][j][x] |= f[i][j][x]
+                    f[i + 1][j + 1][x | nums[i]] |= f[i][j][x]
+
+        g = [[[False] * m for _ in range(k + 2)] for _ in range(n + 1)]
+        g[n][0][0] = True
+        for i in range(n, 0, -1):
+            for j in range(k + 1):
+                for y in range(m):
+                    g[i - 1][j][y] |= g[i][j][y]
+                    g[i - 1][j + 1][y | nums[i - 1]] |= g[i][j][y]
+
+        ans = 0
+        for i in range(k, n - k + 1):
+            for x in range(m):
+                if f[i][k][x]:
+                    for y in range(m):
+                        if g[i][k][y]:
+                            ans = max(ans, x ^ y)
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public int maxValue(int[] nums, int k) {
+        int m = 1 << 7;
+        int n = nums.length;
+        boolean[][][] f = new boolean[n + 1][k + 2][m];
+        f[0][0][0] = true;
+
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j <= k; j++) {
+                for (int x = 0; x < m; x++) {
+                    if (f[i][j][x]) {
+                        f[i + 1][j][x] = true;
+                        f[i + 1][j + 1][x | nums[i]] = true;
+                    }
+                }
+            }
+        }
+
+        boolean[][][] g = new boolean[n + 1][k + 2][m];
+        g[n][0][0] = true;
+
+        for (int i = n; i > 0; i--) {
+            for (int j = 0; j <= k; j++) {
+                for (int y = 0; y < m; y++) {
+                    if (g[i][j][y]) {
+                        g[i - 1][j][y] = true;
+                        g[i - 1][j + 1][y | nums[i - 1]] = true;
+                    }
+                }
+            }
+        }
+
+        int ans = 0;
+
+        for (int i = k; i <= n - k; i++) {
+            for (int x = 0; x < m; x++) {
+                if (f[i][k][x]) {
+                    for (int y = 0; y < m; y++) {
+                        if (g[i][k][y]) {
+                            ans = Math.max(ans, x ^ y);
+                        }
+                    }
+                }
+            }
+        }
+
+        return ans;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    int maxValue(vector<int>& nums, int k) {
+        int m = 1 << 7;
+        int n = nums.size();
+
+        vector<vector<vector<bool>>> f(n + 1, vector<vector<bool>>(k + 2, vector<bool>(m, false)));
+        f[0][0][0] = true;
+
+        for (int i = 0; i < n; i++) {
+            for (int j = 0; j <= k; j++) {
+                for (int x = 0; x < m; x++) {
+                    if (f[i][j][x]) {
+                        f[i + 1][j][x] = true;
+                        f[i + 1][j + 1][x | nums[i]] = true;
+                    }
+                }
+            }
+        }
+
+        vector<vector<vector<bool>>> g(n + 1, vector<vector<bool>>(k + 2, vector<bool>(m, false)));
+        g[n][0][0] = true;
+
+        for (int i = n; i > 0; i--) {
+            for (int j = 0; j <= k; j++) {
+                for (int y = 0; y < m; y++) {
+                    if (g[i][j][y]) {
+                        g[i - 1][j][y] = true;
+                        g[i - 1][j + 1][y | nums[i - 1]] = true;
+                    }
+                }
+            }
+        }
+
+        int ans = 0;
+
+        for (int i = k; i <= n - k; i++) {
+            for (int x = 0; x < m; x++) {
+                if (f[i][k][x]) {
+                    for (int y = 0; y < m; y++) {
+                        if (g[i][k][y]) {
+                            ans = max(ans, x ^ y);
+                        }
+                    }
+                }
+            }
+        }
+
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func maxValue(nums []int, k int) int {
+	m := 1 << 7
+	n := len(nums)
+
+	f := make([][][]bool, n+1)
+	for i := range f {
+		f[i] = make([][]bool, k+2)
+		for j := range f[i] {
+			f[i][j] = make([]bool, m)
+		}
+	}
+	f[0][0][0] = true
+
+	for i := 0; i < n; i++ {
+		for j := 0; j <= k; j++ {
+			for x := 0; x < m; x++ {
+				if f[i][j][x] {
+					f[i+1][j][x] = true
+					f[i+1][j+1][x|nums[i]] = true
+				}
+			}
+		}
+	}
+
+	g := make([][][]bool, n+1)
+	for i := range g {
+		g[i] = make([][]bool, k+2)
+		for j := range g[i] {
+			g[i][j] = make([]bool, m)
+		}
+	}
+	g[n][0][0] = true
+
+	for i := n; i > 0; i-- {
+		for j := 0; j <= k; j++ {
+			for y := 0; y < m; y++ {
+				if g[i][j][y] {
+					g[i-1][j][y] = true
+					g[i-1][j+1][y|nums[i-1]] = true
+				}
+			}
+		}
+	}
+
+	ans := 0
+
+	for i := k; i <= n-k; i++ {
+		for x := 0; x < m; x++ {
+			if f[i][k][x] {
+				for y := 0; y < m; y++ {
+					if g[i][k][y] {
+						ans = max(ans, x^y)
+					}
+				}
+			}
+		}
+	}
+
+	return ans
+}
+```
 
+#### TypeScript
+
+```ts
+function maxValue(nums: number[], k: number): number {
+    const m = 1 << 7;
+    const n = nums.length;
+
+    const f: boolean[][][] = Array.from({ length: n + 1 }, () =>
+        Array.from({ length: k + 2 }, () => Array(m).fill(false)),
+    );
+    f[0][0][0] = true;
+
+    for (let i = 0; i < n; i++) {
+        for (let j = 0; j <= k; j++) {
+            for (let x = 0; x < m; x++) {
+                if (f[i][j][x]) {
+                    f[i + 1][j][x] = true;
+                    f[i + 1][j + 1][x | nums[i]] = true;
+                }
+            }
+        }
+    }
+
+    const g: boolean[][][] = Array.from({ length: n + 1 }, () =>
+        Array.from({ length: k + 2 }, () => Array(m).fill(false)),
+    );
+    g[n][0][0] = true;
+
+    for (let i = n; i > 0; i--) {
+        for (let j = 0; j <= k; j++) {
+            for (let y = 0; y < m; y++) {
+                if (g[i][j][y]) {
+                    g[i - 1][j][y] = true;
+                    g[i - 1][j + 1][y | nums[i - 1]] = true;
+                }
+            }
+        }
+    }
+
+    let ans = 0;
+
+    for (let i = k; i <= n - k; i++) {
+        for (let x = 0; x < m; x++) {
+            if (f[i][k][x]) {
+                for (let y = 0; y < m; y++) {
+                    if (g[i][k][y]) {
+                        ans = Math.max(ans, x ^ y);
+                    }
+                }
+            }
+        }
+    }
+
+    return ans;
+}
 ```
 
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