feat: add solutions to lc problem: No.2272

No.2272.Substring With Largest Variance

Author: yanglbme

solution/0900-0999/0934.Shortest Bridge/README.md

@@ -78,21 +78,21 @@ class Solution:
         def dfs(i, j):
             q.append((i, j))
             grid[i][j] = 2
-            for a, b in dirs:
+            for a, b in pairwise(dirs):
                 x, y = i + a, j + b
                 if 0 <= x < n and 0 <= y < n and grid[x][y] == 1:
                     dfs(x, y)
 
         n = len(grid)
-        dirs = ((0, 1), (0, -1), (1, 0), (-1, 0))
+        dirs = (-1, 0, 1, 0, -1)
         q = deque()
         i, j = next((i, j) for i in range(n) for j in range(n) if grid[i][j])
         dfs(i, j)
         ans = 0
         while 1:
             for _ in range(len(q)):
                 i, j = q.popleft()
-                for a, b in dirs:
+                for a, b in pairwise(dirs):
                     x, y = i + a, j + b
                     if 0 <= x < n and 0 <= y < n:
                         if grid[x][y] == 1:
@@ -163,15 +163,13 @@ class Solution {
 ### **C++**
 
 ```cpp
-using pii = pair<int, int>;
-
 class Solution {
 public:
     const static inline vector<int> dirs = {-1, 0, 1, 0, -1};
 
     int shortestBridge(vector<vector<int>>& grid) {
         int n = grid.size();
-        queue<pii> q;
+        queue<pair<int, int>> q;
         function<void(int, int)> dfs = [&](int i, int j) {
             grid[i][j] = 2;
             q.emplace(i, j);
@@ -193,11 +191,11 @@ public:
         }
         int ans = 0;
         while (1) {
-            for (int k = q.size(); k; --k) {
+            for (int h = q.size(); h; --h) {
                 auto [i, j] = q.front();
                 q.pop();
-                for (int h = 0; h < 4; ++h) {
-                    int x = i + dirs[h], y = j + dirs[h + 1];
+                for (int k = 0; k < 4; ++k) {
+                    int x = i + dirs[k], y = j + dirs[k + 1];
                     if (x >= 0 && x < n && y >= 0 && y < n) {
                         if (grid[x][y] == 1) return ans;
                         if (grid[x][y] == 0) {
@@ -216,7 +214,7 @@ public:
 ### **Go**
 
 ```go
-func shortestBridge(grid [][]int) int {
+func shortestBridge(grid [][]int) (ans int) {
 	n := len(grid)
 	dirs := []int{-1, 0, 1, 0, -1}
 	type pair struct{ i, j int }
@@ -241,7 +239,6 @@ func shortestBridge(grid [][]int) int {
 			}
 		}
 	}
-	ans := 0
 	for {
 		for i := len(q); i > 0; i-- {
 			p := q[0]
@@ -250,7 +247,7 @@ func shortestBridge(grid [][]int) int {
 				x, y := p.i+dirs[k], p.j+dirs[k+1]
 				if x >= 0 && x < n && y >= 0 && y < n {
 					if grid[x][y] == 1 {
-						return ans
+						return
 					}
 					if grid[x][y] == 0 {
 						grid[x][y] = 2

solution/0900-0999/0934.Shortest Bridge/README_EN.md

@@ -58,21 +58,21 @@ class Solution:
         def dfs(i, j):
             q.append((i, j))
             grid[i][j] = 2
-            for a, b in dirs:
+            for a, b in pairwise(dirs):
                 x, y = i + a, j + b
                 if 0 <= x < n and 0 <= y < n and grid[x][y] == 1:
                     dfs(x, y)
 
         n = len(grid)
-        dirs = ((0, 1), (0, -1), (1, 0), (-1, 0))
+        dirs = (-1, 0, 1, 0, -1)
         q = deque()
         i, j = next((i, j) for i in range(n) for j in range(n) if grid[i][j])
         dfs(i, j)
         ans = 0
         while 1:
             for _ in range(len(q)):
                 i, j = q.popleft()
-                for a, b in dirs:
+                for a, b in pairwise(dirs):
                     x, y = i + a, j + b
                     if 0 <= x < n and 0 <= y < n:
                         if grid[x][y] == 1:
@@ -141,15 +141,13 @@ class Solution {
 ### **C++**
 
 ```cpp
-using pii = pair<int, int>;
-
 class Solution {
 public:
     const static inline vector<int> dirs = {-1, 0, 1, 0, -1};
 
     int shortestBridge(vector<vector<int>>& grid) {
         int n = grid.size();
-        queue<pii> q;
+        queue<pair<int, int>> q;
         function<void(int, int)> dfs = [&](int i, int j) {
             grid[i][j] = 2;
             q.emplace(i, j);
@@ -171,11 +169,11 @@ public:
         }
         int ans = 0;
         while (1) {
-            for (int k = q.size(); k; --k) {
+            for (int h = q.size(); h; --h) {
                 auto [i, j] = q.front();
                 q.pop();
-                for (int h = 0; h < 4; ++h) {
-                    int x = i + dirs[h], y = j + dirs[h + 1];
+                for (int k = 0; k < 4; ++k) {
+                    int x = i + dirs[k], y = j + dirs[k + 1];
                     if (x >= 0 && x < n && y >= 0 && y < n) {
                         if (grid[x][y] == 1) return ans;
                         if (grid[x][y] == 0) {
@@ -194,7 +192,7 @@ public:
 ### **Go**
 
 ```go
-func shortestBridge(grid [][]int) int {
+func shortestBridge(grid [][]int) (ans int) {
 	n := len(grid)
 	dirs := []int{-1, 0, 1, 0, -1}
 	type pair struct{ i, j int }
@@ -219,7 +217,6 @@ func shortestBridge(grid [][]int) int {
 			}
 		}
 	}
-	ans := 0
 	for {
 		for i := len(q); i > 0; i-- {
 			p := q[0]
@@ -228,7 +225,7 @@ func shortestBridge(grid [][]int) int {
 				x, y := p.i+dirs[k], p.j+dirs[k+1]
 				if x >= 0 && x < n && y >= 0 && y < n {
 					if grid[x][y] == 1 {
-						return ans
+						return
 					}
 					if grid[x][y] == 0 {
 						grid[x][y] = 2

solution/0900-0999/0934.Shortest Bridge/Solution.cpp

@@ -1,12 +1,10 @@
-using pii = pair<int, int>;
-
 class Solution {
 public:
     const static inline vector<int> dirs = {-1, 0, 1, 0, -1};
 
     int shortestBridge(vector<vector<int>>& grid) {
         int n = grid.size();
-        queue<pii> q;
+        queue<pair<int, int>> q;
         function<void(int, int)> dfs = [&](int i, int j) {
             grid[i][j] = 2;
             q.emplace(i, j);
@@ -28,11 +26,11 @@ class Solution {
         }
         int ans = 0;
         while (1) {
-            for (int k = q.size(); k; --k) {
+            for (int h = q.size(); h; --h) {
                 auto [i, j] = q.front();
                 q.pop();
-                for (int h = 0; h < 4; ++h) {
-                    int x = i + dirs[h], y = j + dirs[h + 1];
+                for (int k = 0; k < 4; ++k) {
+                    int x = i + dirs[k], y = j + dirs[k + 1];
                     if (x >= 0 && x < n && y >= 0 && y < n) {
                         if (grid[x][y] == 1) return ans;
                         if (grid[x][y] == 0) {

solution/0900-0999/0934.Shortest Bridge/Solution.go

@@ -1,4 +1,4 @@
-func shortestBridge(grid [][]int) int {
+func shortestBridge(grid [][]int) (ans int) {
 	n := len(grid)
 	dirs := []int{-1, 0, 1, 0, -1}
 	type pair struct{ i, j int }
@@ -23,7 +23,6 @@ func shortestBridge(grid [][]int) int {
 			}
 		}
 	}
-	ans := 0
 	for {
 		for i := len(q); i > 0; i-- {
 			p := q[0]
@@ -32,7 +31,7 @@ func shortestBridge(grid [][]int) int {
 				x, y := p.i+dirs[k], p.j+dirs[k+1]
 				if x >= 0 && x < n && y >= 0 && y < n {
 					if grid[x][y] == 1 {
-						return ans
+						return
 					}
 					if grid[x][y] == 0 {
 						grid[x][y] = 2

solution/0900-0999/0934.Shortest Bridge/Solution.py

@@ -3,21 +3,21 @@ def shortestBridge(self, grid: List[List[int]]) -> int:
         def dfs(i, j):
             q.append((i, j))
             grid[i][j] = 2
-            for a, b in dirs:
+            for a, b in pairwise(dirs):
                 x, y = i + a, j + b
                 if 0 <= x < n and 0 <= y < n and grid[x][y] == 1:
                     dfs(x, y)
 
         n = len(grid)
-        dirs = ((0, 1), (0, -1), (1, 0), (-1, 0))
+        dirs = (-1, 0, 1, 0, -1)
         q = deque()
         i, j = next((i, j) for i in range(n) for j in range(n) if grid[i][j])
         dfs(i, j)
         ans = 0
         while 1:
             for _ in range(len(q)):
                 i, j = q.popleft()
-                for a, b in dirs:
+                for a, b in pairwise(dirs):
                     x, y = i + a, j + b
                     if 0 <= x < n and 0 <= y < n:
                         if grid[x][y] == 1:

solution/2200-2299/2272.Substring With Largest Variance/README.md

@@ -50,22 +50,140 @@ s 中没有字母出现超过 1 次，所以 s 中每个子字符串的波动值
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：枚举 + 动态规划**
+
+由于字符集只包含小写字母，我们可以考虑枚举出现次数最多的字符 $a$ 以及出现次数最少的字符 $b$。对于一个子串来说，这两种字符出现的次数之差就是子串的波动值。
+
+具体实现上，我们使用双重循环枚举 $a$ 和 $b$，用 $f[0]$ 记录以当前字符结尾的连续出现字符 $a$ 的个数，用 $f[1]$ 记录以当前字符结尾的，并且包含 $a$ 和 $b$ 的子串的波动值。迭代取 $f[1]$ 的最大值即可。
+
+递推公式如下：
+
+1. 如果当前字符为 $a$，则 $f[0]$ 和 $f[1]$ 都加 $1$；
+1. 如果当前字符为 $b$，则 $f[1]=\max(f[1]-1, f[0]-1)$，而 $f[0]=0$；
+1. 否则，无需考虑。
+
+注意，初始时将 $f[1]$ 赋值为一个负数最大值，可以保证更新答案时是合法的。
+
+时间复杂度 $O(n\times C^2)$，其中 $n$ 表示字符串 $s$ 的长度，而 $C$ 为字符集大小，本题中 $C=26$。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def largestVariance(self, s: str) -> int:
+        ans = 0
+        for a, b in permutations(ascii_lowercase, 2):
+            if a == b:
+                continue
+            f = [0, -inf]
+            for c in s:
+                if c == a:
+                    f[0], f[1] = f[0] + 1, f[1] + 1
+                elif c == b:
+                    f[1] = max(f[1] - 1, f[0] - 1)
+                    f[0] = 0
+                if ans < f[1]:
+                    ans = f[1]
+        return ans
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public int largestVariance(String s) {
+        int n = s.length();
+        int ans = 0;
+        for (char a = 'a'; a <= 'z'; ++a) {
+            for (char b = 'a'; b <= 'z'; ++b) {
+                if (a == b) {
+                    continue;
+                }
+                int[] f = new int[] {0, -n};
+                for (int i = 0; i < n; ++i) {
+                    if (s.charAt(i) == a) {
+                        f[0]++;
+                        f[1]++;
+                    } else if (s.charAt(i) == b) {
+                        f[1] = Math.max(f[0] - 1, f[1] - 1);
+                        f[0] = 0;
+                    }
+                    ans = Math.max(ans, f[1]);
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    int largestVariance(string s) {
+        int n = s.size();
+        int ans = 0;
+        for (char a = 'a'; a <= 'z'; ++a) {
+            for (char b = 'a'; b <= 'z'; ++b) {
+                if (a == b) continue;
+                int f[2] = {0, -n};
+                for (char c : s) {
+                    if (c == a) {
+                        f[0]++;
+                        f[1]++;
+                    } else if (c == b) {
+                        f[1] = max(f[1] - 1, f[0] - 1);
+                        f[0] = 0;
+                    }
+                    ans = max(ans, f[1]);
+                }
+            }
+        }
+        return ans;
+    }
+};
+```
 
+### **Go**
+
+```go
+func largestVariance(s string) int {
+	ans, n := 0, len(s)
+	for a := 'a'; a <= 'z'; a++ {
+		for b := 'a'; b <= 'z'; b++ {
+			if a == b {
+				continue
+			}
+			f := [2]int{0, -n}
+			for _, c := range s {
+				if c == a {
+					f[0]++
+					f[1]++
+				} else if c == b {
+					f[1] = max(f[1]-1, f[0]-1)
+					f[0] = 0
+				}
+				ans = max(ans, f[1])
+			}
+		}
+	}
+	return ans
+}
+
+func max(a, b int) int {
+	if a > b {
+		return a
+	}
+	return b
+}
 ```
 
 ### **TypeScript**