feat: add solutions to lc problem: No.424 (#3029)

No.0424.Longest Repeating Character Replacement

Author: yanglbme

solution/0400-0499/0424.Longest Repeating Character Replacement/README.md

@@ -59,7 +59,15 @@ tags:
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：双指针
+
+我们使用一个哈希表 `cnt` 统计字符串中每个字符出现的次数，使用双指针 `l` 和 `r` 维护一个滑动窗口，使得窗口的大小减去出现次数最多的字符的次数，结果不超过 $k$。
+
+我们遍历字符串，每次更新窗口的右边界 `r`，并更新窗口内的字符出现次数，同时更新出现过的字符的最大出现次数 `mx`。当窗口的大小减去 `mx` 大于 $k$ 时，我们需要缩小窗口的左边界 `l`，同时更新窗口内的字符出现次数，直到窗口的大小减去 `mx` 不大于 $k$。
+
+最后，答案为 $n - l$，其中 $n$ 为字符串的长度。
+
+时间复杂度 $O(n)$，空间复杂度 $O(|\Sigma|)$。其中 $n$ 为字符串的长度，而 $|\Sigma|$ 为字符集的大小，本题中字符集为大写英文字母，所以 $|\Sigma| = 26$。
 
 <!-- tabs:start -->
 
@@ -68,36 +76,32 @@ tags:
 ```python
 class Solution:
     def characterReplacement(self, s: str, k: int) -> int:
-        counter = [0] * 26
-        i = j = maxCnt = 0
-        while i < len(s):
-            counter[ord(s[i]) - ord('A')] += 1
-            maxCnt = max(maxCnt, counter[ord(s[i]) - ord('A')])
-            if i - j + 1 > maxCnt + k:
-                counter[ord(s[j]) - ord('A')] -= 1
-                j += 1
-            i += 1
-        return i - j
+        cnt = Counter()
+        l = mx = 0
+        for r, c in enumerate(s):
+            cnt[c] += 1
+            mx = max(mx, cnt[c])
+            if r - l + 1 - mx > k:
+                cnt[s[l]] -= 1
+                l += 1
+        return len(s) - l
 ```
 
 #### Java
 
 ```java
 class Solution {
     public int characterReplacement(String s, int k) {
-        int[] counter = new int[26];
-        int i = 0;
-        int j = 0;
-        for (int maxCnt = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            ++counter[c - 'A'];
-            maxCnt = Math.max(maxCnt, counter[c - 'A']);
-            if (i - j + 1 - maxCnt > k) {
-                --counter[s.charAt(j) - 'A'];
-                ++j;
+        int[] cnt = new int[26];
+        int l = 0, mx = 0;
+        int n = s.length();
+        for (int r = 0; r < n; ++r) {
+            mx = Math.max(mx, ++cnt[s.charAt(r) - 'A']);
+            if (r - l + 1 - mx > k) {
+                --cnt[s.charAt(l++) - 'A'];
             }
         }
-        return i - j;
+        return n - l;
     }
 }
 ```
@@ -108,18 +112,16 @@ class Solution {
 class Solution {
 public:
     int characterReplacement(string s, int k) {
-        vector<int> counter(26);
-        int i = 0, j = 0, maxCnt = 0;
-        for (char& c : s) {
-            ++counter[c - 'A'];
-            maxCnt = max(maxCnt, counter[c - 'A']);
-            if (i - j + 1 > maxCnt + k) {
-                --counter[s[j] - 'A'];
-                ++j;
+        int cnt[26]{};
+        int l = 0, mx = 0;
+        int n = s.length();
+        for (int r = 0; r < n; ++r) {
+            mx = max(mx, ++cnt[s[r] - 'A']);
+            if (r - l + 1 - mx > k) {
+                --cnt[s[l++] - 'A'];
             }
-            ++i;
         }
-        return i - j;
+        return n - l;
     }
 };
 ```
@@ -128,20 +130,35 @@ public:
 
 ```go
 func characterReplacement(s string, k int) int {
-	counter := make([]int, 26)
-	j, maxCnt := 0, 0
-	for i := range s {
-		c := s[i] - 'A'
-		counter[c]++
-		if maxCnt < counter[c] {
-			maxCnt = counter[c]
-		}
-		if i-j+1 > maxCnt+k {
-			counter[s[j]-'A']--
-			j++
+	cnt := [26]int{}
+	l, mx := 0, 0
+	for r, c := range s {
+		cnt[c-'A']++
+		mx = max(mx, cnt[c-'A'])
+		if r-l+1-mx > k {
+			cnt[s[l]-'A']--
+			l++
 		}
 	}
-	return len(s) - j
+	return len(s) - l
+}
+```
+
+#### TypeScript
+
+```ts
+function characterReplacement(s: string, k: number): number {
+    const idx = (c: string) => c.charCodeAt(0) - 65;
+    const cnt: number[] = Array(26).fill(0);
+    const n = s.length;
+    let [l, mx] = [0, 0];
+    for (let r = 0; r < n; ++r) {
+        mx = Math.max(mx, ++cnt[idx(s[r])]);
+        if (r - l + 1 - mx > k) {
+            --cnt[idx(s[l++])];
+        }
+    }
+    return n - l;
 }
 ```
 

solution/0400-0499/0424.Longest Repeating Character Replacement/README_EN.md

@@ -55,7 +55,15 @@ There may exists other ways to achieve this answer too.</pre>
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers
+
+We use a hash table `cnt` to count the occurrence of each character in the string, and two pointers `l` and `r` to maintain a sliding window, such that the size of the window minus the count of the most frequent character does not exceed $k$.
+
+We iterate through the string, updating the right boundary `r` of the window each time, updating the count of characters within the window, and updating the maximum count `mx` of the characters that have appeared. When the size of the window minus `mx` is greater than $k$, we need to shrink the left boundary `l` of the window, updating the count of characters within the window, until the size of the window minus `mx` is no longer greater than $k$.
+
+Finally, the answer is $n - l$, where $n$ is the length of the string.
+
+The time complexity is $O(n)$, and the space complexity is $O(|\Sigma|)$. Where $n$ is the length of the string, and $|\Sigma|$ is the size of the character set. In this problem, the character set is uppercase English letters, so $|\Sigma| = 26$.
 
 <!-- tabs:start -->
 
@@ -64,36 +72,32 @@ There may exists other ways to achieve this answer too.</pre>
 ```python
 class Solution:
     def characterReplacement(self, s: str, k: int) -> int:
-        counter = [0] * 26
-        i = j = maxCnt = 0
-        while i < len(s):
-            counter[ord(s[i]) - ord('A')] += 1
-            maxCnt = max(maxCnt, counter[ord(s[i]) - ord('A')])
-            if i - j + 1 > maxCnt + k:
-                counter[ord(s[j]) - ord('A')] -= 1
-                j += 1
-            i += 1
-        return i - j
+        cnt = Counter()
+        l = mx = 0
+        for r, c in enumerate(s):
+            cnt[c] += 1
+            mx = max(mx, cnt[c])
+            if r - l + 1 - mx > k:
+                cnt[s[l]] -= 1
+                l += 1
+        return len(s) - l
 ```
 
 #### Java
 
 ```java
 class Solution {
     public int characterReplacement(String s, int k) {
-        int[] counter = new int[26];
-        int i = 0;
-        int j = 0;
-        for (int maxCnt = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            ++counter[c - 'A'];
-            maxCnt = Math.max(maxCnt, counter[c - 'A']);
-            if (i - j + 1 - maxCnt > k) {
-                --counter[s.charAt(j) - 'A'];
-                ++j;
+        int[] cnt = new int[26];
+        int l = 0, mx = 0;
+        int n = s.length();
+        for (int r = 0; r < n; ++r) {
+            mx = Math.max(mx, ++cnt[s.charAt(r) - 'A']);
+            if (r - l + 1 - mx > k) {
+                --cnt[s.charAt(l++) - 'A'];
             }
         }
-        return i - j;
+        return n - l;
     }
 }
 ```
@@ -104,18 +108,16 @@ class Solution {
 class Solution {
 public:
     int characterReplacement(string s, int k) {
-        vector<int> counter(26);
-        int i = 0, j = 0, maxCnt = 0;
-        for (char& c : s) {
-            ++counter[c - 'A'];
-            maxCnt = max(maxCnt, counter[c - 'A']);
-            if (i - j + 1 > maxCnt + k) {
-                --counter[s[j] - 'A'];
-                ++j;
+        int cnt[26]{};
+        int l = 0, mx = 0;
+        int n = s.length();
+        for (int r = 0; r < n; ++r) {
+            mx = max(mx, ++cnt[s[r] - 'A']);
+            if (r - l + 1 - mx > k) {
+                --cnt[s[l++] - 'A'];
             }
-            ++i;
         }
-        return i - j;
+        return n - l;
     }
 };
 ```
@@ -124,20 +126,35 @@ public:
 
 ```go
 func characterReplacement(s string, k int) int {
-	counter := make([]int, 26)
-	j, maxCnt := 0, 0
-	for i := range s {
-		c := s[i] - 'A'
-		counter[c]++
-		if maxCnt < counter[c] {
-			maxCnt = counter[c]
-		}
-		if i-j+1 > maxCnt+k {
-			counter[s[j]-'A']--
-			j++
+	cnt := [26]int{}
+	l, mx := 0, 0
+	for r, c := range s {
+		cnt[c-'A']++
+		mx = max(mx, cnt[c-'A'])
+		if r-l+1-mx > k {
+			cnt[s[l]-'A']--
+			l++
 		}
 	}
-	return len(s) - j
+	return len(s) - l
+}
+```
+
+#### TypeScript
+
+```ts
+function characterReplacement(s: string, k: number): number {
+    const idx = (c: string) => c.charCodeAt(0) - 65;
+    const cnt: number[] = Array(26).fill(0);
+    const n = s.length;
+    let [l, mx] = [0, 0];
+    for (let r = 0; r < n; ++r) {
+        mx = Math.max(mx, ++cnt[idx(s[r])]);
+        if (r - l + 1 - mx > k) {
+            --cnt[idx(s[l++])];
+        }
+    }
+    return n - l;
 }
 ```
 

solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.cpp

@@ -1,17 +1,15 @@
 class Solution {
 public:
     int characterReplacement(string s, int k) {
-        vector<int> counter(26);
-        int i = 0, j = 0, maxCnt = 0;
-        for (char& c : s) {
-            ++counter[c - 'A'];
-            maxCnt = max(maxCnt, counter[c - 'A']);
-            if (i - j + 1 > maxCnt + k) {
-                --counter[s[j] - 'A'];
-                ++j;
+        int cnt[26]{};
+        int l = 0, mx = 0;
+        int n = s.length();
+        for (int r = 0; r < n; ++r) {
+            mx = max(mx, ++cnt[s[r] - 'A']);
+            if (r - l + 1 - mx > k) {
+                --cnt[s[l++] - 'A'];
             }
-            ++i;
         }
-        return i - j;
+        return n - l;
     }
 };

solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.go

@@ -1,16 +1,13 @@
 func characterReplacement(s string, k int) int {
-	counter := make([]int, 26)
-	j, maxCnt := 0, 0
-	for i := range s {
-		c := s[i] - 'A'
-		counter[c]++
-		if maxCnt < counter[c] {
-			maxCnt = counter[c]
-		}
-		if i-j+1 > maxCnt+k {
-			counter[s[j]-'A']--
-			j++
+	cnt := [26]int{}
+	l, mx := 0, 0
+	for r, c := range s {
+		cnt[c-'A']++
+		mx = max(mx, cnt[c-'A'])
+		if r-l+1-mx > k {
+			cnt[s[l]-'A']--
+			l++
 		}
 	}
-	return len(s) - j
+	return len(s) - l
 }

solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.java

@@ -1,17 +1,14 @@
 class Solution {
     public int characterReplacement(String s, int k) {
-        int[] counter = new int[26];
-        int i = 0;
-        int j = 0;
-        for (int maxCnt = 0; i < s.length(); ++i) {
-            char c = s.charAt(i);
-            ++counter[c - 'A'];
-            maxCnt = Math.max(maxCnt, counter[c - 'A']);
-            if (i - j + 1 - maxCnt > k) {
-                --counter[s.charAt(j) - 'A'];
-                ++j;
+        int[] cnt = new int[26];
+        int l = 0, mx = 0;
+        int n = s.length();
+        for (int r = 0; r < n; ++r) {
+            mx = Math.max(mx, ++cnt[s.charAt(r) - 'A']);
+            if (r - l + 1 - mx > k) {
+                --cnt[s.charAt(l++) - 'A'];
             }
         }
-        return i - j;
+        return n - l;
     }
 }

solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.py

@@ -1,12 +1,11 @@
 class Solution:
     def characterReplacement(self, s: str, k: int) -> int:
-        counter = [0] * 26
-        i = j = maxCnt = 0
-        while i < len(s):
-            counter[ord(s[i]) - ord('A')] += 1
-            maxCnt = max(maxCnt, counter[ord(s[i]) - ord('A')])
-            if i - j + 1 > maxCnt + k:
-                counter[ord(s[j]) - ord('A')] -= 1
-                j += 1
-            i += 1
-        return i - j
+        cnt = Counter()
+        l = mx = 0
+        for r, c in enumerate(s):
+            cnt[c] += 1
+            mx = max(mx, cnt[c])
+            if r - l + 1 - mx > k:
+                cnt[s[l]] -= 1
+                l += 1
+        return len(s) - l