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

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105 changes: 61 additions & 44 deletions solution/0400-0499/0424.Longest Repeating Character Replacement/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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 -->

Expand All @@ -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;
}
}
```
Expand All @@ -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;
}
};
```
Expand All @@ -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;
}
```

Expand Down
105 changes: 61 additions & 44 deletions solution/0400-0499/0424.Longest Repeating Character Replacement/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -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 -->

Expand All @@ -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;
}
}
```
Expand All @@ -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;
}
};
```
Expand All @@ -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;
}
```

Expand Down
18 changes: 8 additions & 10 deletions solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.cpp
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Original file line number Diff line number Diff line change
@@ -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;
}
};
21 changes: 9 additions & 12 deletions solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.go
View file
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Original file line number Diff line number Diff line change
@@ -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
}
19 changes: 8 additions & 11 deletions solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.java
View file
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Original file line number Diff line number Diff line change
@@ -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;
}
}
19 changes: 9 additions & 10 deletions solution/0400-0499/0424.Longest Repeating Character Replacement/Solution.py
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -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
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