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feat: add solutions to lc problem: No.2516 #2581

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Apr 13, 2024
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feat: add solutions to lc problem: No.2516 #2581

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116 changes: 64 additions & 52 deletions solution/2500-2599/2516.Take K of Each Character From Left and Right/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -48,15 +48,15 @@

### 方法一：滑动窗口

我们先用哈希表或者一个长度为 $3$ 的数组 `cnt` 统计字符串 $s$ 中每个字符的个数，如果有字符的个数小于 $k$ 个，则无法取到，提前返回 $-1$。
我们先用哈希表或者一个长度为 $3$ 的数组 $cnt$ 统计字符串 $s$ 中每个字符的个数，如果有字符的个数小于 $k$ 个，则无法取到，提前返回 $-1$。

题目要我们在字符串左侧以及右侧取走字符，最终取到的每种字符的个数都不少于 $k$ 个。我们不妨反着考虑问题，取走中间某个窗口大小的字符串，使得剩下的两侧字符串中，每种字符的个数都不少于 $k$ 个。

因此，我们维护一个滑动窗口，用指针 $j$ 和 $i$ 分别表示窗口的左右边界，窗口内的字符串是我们要取走的。我们每一次移动右边界 $i$，将对应的字符 $s[i]$ 加入到窗口中（也即取走一个字符 $s[i]$），此时如果 $cnt[s[i]]$ 个数小于 $k$，那么我们循环移动左边界 $j$，直到 $cnt[s[i]]$ 个数不小于 $k$ 为止。此时的窗口大小为 $i - j + 1$，更新最大窗口。

最终的答案就是字符串 $s$ 的长度减去最大窗口的大小。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为字符串 $s$ 的长度。
时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。空间复杂度 $O(1)$。

<!-- tabs:start -->

Expand All @@ -66,14 +66,14 @@ class Solution:
cnt = Counter(s)
if any(cnt[c] < k for c in "abc"):
return -1
ans = j = 0
mx = j = 0
for i, c in enumerate(s):
cnt[c] -= 1
while cnt[c] < k:
cnt[s[j]] += 1
j += 1
ans = max(ans, i - j + 1)
return len(s) - ans
mx = max(mx, i - j + 1)
return len(s) - mx
```

```java
Expand All @@ -84,19 +84,21 @@ class Solution {
for (int i = 0; i < n; ++i) {
++cnt[s.charAt(i) - 'a'];
}
if (cnt[0] < k || cnt[1] < k || cnt[2] < k) {
return -1;
for (int x : cnt) {
if (x < k) {
return -1;
}
}
int ans = 0, j = 0;
int mx = 0, j = 0;
for (int i = 0; i < n; ++i) {
int c = s.charAt(i) - 'a';
--cnt[c];
while (cnt[c] < k) {
++cnt[s.charAt(j++) - 'a'];
}
ans = Math.max(ans, i - j + 1);
mx = Math.max(mx, i - j + 1);
}
return n - ans;
return n - mx;
}
}
```
Expand All @@ -105,20 +107,26 @@ class Solution {
class Solution {
public:
int takeCharacters(string s, int k) {
int cnt[3] = {0};
for (char& c : s) ++cnt[c - 'a'];
if (cnt[0] < k || cnt[1] < k || cnt[2] < k) return -1;
int ans = 0, j = 0;
int n = s.size();
int cnt[3]{};
int n = s.length();
for (int i = 0; i < n; ++i) {
++cnt[s[i] - 'a'];
}
for (int x : cnt) {
if (x < k) {
return -1;
}
}
int mx = 0, j = 0;
for (int i = 0; i < n; ++i) {
int c = s[i] - 'a';
--cnt[c];
while (cnt[c] < k) {
++cnt[s[j++] - 'a'];
}
ans = max(ans, i - j + 1);
mx = max(mx, i - j + 1);
}
return n - ans;
return n - mx;
}
};
```
Expand All @@ -129,70 +137,74 @@ func takeCharacters(s string, k int) int {
for _, c := range s {
cnt[c-'a']++
}
if cnt[0] < k || cnt[1] < k || cnt[2] < k {
return -1
for _, x := range cnt {
if x < k {
return -1
}
}
ans, j := 0, 0
mx, j := 0, 0
for i, c := range s {
c -= 'a'
cnt[c]--
for cnt[c] < k {
for cnt[c]--; cnt[c] < k; j++ {
cnt[s[j]-'a']++
j++
}
ans = max(ans, i-j+1)
mx = max(mx, i-j+1)
}
return len(s) - ans
return len(s) - mx
}
```

```ts
function takeCharacters(s: string, k: number): number {
const getIndex = (c: string) => c.charCodeAt(0) - 'a'.charCodeAt(0);
const count = [0, 0, 0];
const idx = (c: string) => c.charCodeAt(0) - 97;
const cnt: number[] = Array(3).fill(0);
for (const c of s) {
count[getIndex(c)]++;
++cnt[idx(c)];
}
if (count.some(v => v < k)) {
if (cnt.some(v => v < k)) {
return -1;
}
const n = s.length;
let ans = 0;
for (let i = 0, j = 0; j < n; j++) {
count[getIndex(s[j])]--;
while (count[getIndex(s[j])] < k) {
count[getIndex(s[i])]++;
i++;
let [mx, j] = [0, 0];
for (let i = 0; i < n; ++i) {
const c = idx(s[i]);
--cnt[c];
while (cnt[c] < k) {
++cnt[idx(s[j++])];
}
ans = Math.max(ans, j - i + 1);
mx = Math.max(mx, i - j + 1);
}
return n - ans;
return n - mx;
}
```

```rust
use std::collections::HashMap;

impl Solution {
pub fn take_characters(s: String, k: i32) -> i32 {
let s = s.as_bytes();
let mut count = vec![0; 3];
for c in s.iter() {
count[(c - b'a') as usize] += 1;
let mut cnt: HashMap<char, i32> = HashMap::new();
for c in s.chars() {
*cnt.entry(c).or_insert(0) += 1;
}
if count.iter().any(|v| *v < k) {

if "abc".chars().any(|c| cnt.get(&c).unwrap_or(&0) < &k) {
return -1;
}
let n = s.len();
let mut ans = 0;
let mut i = 0;
for j in 0..n {
count[(s[j] - b'a') as usize] -= 1;
while count[(s[j] - b'a') as usize] < k {
count[(s[i] - b'a') as usize] += 1;
i += 1;

let mut mx = 0;
let mut j = 0;
let mut cs = s.chars().collect::<Vec<char>>();
for i in 0..cs.len() {
let c = cs[i];
*cnt.get_mut(&c).unwrap() -= 1;
while cnt.get(&c).unwrap() < &k {
*cnt.get_mut(&cs[j]).unwrap() += 1;
j += 1;
}
ans = ans.max(j - i + 1);
mx = mx.max(i - j + 1);
}
(n - ans) as i32
(cs.len() as i32) - (mx as i32)
}
}
```
Expand Down
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