feat: add solutions to lc problem: No.2953 #2058

Original file line number	Diff line number	Diff line change
Expand Up		@@ -58,27 +58,271 @@
		<!-- 这里可写当前语言的特殊实现逻辑 -->

		```python

	class Solution:
	def countCompleteSubstrings(self, word: str, k: int) -> int:
	def f(s: str) -> int:
	m = len(s)
	ans = 0
	for i in range(1, 27):
	l = i * k
	if l > m:
	break
	cnt = Counter(s[:l])
	freq = Counter(cnt.values())
	ans += freq[k] == i
	for j in range(l, m):
	freq[cnt[s[j]]] -= 1
	cnt[s[j]] += 1
	freq[cnt[s[j]]] += 1

	freq[cnt[s[j - l]]] -= 1
	cnt[s[j - l]] -= 1
	freq[cnt[s[j - l]]] += 1

	ans += freq[k] == i
	return ans

	n = len(word)
	ans = i = 0
	while i < n:
	j = i + 1
	while j < n and abs(ord(word[j]) - ord(word[j - 1])) <= 2:
	j += 1
	ans += f(word[i:j])
	i = j
	return ans
		```

		### Java

		<!-- 这里可写当前语言的特殊实现逻辑 -->

		```java

	class Solution {
	public int countCompleteSubstrings(String word, int k) {
	int n = word.length();
	int ans = 0;
	for (int i = 0; i < n;) {
	int j = i + 1;
	while (j < n && Math.abs(word.charAt(j) - word.charAt(j - 1)) <= 2) {
	++j;
	}
	ans += f(word.substring(i, j), k);
	i = j;
	}
	return ans;
	}

	private int f(String s, int k) {
	int m = s.length();
	int ans = 0;
	for (int i = 1; i <= 26 && i * k <= m; ++i) {
	int l = i * k;
	int[] cnt = new int[26];
	for (int j = 0; j < l; ++j) {
	++cnt[s.charAt(j) - 'a'];
	}
	Map<Integer, Integer> freq = new HashMap<>();
	for (int x : cnt) {
	if (x > 0) {
	freq.merge(x, 1, Integer::sum);
	}
	}
	if (freq.getOrDefault(k, 0) == i) {
	++ans;
	}
	for (int j = l; j < m; ++j) {
	int a = s.charAt(j) - 'a';
	int b = s.charAt(j - l) - 'a';
	freq.merge(cnt[a], -1, Integer::sum);
	++cnt[a];
	freq.merge(cnt[a], 1, Integer::sum);

	freq.merge(cnt[b], -1, Integer::sum);
	--cnt[b];
	freq.merge(cnt[b], 1, Integer::sum);
	if (freq.getOrDefault(k, 0) == i) {
	++ans;
	}
	}
	}
	return ans;
	}
	}
		```

		### C++

		```cpp

	class Solution {
	public:
	int countCompleteSubstrings(string word, int k) {
	int n = word.length();
	int ans = 0;
	auto f = [&](string s) {
	int m = s.length();
	int ans = 0;
	for (int i = 1; i <= 26 && i * k <= m; ++i) {
	int l = i * k;
	int cnt[26]{};
	for (int j = 0; j < l; ++j) {
	++cnt[s[j] - 'a'];
	}
	unordered_map<int, int> freq;
	for (int x : cnt) {
	if (x > 0) {
	freq[x]++;
	}
	}
	if (freq[k] == i) {
	++ans;
	}
	for (int j = l; j < m; ++j) {
	int a = s[j] - 'a';
	int b = s[j - l] - 'a';
	freq[cnt[a]]--;
	cnt[a]++;
	freq[cnt[a]]++;

	freq[cnt[b]]--;
	cnt[b]--;
	freq[cnt[b]]++;

	if (freq[k] == i) {
	++ans;
	}
	}
	}
	return ans;
	};
	for (int i = 0; i < n;) {
	int j = i + 1;
	while (j < n && abs(word[j] - word[j - 1]) <= 2) {
	++j;
	}
	ans += f(word.substr(i, j - i));
	i = j;
	}
	return ans;
	}
	};
		```

		### Go

		```go
	func countCompleteSubstrings(word string, k int) (ans int) {
	n := len(word)
	f := func(s string) (ans int) {
	m := len(s)
	for i := 1; i <= 26 && i*k <= m; i++ {
	l := i * k
	cnt := [26]int{}
	for j := 0; j < l; j++ {
	cnt[int(s[j]-'a')]++
	}
	freq := map[int]int{}
	for _, x := range cnt {
	if x > 0 {
	freq[x]++
	}
	}
	if freq[k] == i {
	ans++
	}
	for j := l; j < m; j++ {
	a := int(s[j] - 'a')
	b := int(s[j-l] - 'a')
	freq[cnt[a]]--
	cnt[a]++
	freq[cnt[a]]++

	freq[cnt[b]]--
	cnt[b]--
	freq[cnt[b]]++

	if freq[k] == i {
	ans++
	}
	}
	}
	return
	}
	for i := 0; i < n; {
	j := i + 1
	for j < n && abs(int(word[j])-int(word[j-1])) <= 2 {
	j++
	}
	ans += f(word[i:j])
	i = j
	}
	return
	}

	func abs(x int) int {
	if x < 0 {
	return -x
	}
	return x
	}
	```

	### TypeScript

	```ts
	function countCompleteSubstrings(word: string, k: number): number {
	const f = (s: string): number => {
	const m = s.length;
	let ans = 0;
	for (let i = 1; i <= 26 && i * k <= m; i++) {
	const l = i * k;
	const cnt: number[] = new Array(26).fill(0);
	for (let j = 0; j < l; j++) {
	cnt[s.charCodeAt(j) - 'a'.charCodeAt(0)]++;
	}
	const freq: { [key: number]: number } = {};
	for (const x of cnt) {
	if (x > 0) {
	freq[x] = (freq[x] \|\| 0) + 1;
	}
	}
	if (freq[k] === i) {
	ans++;
	}

	for (let j = l; j < m; j++) {
	const a = s.charCodeAt(j) - 'a'.charCodeAt(0);
	const b = s.charCodeAt(j - l) - 'a'.charCodeAt(0);

	freq[cnt[a]]--;
	cnt[a]++;
	freq[cnt[a]] = (freq[cnt[a]] \|\| 0) + 1;

	freq[cnt[b]]--;
	cnt[b]--;
	freq[cnt[b]] = (freq[cnt[b]] \|\| 0) + 1;

	if (freq[k] === i) {
	ans++;
	}
	}
	}

	return ans;
	};

	let n = word.length;
	let ans = 0;
	for (let i = 0; i < n; ) {
	let j = i + 1;
	while (j < n && Math.abs(word.charCodeAt(j) - word.charCodeAt(j - 1)) <= 2) {
	j++;
	}
	ans += f(word.substring(i, j));
	i = j;
	}
	return ans;
	}
		```

		### ...
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