Skip to content

feat: add solutions to lc problem: No.3234 #4825

New issue

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

Sign up for GitHub

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

Merged
merged 1 commit into from
Nov 9, 2025
Merged

feat: add solutions to lc problem: No.3234 #4825

Show file tree
Hide file tree
Changes from all commits
Commits
Show all changes
1 commit
Select commit
File filter

Filter by extension

Filter by extension
Conversations
Failed to load comments.
Loading
Jump to
Jump to file
Failed to load files.
Loading
Diff view
Diff view
146 changes: 142 additions & 4 deletions ...tion/3200-3299/3234.Count the Number of Substrings With Dominant Ones/README.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -167,32 +167,170 @@ tags:

<!-- solution:start -->

### 方法一
### 方法一：预处理 + 枚举

根据题目描述，显著字符串满足 $\textit{cnt}_1 \geq \textit{cnt}_0^2$，那么 $\textit{cnt}_0$ 的最大值不超过 $\sqrt{n}$，其中 $n$ 是字符串的长度。因此我们可以枚举 $\textit{cnt}_0$ 的值，然后计算满足条件的子字符串数量。

我们首先预处理字符串中每个位置开始的第一个 $0$ 的位置，存储在数组 $\textit{nxt}$ 中，其中 $\textit{nxt}[i]$ 表示从位置 $i$ 开始的第一个 $0$ 的位置，如果不存在则为 $n$。

接下来，我们遍历字符串的每个位置 $i$ 作为子字符串的起始位置，初始化 $\textit{cnt}_0$ 的值为 $0$ 或 $1$（取决于当前位置是否为 $0$）。然后我们使用一个指针 $j$ 从位置 $i$ 开始，逐步跳转到下一个 $0$ 的位置，同时更新 $\textit{cnt}_0$ 的值。

对于从位置 $i$ 开始，且包含 $\textit{cnt}_0$ 个 $0$ 的子字符串，最多可以包含 $\textit{nxt}[j + 1] - i - \textit{cnt}_0$ 个 $1$。如果这个数量大于等于 $\textit{cnt}_0^2$，则说明存在满足条件的子字符串。此时需要判断字符串从右端点 $\textit{nxt}[j + 1] - 1$ 向左移动多少个位置，可以使得子字符串仍然满足条件。具体来说，右端点一共可以有 $\min(\textit{nxt}[j + 1] - j, \textit{cnt}_1 - \textit{cnt}_0^2 + 1)$ 种选择方式。我们将这些数量累加到答案中。然后将指针 $j$ 移动到下一个 $0$ 的位置，继续枚举下一个 $\textit{cnt}_0$ 的值，直到 $\textit{cnt}_0^2$ 超过字符串长度为止。

时间复杂度 $O(n \times \sqrt{n})$，空间复杂度 $O(n)$，其中 $n$ 是字符串的长度。

<!-- tabs:start -->

#### Python3

```python

class Solution:
def numberOfSubstrings(self, s: str) -> int:
n = len(s)
nxt = [n] * (n + 1)
for i in range(n - 1, -1, -1):
nxt[i] = nxt[i + 1]
if s[i] == "0":
nxt[i] = i
ans = 0
for i in range(n):
cnt0 = int(s[i] == "0")
j = i
while j < n and cnt0 * cnt0 <= n:
cnt1 = (nxt[j + 1] - i) - cnt0
if cnt1 >= cnt0 * cnt0:
ans += min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1)
j = nxt[j + 1]
cnt0 += 1
return ans
```

#### Java

```java

class Solution {
public int numberOfSubstrings(String s) {
int n = s.length();
int[] nxt = new int[n + 1];
nxt[n] = n;
for (int i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s.charAt(i) == '0') {
nxt[i] = i;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt0 = s.charAt(i) == '0' ? 1 : 0;
int j = i;
while (j < n && 1L * cnt0 * cnt0 <= n) {
int cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += Math.min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
}
```

#### C++

```cpp

class Solution {
public:
int numberOfSubstrings(string s) {
int n = s.size();
vector<int> nxt(n + 1);
nxt[n] = n;
for (int i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s[i] == '0') {
nxt[i] = i;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt0 = s[i] == '0' ? 1 : 0;
int j = i;
while (j < n && 1LL * cnt0 * cnt0 <= n) {
int cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
};
```

#### Go

```go
func numberOfSubstrings(s string) int {
n := len(s)
nxt := make([]int, n+1)
nxt[n] = n
for i := n - 1; i >= 0; i-- {
nxt[i] = nxt[i+1]
if s[i] == '0' {
nxt[i] = i
}
}
ans := 0
for i := 0; i < n; i++ {
cnt0 := 0
if s[i] == '0' {
cnt0 = 1
}
j := i
for j < n && int64(cnt0*cnt0) <= int64(n) {
cnt1 := nxt[j+1] - i - cnt0
if cnt1 >= cnt0*cnt0 {
ans += min(nxt[j+1]-j, cnt1-cnt0*cnt0+1)
}
j = nxt[j+1]
cnt0++
}
}
return ans
}
```

#### TypeScript

```ts
function numberOfSubstrings(s: string): number {
const n = s.length;
const nxt: number[] = Array(n + 1).fill(0);
nxt[n] = n;
for (let i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s[i] === '0') {
nxt[i] = i;
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
let cnt0 = s[i] === '0' ? 1 : 0;
let j = i;
while (j < n && cnt0 * cnt0 <= n) {
const cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += Math.min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
```

<!-- tabs:end -->
Expand Down
146 changes: 142 additions & 4 deletions ...n/3200-3299/3234.Count the Number of Substrings With Dominant Ones/README_EN.md
View file
Open in desktop
Original file line number Diff line number Diff line change
Expand Up @@ -165,32 +165,170 @@ tags:

<!-- solution:start -->

### Solution 1
### Solution 1: Preprocessing + Enumeration

According to the problem description, a dominant string satisfies $\textit{cnt}_1 \geq \textit{cnt}_0^2$, which means the maximum value of $\textit{cnt}_0$ does not exceed $\sqrt{n}$, where $n$ is the length of the string. Therefore, we can enumerate the value of $\textit{cnt}_0$ and then calculate the number of substrings that satisfy the condition.

We first preprocess the position of the first $0$ starting from each position in the string, storing it in the array $\textit{nxt}$, where $\textit{nxt}[i]$ represents the position of the first $0$ starting from position $i$, or $n$ if it doesn't exist.

Next, we iterate through each position $i$ in the string as the starting position of the substring, initializing $\textit{cnt}_0$ to $0$ or $1$ (depending on whether the current position is $0$). Then we use a pointer $j$ starting from position $i$, jumping step by step to the position of the next $0$, while updating the value of $\textit{cnt}_0$.

For a substring starting at position $i$ and containing $\textit{cnt}_0$ zeros, it can contain at most $\textit{nxt}[j + 1] - i - \textit{cnt}_0$ ones. If this quantity is greater than or equal to $\textit{cnt}_0^2$, it means there exists a substring that satisfies the condition. At this point, we need to determine how many positions the right endpoint $\textit{nxt}[j + 1] - 1$ can move left while still satisfying the condition. Specifically, the right endpoint has a total of $\min(\textit{nxt}[j + 1] - j, \textit{cnt}_1 - \textit{cnt}_0^2 + 1)$ possible choices. We accumulate these counts to the answer. Then we move pointer $j$ to the position of the next $0$ and continue enumerating the next value of $\textit{cnt}_0$ until $\textit{cnt}_0^2$ exceeds the string length.

The time complexity is $O(n \times \sqrt{n})$, and the space complexity is $O(n)$, where $n$ is the length of the string.

<!-- tabs:start -->

#### Python3

```python

class Solution:
def numberOfSubstrings(self, s: str) -> int:
n = len(s)
nxt = [n] * (n + 1)
for i in range(n - 1, -1, -1):
nxt[i] = nxt[i + 1]
if s[i] == "0":
nxt[i] = i
ans = 0
for i in range(n):
cnt0 = int(s[i] == "0")
j = i
while j < n and cnt0 * cnt0 <= n:
cnt1 = (nxt[j + 1] - i) - cnt0
if cnt1 >= cnt0 * cnt0:
ans += min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1)
j = nxt[j + 1]
cnt0 += 1
return ans
```

#### Java

```java

class Solution {
public int numberOfSubstrings(String s) {
int n = s.length();
int[] nxt = new int[n + 1];
nxt[n] = n;
for (int i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s.charAt(i) == '0') {
nxt[i] = i;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt0 = s.charAt(i) == '0' ? 1 : 0;
int j = i;
while (j < n && 1L * cnt0 * cnt0 <= n) {
int cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += Math.min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
}
```

#### C++

```cpp

class Solution {
public:
int numberOfSubstrings(string s) {
int n = s.size();
vector<int> nxt(n + 1);
nxt[n] = n;
for (int i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s[i] == '0') {
nxt[i] = i;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt0 = s[i] == '0' ? 1 : 0;
int j = i;
while (j < n && 1LL * cnt0 * cnt0 <= n) {
int cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
};
```

#### Go

```go
func numberOfSubstrings(s string) int {
n := len(s)
nxt := make([]int, n+1)
nxt[n] = n
for i := n - 1; i >= 0; i-- {
nxt[i] = nxt[i+1]
if s[i] == '0' {
nxt[i] = i
}
}
ans := 0
for i := 0; i < n; i++ {
cnt0 := 0
if s[i] == '0' {
cnt0 = 1
}
j := i
for j < n && int64(cnt0*cnt0) <= int64(n) {
cnt1 := nxt[j+1] - i - cnt0
if cnt1 >= cnt0*cnt0 {
ans += min(nxt[j+1]-j, cnt1-cnt0*cnt0+1)
}
j = nxt[j+1]
cnt0++
}
}
return ans
}
```

#### TypeScript

```ts
function numberOfSubstrings(s: string): number {
const n = s.length;
const nxt: number[] = Array(n + 1).fill(0);
nxt[n] = n;
for (let i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s[i] === '0') {
nxt[i] = i;
}
}
let ans = 0;
for (let i = 0; i < n; ++i) {
let cnt0 = s[i] === '0' ? 1 : 0;
let j = i;
while (j < n && cnt0 * cnt0 <= n) {
const cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += Math.min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
```

<!-- tabs:end -->
Expand Down
28 changes: 28 additions & 0 deletions solution/3200-3299/3234.Count the Number of Substrings With Dominant Ones/Solution.cpp
View file
Open in desktop
Original file line number Diff line number Diff line change
@@ -0,0 +1,28 @@
class Solution {
public:
int numberOfSubstrings(string s) {
int n = s.size();
vector<int> nxt(n + 1);
nxt[n] = n;
for (int i = n - 1; i >= 0; --i) {
nxt[i] = nxt[i + 1];
if (s[i] == '0') {
nxt[i] = i;
}
}
int ans = 0;
for (int i = 0; i < n; ++i) {
int cnt0 = s[i] == '0' ? 1 : 0;
int j = i;
while (j < n && 1LL * cnt0 * cnt0 <= n) {
int cnt1 = nxt[j + 1] - i - cnt0;
if (cnt1 >= cnt0 * cnt0) {
ans += min(nxt[j + 1] - j, cnt1 - cnt0 * cnt0 + 1);
}
j = nxt[j + 1];
++cnt0;
}
}
return ans;
}
};
Loading