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Dec 8, 2023
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feat: add solutions to lc problem: No.1545 #2071

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112 changes: 111 additions & 1 deletion solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README.md
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Expand Up @@ -71,22 +71,132 @@

<!-- 这里可写通用的实现逻辑 -->

**方法一：分类讨论 + 递归**

我们可以发现，对于 $S_n$，其前半部分和 $S_{n-1}$ 是一样的，而后半部分是 $S_{n-1}$ 的反转取反。因此我们可以设计一个函数 $dfs(n, k)$，表示第 $n$ 个字符串的第 $k$ 位字符。答案即为 $dfs(n, k)$。

函数 $dfs(n, k)$ 的计算过程如下：

- 如果 $k = 1$，那么答案为 $0$；
- 如果 $k$ 是 $2$ 的幂次方，那么答案为 $1$；
- 如果 $k \times 2 \lt 2^n - 1$，说明 $k$ 在前半部分，答案为 $dfs(n - 1, k)$；
- 否则，答案为 $dfs(n - 1, 2^n - k) \oplus 1$，其中 $\oplus$ 表示异或运算。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为题目给定的 $n$。

<!-- tabs:start -->

### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python

class Solution:
def findKthBit(self, n: int, k: int) -> str:
def dfs(n: int, k: int) -> int:
if k == 1:
return 0
if (k & (k - 1)) == 0:
return 1
m = 1 << n
if k * 2 < m - 1:
return dfs(n - 1, k)
return dfs(n - 1, m - k) ^ 1

return str(dfs(n, k))
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```java
class Solution {
public char findKthBit(int n, int k) {
return (char) ('0' + dfs(n, k));
}

private int dfs(int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
}
}
```

### **C++**

```cpp
class Solution {
public:
char findKthBit(int n, int k) {
function<int(int, int)> dfs = [&](int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return '0' + dfs(n, k);
}
};
```

### **Go**

```go
func findKthBit(n int, k int) byte {
var dfs func(n, k int) int
dfs = func(n, k int) int {
if k == 1 {
return 0
}
if k&(k-1) == 0 {
return 1
}
m := 1 << n
if k*2 < m-1 {
return dfs(n-1, k)
}
return dfs(n-1, m-k) ^ 1
}
return byte('0' + dfs(n, k))
}
```

### **TypeScript**

```ts
function findKthBit(n: number, k: number): string {
const dfs = (n: number, k: number): number => {
if (k === 1) {
return 0;
}
if ((k & (k - 1)) === 0) {
return 1;
}
const m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return dfs(n, k).toString();
}
```

### **...**
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112 changes: 111 additions & 1 deletion solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README_EN.md
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Expand Up @@ -53,18 +53,128 @@ The 11<sup>th</sup> bit is &quot;1&quot;.

## Solutions

**Solution 1: Case Analysis + Recursion**

We can observe that for $S_n$, the first half is the same as $S_{n-1}$, and the second half is the reverse and negation of $S_{n-1}$. Therefore, we can design a function $dfs(n, k)$, which represents the $k$-th character of the $n$-th string. The answer is $dfs(n, k)$.

The calculation process of the function $dfs(n, k)$ is as follows:

- If $k = 1$, then the answer is $0$;
- If $k$ is a power of $2$, then the answer is $1$;
- If $k \times 2 < 2^n - 1$, it means that $k$ is in the first half, and the answer is $dfs(n - 1, k)$;
- Otherwise, the answer is $dfs(n - 1, 2^n - k) \oplus 1$, where $\oplus$ represents the XOR operation.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the given $n$ in the problem.

<!-- tabs:start -->

### **Python3**

```python

class Solution:
def findKthBit(self, n: int, k: int) -> str:
def dfs(n: int, k: int) -> int:
if k == 1:
return 0
if (k & (k - 1)) == 0:
return 1
m = 1 << n
if k * 2 < m - 1:
return dfs(n - 1, k)
return dfs(n - 1, m - k) ^ 1

return str(dfs(n, k))
```

### **Java**

```java
class Solution {
public char findKthBit(int n, int k) {
return (char) ('0' + dfs(n, k));
}

private int dfs(int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
}
}
```

### **C++**

```cpp
class Solution {
public:
char findKthBit(int n, int k) {
function<int(int, int)> dfs = [&](int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return '0' + dfs(n, k);
}
};
```

### **Go**

```go
func findKthBit(n int, k int) byte {
var dfs func(n, k int) int
dfs = func(n, k int) int {
if k == 1 {
return 0
}
if k&(k-1) == 0 {
return 1
}
m := 1 << n
if k*2 < m-1 {
return dfs(n-1, k)
}
return dfs(n-1, m-k) ^ 1
}
return byte('0' + dfs(n, k))
}
```

### **TypeScript**

```ts
function findKthBit(n: number, k: number): string {
const dfs = (n: number, k: number): number => {
if (k === 1) {
return 0;
}
if ((k & (k - 1)) === 0) {
return 1;
}
const m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return dfs(n, k).toString();
}
```

### **...**
Expand Down
19 changes: 19 additions & 0 deletions solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.cpp
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@@ -0,0 +1,19 @@
class Solution {
public:
char findKthBit(int n, int k) {
function<int(int, int)> dfs = [&](int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return '0' + dfs(n, k);
}
};
17 changes: 17 additions & 0 deletions solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.go
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Original file line number Diff line number Diff line change
@@ -0,0 +1,17 @@
func findKthBit(n int, k int) byte {
var dfs func(n, k int) int
dfs = func(n, k int) int {
if k == 1 {
return 0
}
if k&(k-1) == 0 {
return 1
}
m := 1 << n
if k*2 < m-1 {
return dfs(n-1, k)
}
return dfs(n-1, m-k) ^ 1
}
return byte('0' + dfs(n, k))
}
55 changes: 18 additions & 37 deletions solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.java
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Original file line number Diff line number Diff line change
@@ -1,38 +1,19 @@
class Solution {
public char findKthBit(int n, int k) {
if (k == 1 || n == 1) {
return '0';
}
Set<Integer> set = new HashSet<>();
int len = calcLength(n, set);
if (set.contains(k)) {
return '1';
}
// 中间，返回1
if (k < len / 2) {
return findKthBit(n - 1, k);
} else {
if (set.contains(len - k)) {
return '1';
}
return r(findKthBit(n - 1, len - k + 1));
}
}

private char r(char b) {
if (b == '0') {
return '1';
}
return '0';
}

private int calcLength(int n, Set<Integer> set) {
if (n == 1) {
return 1;
}

int ans = 2 * calcLength(n - 1, set) + 1;
set.add(ans + 1);
return ans;
}
class Solution {
public char findKthBit(int n, int k) {
return (char) ('0' + dfs(n, k));
}

private int dfs(int n, int k) {
if (k == 1) {
return 0;
}
if ((k & (k - 1)) == 0) {
return 1;
}
int m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
}
}
13 changes: 13 additions & 0 deletions solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.py
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Original file line number Diff line number Diff line change
@@ -0,0 +1,13 @@
class Solution:
def findKthBit(self, n: int, k: int) -> str:
def dfs(n: int, k: int) -> int:
if k == 1:
return 0
if (k & (k - 1)) == 0:
return 1
m = 1 << n
if k * 2 < m - 1:
return dfs(n - 1, k)
return dfs(n - 1, m - k) ^ 1

return str(dfs(n, k))
16 changes: 16 additions & 0 deletions solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.ts
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@@ -0,0 +1,16 @@
function findKthBit(n: number, k: number): string {
const dfs = (n: number, k: number): number => {
if (k === 1) {
return 0;
}
if ((k & (k - 1)) === 0) {
return 1;
}
const m = 1 << n;
if (k * 2 < m - 1) {
return dfs(n - 1, k);
}
return dfs(n - 1, m - k) ^ 1;
};
return dfs(n, k).toString();
}
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