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feat: add solutions to lc problem: No.1545 (doocs#2071)
No.1545.Find Kth Bit in Nth Binary String
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‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README.md

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Original file line numberDiff line numberDiff line change
@@ -71,22 +71,132 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：分类讨论 + 递归**
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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 $dfs(n, k)$。
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函数 $dfs(n, k)$ 的计算过程如下：
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- 如果 $k = 1$，那么答案为 $0$；
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- 如果 $k$ 是 $2$ 的幂次方，那么答案为 $1$；
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- 如果 $k \times 2 \lt 2^n - 1$，说明 $k$ 在前半部分，答案为 $dfs(n - 1, k)$；
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- 否则，答案为 $dfs(n - 1, 2^n - k) \oplus 1$，其中 $\oplus$ 表示异或运算。
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时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为题目给定的 $n$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findKthBit(self, n: int, k: int) -> str:
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def dfs(n: int, k: int) -> int:
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if k == 1:
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return 0
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if (k & (k - 1)) == 0:
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return 1
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m = 1 << n
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if k * 2 < m - 1:
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return dfs(n - 1, k)
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return dfs(n - 1, m - k) ^ 1
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return str(dfs(n, k))
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public char findKthBit(int n, int k) {
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return (char) ('0' + dfs(n, k));
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}
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private int dfs(int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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char findKthBit(int n, int k) {
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function<int(int, int)> dfs = [&](int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return '0' + dfs(n, k);
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}
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};
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```
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### **Go**
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```go
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func findKthBit(n int, k int) byte {
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var dfs func(n, k int) int
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dfs = func(n, k int) int {
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if k == 1 {
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return 0
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}
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if k&(k-1) == 0 {
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return 1
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}
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m := 1 << n
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if k*2 < m-1 {
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return dfs(n-1, k)
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}
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return dfs(n-1, m-k) ^ 1
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}
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return byte('0' + dfs(n, k))
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}
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```
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### **TypeScript**
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```ts
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function findKthBit(n: number, k: number): string {
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const dfs = (n: number, k: number): number => {
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if (k === 1) {
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return 0;
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}
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if ((k & (k - 1)) === 0) {
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return 1;
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}
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const m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return dfs(n, k).toString();
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}
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```
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### **...**

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/README_EN.md

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@@ -53,18 +53,128 @@ The 11<sup>th</sup> bit is &quot;1&quot;.
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## Solutions
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**Solution 1: Case Analysis + Recursion**
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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 $dfs(n, k)$.
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The calculation process of the function $dfs(n, k)$ is as follows:
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- If $k = 1$, then the answer is $0$;
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- If $k$ is a power of $2$, then the answer is $1$;
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- If $k \times 2 < 2^n - 1$, it means that $k$ is in the first half, and the answer is $dfs(n - 1, k)$;
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- Otherwise, the answer is $dfs(n - 1, 2^n - k) \oplus 1$, where $\oplus$ represents the XOR operation.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the given $n$ in the problem.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findKthBit(self, n: int, k: int) -> str:
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def dfs(n: int, k: int) -> int:
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if k == 1:
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return 0
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if (k & (k - 1)) == 0:
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return 1
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m = 1 << n
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if k * 2 < m - 1:
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return dfs(n - 1, k)
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return dfs(n - 1, m - k) ^ 1
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return str(dfs(n, k))
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```
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### **Java**
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```java
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class Solution {
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public char findKthBit(int n, int k) {
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return (char) ('0' + dfs(n, k));
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}
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private int dfs(int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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char findKthBit(int n, int k) {
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function<int(int, int)> dfs = [&](int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return '0' + dfs(n, k);
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}
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};
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```
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### **Go**
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```go
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func findKthBit(n int, k int) byte {
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var dfs func(n, k int) int
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dfs = func(n, k int) int {
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if k == 1 {
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return 0
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}
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if k&(k-1) == 0 {
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return 1
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}
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m := 1 << n
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if k*2 < m-1 {
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return dfs(n-1, k)
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}
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return dfs(n-1, m-k) ^ 1
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}
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return byte('0' + dfs(n, k))
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}
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```
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### **TypeScript**
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```ts
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function findKthBit(n: number, k: number): string {
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const dfs = (n: number, k: number): number => {
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if (k === 1) {
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return 0;
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}
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if ((k & (k - 1)) === 0) {
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return 1;
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}
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const m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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};
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return dfs(n, k).toString();
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}
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```
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### **...**

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.cpp

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

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.go

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

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.java

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1-
class Solution {
2-
public char findKthBit(int n, int k) {
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if (k == 1 || n == 1) {
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return '0';
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}
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Set<Integer> set = new HashSet<>();
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int len = calcLength(n, set);
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if (set.contains(k)) {
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return '1';
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}
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// 中间，返回1
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if (k < len / 2) {
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return findKthBit(n - 1, k);
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} else {
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if (set.contains(len - k)) {
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return '1';
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}
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return r(findKthBit(n - 1, len - k + 1));
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}
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}
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private char r(char b) {
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if (b == '0') {
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return '1';
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}
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return '0';
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}
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private int calcLength(int n, Set<Integer> set) {
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if (n == 1) {
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return 1;
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}
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int ans = 2 * calcLength(n - 1, set) + 1;
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set.add(ans + 1);
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return ans;
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}
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class Solution {
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public char findKthBit(int n, int k) {
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return (char) ('0' + dfs(n, k));
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}
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private int dfs(int n, int k) {
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if (k == 1) {
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return 0;
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}
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if ((k & (k - 1)) == 0) {
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return 1;
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}
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int m = 1 << n;
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if (k * 2 < m - 1) {
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return dfs(n - 1, k);
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}
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return dfs(n - 1, m - k) ^ 1;
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}
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}

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.py

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

‎solution/1500-1599/1545.Find Kth Bit in Nth Binary String/Solution.ts

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

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