feat: add solutions to lc problem: No.3307

No.3307.Find the K-th Character in String Game II

Author: yanglbme

solution/3300-3399/3307.Find the K-th Character in String Game II/README.md

@@ -86,32 +86,138 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3307.Fi
 
 <!-- solution:start -->
 
-### 方法一
+### 方法一：递推
+
+由于每次操作后，字符串的长度都会翻倍，因此，如果进行 $i$ 次操作，字符串的长度将会是 $2^i$。
+
+我们可以模拟这个过程，找到第一个大于等于 $k$ 的字符串长度 $n$。
+
+接下来，我们再往回推，分情况讨论：
+
+-   如果 $k \gt n / 2$，说明 $k$ 在后半部分，如果此时 $\textit{operations}[i - 1] = 1$，说明 $k$ 所在的字符是由前半部分的字符加上 $1$ 得到的，我们加上 $1$。然后我们更新 $k$ 为 $k - n / 2$。
+-   如果 $k \le n / 2$，说明 $k$ 在前半部分，不会受到 $\textit{operations}[i - 1]$ 的影响。
+-   接下来，我们更新 $n$ 为 $n / 2$，继续往前推，直到 $n = 1$。
+
+最后，我们将得到的数字对 $26$ 取模，加上 `'a'` 的 ASCII 码，即可得到答案。
+
+时间复杂度 $O(\log k)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def kthCharacter(self, k: int, operations: List[int]) -> str:
+        n, i = 1, 0
+        while n < k:
+            n *= 2
+            i += 1
+        d = 0
+        while n > 1:
+            if k > n // 2:
+                k -= n // 2
+                d += operations[i - 1]
+            n //= 2
+            i -= 1
+        return chr(d % 26 + ord("a"))
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public char kthCharacter(long k, int[] operations) {
+        long n = 1;
+        int i = 0;
+        while (n < k) {
+            n *= 2;
+            ++i;
+        }
+        int d = 0;
+        while (n > 1) {
+            if (k > n / 2) {
+                k -= n / 2;
+                d += operations[i - 1];
+            }
+            n /= 2;
+            --i;
+        }
+        return (char) ('a' + (d % 26));
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    char kthCharacter(long long k, vector<int>& operations) {
+        long long n = 1;
+        int i = 0;
+        while (n < k) {
+            n *= 2;
+            ++i;
+        }
+        int d = 0;
+        while (n > 1) {
+            if (k > n / 2) {
+                k -= n / 2;
+                d += operations[i - 1];
+            }
+            n /= 2;
+            --i;
+        }
+        return 'a' + (d % 26);
+    }
+};
 ```
 
 #### Go
 
 ```go
+func kthCharacter(k int64, operations []int) byte {
+	n := int64(1)
+	i := 0
+	for n < k {
+		n *= 2
+		i++
+	}
+	d := 0
+	for n > 1 {
+		if k > n/2 {
+			k -= n / 2
+			d += operations[i-1]
+		}
+		n /= 2
+		i--
+	}
+	return byte('a' + (d % 26))
+}
+```
 
+#### TypeScript
+
+```ts
+function kthCharacter(k: number, operations: number[]): string {
+    let n = 1;
+    let i = 0;
+    while (n < k) {
+        n *= 2;
+        i++;
+    }
+    let d = 0;
+    while (n > 1) {
+        if (k > n / 2) {
+            k -= n / 2;
+            d += operations[i - 1];
+        }
+        n /= 2;
+        i--;
+    }
+    return String.fromCharCode('a'.charCodeAt(0) + (d % 26));
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3307.Find the K-th Character in String Game II/README_EN.md

@@ -83,32 +83,138 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3300-3399/3307.Fi
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Recurrence
+
+Since the length of the string doubles after each operation, if we perform $i$ operations, the length of the string will be $2^i$.
+
+We can simulate this process to find the first string length $n$ that is greater than or equal to $k$.
+
+Next, we backtrack and discuss the following cases:
+
+-   If $k \gt n / 2$, it means $k$ is in the second half. If $\textit{operations}[i - 1] = 1$, it means the character at position $k$ is obtained by adding $1$ to the character in the first half. We add $1$ to it. Then we update $k$ to $k - n / 2$.
+-   If $k \le n / 2$, it means $k$ is in the first half and is not affected by $\textit{operations}[i - 1]$.
+-   Next, we update $n$ to $n / 2$ and continue backtracking until $n = 1$.
+
+Finally, we take the resulting number modulo $26$ and add the ASCII code of `'a'` to get the answer.
+
+The time complexity is $O(\log k)$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
 #### Python3
 
 ```python
-
+class Solution:
+    def kthCharacter(self, k: int, operations: List[int]) -> str:
+        n, i = 1, 0
+        while n < k:
+            n *= 2
+            i += 1
+        d = 0
+        while n > 1:
+            if k > n // 2:
+                k -= n // 2
+                d += operations[i - 1]
+            n //= 2
+            i -= 1
+        return chr(d % 26 + ord("a"))
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public char kthCharacter(long k, int[] operations) {
+        long n = 1;
+        int i = 0;
+        while (n < k) {
+            n *= 2;
+            ++i;
+        }
+        int d = 0;
+        while (n > 1) {
+            if (k > n / 2) {
+                k -= n / 2;
+                d += operations[i - 1];
+            }
+            n /= 2;
+            --i;
+        }
+        return (char) ('a' + (d % 26));
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    char kthCharacter(long long k, vector<int>& operations) {
+        long long n = 1;
+        int i = 0;
+        while (n < k) {
+            n *= 2;
+            ++i;
+        }
+        int d = 0;
+        while (n > 1) {
+            if (k > n / 2) {
+                k -= n / 2;
+                d += operations[i - 1];
+            }
+            n /= 2;
+            --i;
+        }
+        return 'a' + (d % 26);
+    }
+};
 ```
 
 #### Go
 
 ```go
+func kthCharacter(k int64, operations []int) byte {
+	n := int64(1)
+	i := 0
+	for n < k {
+		n *= 2
+		i++
+	}
+	d := 0
+	for n > 1 {
+		if k > n/2 {
+			k -= n / 2
+			d += operations[i-1]
+		}
+		n /= 2
+		i--
+	}
+	return byte('a' + (d % 26))
+}
+```
 
+#### TypeScript
+
+```ts
+function kthCharacter(k: number, operations: number[]): string {
+    let n = 1;
+    let i = 0;
+    while (n < k) {
+        n *= 2;
+        i++;
+    }
+    let d = 0;
+    while (n > 1) {
+        if (k > n / 2) {
+            k -= n / 2;
+            d += operations[i - 1];
+        }
+        n /= 2;
+        i--;
+    }
+    return String.fromCharCode('a'.charCodeAt(0) + (d % 26));
+}
 ```
 
 <!-- tabs:end -->

solution/3300-3399/3307.Find the K-th Character in String Game II/Solution.cpp

@@ -0,0 +1,21 @@
+class Solution {
+public:
+    char kthCharacter(long long k, vector<int>& operations) {
+        long long n = 1;
+        int i = 0;
+        while (n < k) {
+            n *= 2;
+            ++i;
+        }
+        int d = 0;
+        while (n > 1) {
+            if (k > n / 2) {
+                k -= n / 2;
+                d += operations[i - 1];
+            }
+            n /= 2;
+            --i;
+        }
+        return 'a' + (d % 26);
+    }
+};

solution/3300-3399/3307.Find the K-th Character in String Game II/Solution.go

@@ -0,0 +1,18 @@
+func kthCharacter(k int64, operations []int) byte {
+	n := int64(1)
+	i := 0
+	for n < k {
+		n *= 2
+		i++
+	}
+	d := 0
+	for n > 1 {
+		if k > n/2 {
+			k -= n / 2
+			d += operations[i-1]
+		}
+		n /= 2
+		i--
+	}
+	return byte('a' + (d % 26))
+}

solution/3300-3399/3307.Find the K-th Character in String Game II/Solution.java

@@ -0,0 +1,20 @@
+class Solution {
+    public char kthCharacter(long k, int[] operations) {
+        long n = 1;
+        int i = 0;
+        while (n < k) {
+            n *= 2;
+            ++i;
+        }
+        int d = 0;
+        while (n > 1) {
+            if (k > n / 2) {
+                k -= n / 2;
+                d += operations[i - 1];
+            }
+            n /= 2;
+            --i;
+        }
+        return (char) ('a' + (d % 26));
+    }
+}

solution/3300-3399/3307.Find the K-th Character in String Game II/Solution.py

@@ -0,0 +1,14 @@
+class Solution:
+    def kthCharacter(self, k: int, operations: List[int]) -> str:
+        n, i = 1, 0
+        while n < k:
+            n *= 2
+            i += 1
+        d = 0
+        while n > 1:
+            if k > n // 2:
+                k -= n // 2
+                d += operations[i - 1]
+            n //= 2
+            i -= 1
+        return chr(d % 26 + ord("a"))

solution/3300-3399/3307.Find the K-th Character in String Game II/Solution.ts

@@ -0,0 +1,18 @@
+function kthCharacter(k: number, operations: number[]): string {
+    let n = 1;
+    let i = 0;
+    while (n < k) {
+        n *= 2;
+        i++;
+    }
+    let d = 0;
+    while (n > 1) {
+        if (k > n / 2) {
+            k -= n / 2;
+            d += operations[i - 1];
+        }
+        n /= 2;
+        i--;
+    }
+    return String.fromCharCode('a'.charCodeAt(0) + (d % 26));
+}