Merge pull request #1268 from shreyash3087/add/leetcode-541

Docs: Added Solutions to Leetcode 541

Author: ajay-dhangar

dsa-problems/leetcode-problems/0500-0599.md

@@ -242,7 +242,7 @@ export const problems =[
     "problemName": "541. Reverse String II",
     "difficulty": "Easy",
     "leetCodeLink": "https://leetcode.com/problems/reverse-string-ii",
-    "solutionLink": "#"
+    "solutionLink": "/dsa-solutions/lc-solutions/0500-0599/reverse-string-ii"
   },
   {
     "problemName": "542. 01 Matrix",

dsa-solutions/lc-solutions/0500-0599/0541-reverse-string-ii.md

@@ -0,0 +1,127 @@
+---
+id: reverse-string-ii
+title: Reverse String II
+sidebar_label: 0541 - Reverse String II
+tags:
+  - String
+  - Two Pointers
+  - Recursion
+description: "This is a solution to the Reverse String II problem on LeetCode."
+---
+
+## Problem Description
+
+Given a string `s` and an integer `k`, reverse the first `k` characters for every `2k` characters counting from the start of the string.
+
+If there are fewer than `k` characters left, reverse all of them. If there are less than `2k` but greater than or equal to `k` characters, then reverse the first `k` characters and leave the other as original.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: s = "abcdefg", k = 2
+Output: "bacdfeg"
+```
+
+**Example 2:**
+
+```
+Input: s = "abcd", k = 2
+Output: "bacd"
+```
+
+### Constraints
+
+- $1 \leq s.length \leq 10^4$
+- `s` consists of only lowercase English letters.
+- $1 \leq k \leq 10^4$
+
+## Solution for Reverse String II
+
+### Approach 
+#### Intuition and Algorithm
+
+We will reverse each block of 2k characters directly.
+
+Each block starts at a multiple of 2k: for example, 0, 2k, 4k, 6k, .... One thing to be careful about is we may not reverse each block if there aren't enough characters.
+
+To reverse a block of characters from i to j, we can swap characters in positions i++ and j--.
+
+## Code in Different Languages
+
+<Tabs>
+<TabItem value="cpp" label="C++">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```cpp
+#include <iostream>
+#include <string>
+#include <algorithm>
+
+class Solution {
+public:
+    std::string reverseStr(std::string s, int k) {
+        for (int start = 0; start < s.length(); start += 2 * k) {
+            int i = start;
+            int j = std::min(start + k - 1, static_cast<int>(s.length()) - 1);
+            while (i < j) {
+                std::swap(s[i++], s[j--]);
+            }
+        }
+        return s;
+    }
+};
+
+```
+</TabItem>
+<TabItem value="java" label="Java">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```java
+class Solution {
+    public String reverseStr(String s, int k) {
+        char[] a = s.toCharArray();
+        for (int start = 0; start < a.length; start += 2 * k) {
+            int i = start, j = Math.min(start + k - 1, a.length - 1);
+            while (i < j) {
+                char tmp = a[i];
+                a[i++] = a[j];
+                a[j--] = tmp;
+            }
+        }
+        return new String(a);
+    }
+}
+```
+
+</TabItem>
+<TabItem value="python" label="Python">
+  <SolutionAuthor name="@Shreyash3087"/>
+
+```python
+class Solution(object):
+    def reverseStr(self, s, k):
+        a = list(s)
+        for i in xrange(0, len(a), 2*k):
+            a[i:i+k] = reversed(a[i:i+k])
+        return "".join(a)
+```
+</TabItem>
+</Tabs>
+
+## Complexity Analysis
+
+### Time Complexity: $O(N)$
+
+> **Reason**:  where N is the size of `s`. We build a helper array, plus reverse about half the characters in `s`.
+
+### Space Complexity: $O(N)$
+
+> **Reason**: the size of `a`.
+
+## References
+
+- **LeetCode Problem**: [Reverse String II](https://leetcode.com/problems/reverse-string-ii/description/)
+
+- **Solution Link**: [Reverse String II](https://leetcode.com/problems/reverse-string-ii/solutions/)