Merge pull request #546 from manishh12/shuffle-string

Added Shuffle String Solution

Author: ajay-dhangar

dsa-solutions/lc-solutions/1500-1599/1528-Shuffle-String.md

@@ -0,0 +1,332 @@
+---
+
+id: shuffle-string
+title: Shuffle String Solution
+sidebar_label: 1528-Shuffle-String
+tags:
+  - Array
+  - String
+  - LeetCode
+  - JavaScript
+  - TypeScript
+  - Python
+  - Java
+  - C++
+description: "This is a solution to the Shuffle String problem on LeetCode."
+
+---
+
+In this page, we will solve the Shuffle String problem using different approaches: simple iteration and an optimized version using map. We will provide the implementation of the solution in JavaScript, TypeScript, Python, Java, C++, and more.
+
+## Problem Description
+
+You are given a string `s` and an integer array `indices` of the same length. The string `s` will be shuffled such that the character at the `i`th position moves to `indices[i]` in the shuffled string.
+
+Return the shuffled string.
+
+### Examples
+
+**Example 1:**
+
+```plaintext
+Input: s = "codeleet", indices = [4,5,6,7,0,2,1,3]
+Output: "leetcode"
+Explanation: As shown, "codeleet" becomes "leetcode" after shuffling.
+```
+
+**Example 2:**
+
+```plaintext
+Input: s = "abc", indices = [0,1,2]
+Output: "abc"
+Explanation: After shuffling, each character remains in its position.
+```
+
+### Constraints
+
+- `s.length == indices.length == n`
+- `1 <= n <= 100`
+- `s` consists of only lowercase English letters.
+- `0 <= indices[i] < n`
+- All values of `indices` are unique.
+
+---
+
+## Solution for Shuffle String Problem
+
+### Intuition and Approach
+
+The problem can be solved by creating a new array for the shuffled string, placing each character at the position specified by the `indices` array. We will demonstrate a simple iteration approach and an optimized version using map.
+
+<Tabs>
+ <tabItem value="Simple Iteration" label="Simple Iteration">
+
+### Approach 1: Simple Iteration
+
+We iterate through the string and place each character at the position specified by the `indices` array.
+
+#### Implementation
+
+```jsx live
+function shuffleString() {
+  const s = "codeleet";
+  const indices = [4, 5, 6, 7, 0, 2, 1, 3];
+
+  const restoreString = function(s, indices) {
+    let shuffled = new Array(s.length);
+    for (let i = 0; i < s.length; i++) {
+      shuffled[indices[i]] = s[i];
+    }
+    return shuffled.join('');
+  };
+
+  const result = restoreString(s, indices);
+  return (
+    <div>
+      <p>
+        <b>Input:</b> s = "{s}", indices = {JSON.stringify(indices)}
+      </p>
+      <p>
+        <b>Output:</b> {result}
+      </p>
+    </div>
+  );
+}
+```
+
+#### Code in Different Languages
+
+<Tabs>
+  <TabItem value="JavaScript" label="JavaScript" default>
+  <SolutionAuthor name="@manishh12"/>
+   ```javascript
+    function restoreString(s, indices) {
+      let shuffled = new Array(s.length);
+      for (let i = 0; i < s.length; i++) {
+        shuffled[indices[i]] = s[i];
+      }
+      return shuffled.join('');
+    }
+    ```
+
+  </TabItem>
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@manishh12"/>
+   ```typescript
+    function restoreString(s: string, indices: number[]): string {
+      let shuffled: string[] = new Array(s.length);
+      for (let i = 0; i < s.length; i++) {
+        shuffled[indices[i]] = s[i];
+      }
+      return shuffled.join('');
+    }
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python">
+  <SolutionAuthor name="@manishh12"/>
+   ```python
+    class Solution:
+        def restoreString(self, s: str, indices: List[int]) -> str:
+            shuffled = [''] * len(s)
+            for i, index in enumerate(indices):
+                shuffled[index] = s[i]
+            return ''.join(shuffled)
+    ```
+
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@manishh12"/>
+   ```java
+    class Solution {
+        public String restoreString(String s, int[] indices) {
+            char[] shuffled = new char[s.length()];
+            for (int i = 0; i < s.length(); i++) {
+                shuffled[indices[i]] = s.charAt(i);
+            }
+            return new String(shuffled);
+        }
+    }
+    ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@manishh12"/>
+   ```cpp
+    class Solution {
+    public:
+        string restoreString(string s, vector<int>& indices) {
+            string shuffled(s.length(), ' ');
+            for (int i = 0; i < s.length(); i++) {
+                shuffled[indices[i]] = s[i];
+            }
+            return shuffled;
+        }
+    };
+    ```
+
+  </TabItem>  
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $$O(n)$$, where `n` is the length of the string.
+- Space Complexity: $$O(n)$$, as we are using an additional array to store the shuffled string.
+
+</tabItem>
+
+<tabItem value="Using Map" label="Using Map">
+
+### Approach 2: Using Map
+
+We can use a map to store the characters with their respective indices and then reconstruct the shuffled string.
+
+#### Implementation
+
+```jsx live
+function shuffleString() {
+  const s = "codeleet";
+  const indices = [4, 5, 6, 7, 0, 2, 1, 3];
+
+  const restoreString = function(s, indices) {
+    let map = new Map();
+    for (let i = 0; i < s.length; i++) {
+      map.set(indices[i], s[i]);
+    }
+    let shuffled = '';
+    for (let i = 0; i < s.length; i++) {
+      shuffled += map.get(i);
+    }
+    return shuffled;
+  };
+
+  const result = restoreString(s, indices);
+  return (
+    <div>
+      <p>
+        <b>Input:</b> s = "{s}", indices = {JSON.stringify(indices)}
+      </p>
+      <p>
+        <b>Output:</b> {result}
+      </p>
+    </div>
+  );
+}
+```
+
+#### Code in Different Languages
+
+<Tabs>
+  <TabItem value="JavaScript" label="JavaScript" default>
+  <SolutionAuthor name="@manishh12"/>
+   ```javascript
+    function restoreString(s, indices) {
+      let map = new Map();
+      for (let i = 0; i < s.length; i++) {
+        map.set(indices[i], s[i]);
+      }
+      let shuffled = '';
+      for (let i = 0; i < s.length; i++) {
+        shuffled += map.get(i);
+      }
+      return shuffled;
+    }
+    ```
+
+  </TabItem>
+  <TabItem value="TypeScript" label="TypeScript">
+  <SolutionAuthor name="@manishh12"/>
+   ```typescript
+    function restoreString(s: string, indices: number[]): string {
+      let map: Map<number, string> = new Map();
+      for (let i = 0; i < s.length; i++) {
+        map.set(indices[i], s[i]);
+      }
+      let shuffled: string = '';
+      for (let i = 0; i < s.length; i++) {
+        shuffled += map.get(i);
+      }
+      return shuffled;
+    }
+    ```
+
+  </TabItem>
+  <TabItem value="Python" label="Python">
+  <SolutionAuthor name="@manishh12"/>
+   ```python
+    class Solution:
+        def restoreString(self, s: str, indices: List[int]) -> str:
+            map = {indices[i]: s[i] for i in range(len(s))}
+            shuffled = ''.join(map[i] for i in range(len(s)))
+            return shuffled
+    ```
+
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@manishh12"/>
+   ```java
+    import java.util.HashMap;
+    import java.util.Map;
+
+    class Solution {
+        public String restoreString(String s, int[] indices) {
+            Map<Integer, Character> map = new HashMap<>();
+            for (int i = 0; i < s.length(); i++) {
+                map.put(indices[i], s.charAt(i));
+            }
+            StringBuilder shuffled = new StringBuilder();
+            for (int i = 0; i < s.length(); i++) {
+                shuffled.append(map.get(i));
+            }
+            return shuffled.toString();
+        }
+    }
+    ```
+
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@manishh12"/>
+   ```cpp
+    #include <unordered_map>
+    #include <string>
+
+    class Solution {
+    public:
+        string restoreString(string s, vector<int>& indices) {
+            unordered_map<int, char> map;
+            for (int i = 0; i < s.length(); i++) {
+                map[indices[i]] = s[i];
+            }
+            string shuffled(s.length(), ' ');
+            for (int i = 0; i < s.length(); i++) {
+                shuffled[i] = map[i];
+            }
+            return shuffled;
+        }
+    };
+    ```
+
+  </TabItem>  
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $$O(n)$$, where `n` is the length of the string.
+- Space Complexity: $$O(n)$$, as we are using a map to store the characters.
+
+</tabItem>
+</Tabs>
+
+:::tip Note
+
+By using both simple iteration and map-based approaches, we can efficiently solve the Shuffle String problem. The choice between the two approaches depends on the specific requirements and constraints of the problem.
+
+:::
+
+## References
+
+- **LeetCode Problem:** [Shuffle String](https://leetcode.com/problems/shuffle-string/)
+- **Solution Link:** [Shuffle String Solution on LeetCode](https://leetcode.com/problems/shuffle-string/solution/)
+- **Authors LeetCode Profile:** [Manish Kumar Gupta](https://leetcode.com/_manishh12/)
+
+