Added lc solution for 214

dsa-solutions/lc-solutions/0200-0299/0214-Shortest-Palindrome.md

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+---
+id: shortest-palindrome
+title: Shortest Palindrome
+sidebar_label: 0214 Shortest Palindrome
+tags:
+- String
+- KMP Algorithm
+- C++
+- Java
+- Python
+description: "This document provides a solution to finding the shortest palindrome by adding characters in front of the given string."
+---
+
+## Problem
+Given a string `s`, you can convert it to a palindrome by adding characters in front of it. Return the shortest palindrome you can find by performing this transformation.
+
+### Example 1:
+Input: s = "aacecaaa"  
+Output: "aaacecaaa"
+
+### Example 2:
+Input: s = "abcd"  
+Output: "dcbabcd"
+
+### Constraints:
+- $0 \leq \text{s.length} \leq 5 \times 10^4$
+- s consists of lowercase English letters only.
+
+## Solution
+To solve this problem efficiently, we can leverage the KMP (Knuth-Morris-Pratt) algorithm to find the longest prefix of the reversed string that matches the suffix of the original string. This allows us to determine the shortest palindrome.
+
+### Steps:
+1. Reverse the string `s` to get `rev_s`.
+2. Construct a new string `new_s = s + "#" + rev_s`.
+3. Compute the KMP table (prefix function) for `new_s` to find the longest prefix which is also a suffix.
+4. The length of this longest prefix gives us the length of characters from `rev_s` that we need to prepend to `s` to form the shortest palindrome.
+
+#### Code in Different Languages
+
+<Tabs>
+  <TabItem value="Python" label="Python" default>
+  <SolutionAuthor name="@mahek0620"/>
+   ```python
+
+    class Solution(object):
+    def shortestPalindrome(self, s):
+        """
+        :type s: str
+        :rtype: str
+        """
+        if not s:
+            return ""
+
+        # Step 1: Reverse the string
+        rev_s = s[::-1]
+
+        # Step 2: Combine s and rev_s with a separator
+        new_s = s + "#" + rev_s
+
+        # Step 3: Compute KMP table (prefix function) for new_s
+        n = len(new_s)
+        kmp = [0] * n
+
+        for i in range(1, n):
+            j = kmp[i - 1]
+            while j > 0 and new_s[i] != new_s[j]:
+                j = kmp[j - 1]
+            if new_s[i] == new_s[j]:
+                j += 1
+            kmp[i] = j
+
+        # Step 4: Calculate the overlap length
+        overlap = kmp[-1]
+
+        # Step 5: Form the result palindrome
+        return rev_s[:len(s) - overlap] + s
+
+    ```
+  </TabItem>
+  <TabItem value="Java" label="Java">
+  <SolutionAuthor name="@mahek0620"/>
+   ```java
+
+    class Solution {
+    public String shortestPalindrome(String s) {
+        String rev_s = new StringBuilder(s).reverse().toString();
+
+        // Form the new string to check
+        String new_s = s + "#" + rev_s;
+        int n = new_s.length();
+
+        // Compute the KMP table (prefix function)
+        int[] kmp = new int[n];
+        for (int i = 1, j = 0; i < n; ++i) {
+            while (j > 0 && new_s.charAt(i) != new_s.charAt(j)) {
+                j = kmp[j - 1];
+            }
+            if (new_s.charAt(i) == new_s.charAt(j)) {
+                ++j;
+            }
+            kmp[i] = j;
+        }
+
+        // The length of the longest prefix which is also suffix
+        int overlap = kmp[n - 1];
+
+        // Construct the result
+        return rev_s.substring(0, rev_s.length() - overlap) + s;
+    }
+    }
+
+    ```
+  </TabItem>
+  <TabItem value="C++" label="C++">
+  <SolutionAuthor name="@mahek0620"/>
+   ```cpp
+
+    #include <string>
+    #include <vector>
+
+    using namespace std;
+
+    class Solution {
+    public:
+    string shortestPalindrome(string s) {
+        string rev_s = s;
+        reverse(rev_s.begin(), rev_s.end());
+
+        // Form the new string to check
+        string new_s = s + "#" + rev_s;
+        int n = new_s.size();
+
+        // Compute the KMP table (prefix function)
+        vector<int> kmp(n, 0);
+        for (int i = 1, j = 0; i < n; ++i) {
+            while (j > 0 && new_s[i] != new_s[j]) {
+                j = kmp[j - 1];
+            }
+            if (new_s[i] == new_s[j]) {
+                ++j;
+            }
+            kmp[i] = j;
+        }
+
+        // The length of the longest prefix which is also suffix
+        int overlap = kmp[n - 1];
+
+        // Construct the result
+        return rev_s.substr(0, rev_s.size() - overlap) + s;
+    }
+    };
+
+    ```
+
+  </TabItem>
+</Tabs>
+
+
+## References
+
+- **LeetCode Problem:** [LeetCode Problem](https://leetcode.com/problems/shortest-palindrome/)
+- **Solution Link:** [Shortest Palindrome Solution on LeetCode](https://leetcode.com/problems/shortest-palindrome/solutions/)
+- **Authors GeeksforGeeks Profile:** [Mahek Patel](https://leetcode.com/u/mahekrpatel611/)