added solution to 647

dsa-solutions/lc-solutions/0600-0699/647-palindromic-substrings.md

@@ -0,0 +1,182 @@
+---
+id: palindromic-substrings
+title: Palindromic Substrings
+sidebar_label: 647-Palindromic Substrings
+tags:
+  - String Manipulation
+  - Dynamic Programming
+  - LeetCode
+  - Java
+  - Python
+  - C++
+description: "This is a solution to the Palindromic Substrings problem on LeetCode."
+sidebar_position: 11
+---
+
+## Problem Description
+
+Given a string `s`, return the number of palindromic substrings in it.A string is a palindrome when it reads the same backward as forward.
+
+A substring is a contiguous sequence of characters within the string.
+
+### Examples
+
+**Example 1:**
+
+```
+Input: s = "abc"
+Output: 3
+Explanation: Three palindromic strings: "a", "b", "c".
+```
+
+**Example 2:**
+
+```
+Input: s = "aaa"
+Output: 6
+Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".
+```
+
+### Constraints
+
+- `1 <= s.length <= 1000`
+- `s` consists of lowercase English letters.
+
+---
+
+## Solution for Count Palindromic Substrings Problem
+
+To solve this problem, we need to count all the palindromic substrings in the given string `s`.
+
+### Approach: Expand Around Center
+
+The idea is to consider each possible center of a palindrome and expand outwards as long as we have a valid palindrome. For each center, count the palindromic substrings.
+
+#### Steps:
+
+1. **Identify Centers:**
+   - There are 2*n - 1 centers for palindromes in a string of length `n` (including the space between characters for even-length palindromes).
+
+2. **Expand Around Center:**
+   - For each center, expand outwards to check if the substring is a palindrome. If it is, increment the count.
+
+### Brute Force Approach
+
+The brute force approach involves checking all substrings and verifying if each substring is a palindrome. This is inefficient and has a time complexity of $O(n^3)$.
+
+### Optimized Approach
+
+The optimized approach, using the Expand Around Center method, has a time complexity of $O(n^2)$.
+
+### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```cpp
+class Solution {
+public:
+    int countSubstrings(string s) {
+        int n = s.size();
+        int count = 0;
+
+        for (int i = 0; i < n; ++i) {
+            count += countPalindromesAroundCenter(s, i, i); // Odd length
+            count += countPalindromesAroundCenter(s, i, i + 1); // Even length
+        }
+
+        return count;
+    }
+
+private:
+    int countPalindromesAroundCenter(const string& s, int left, int right) {
+        int count = 0;
+
+        while (left >= 0 && right < s.size() && s[left] == s[right]) {
+            ++count;
+            --left;
+            ++right;
+        }
+
+        return count;
+    }
+};
+```
+
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```java
+class Solution {
+    public int countSubstrings(String s) {
+        int n = s.length();
+        int count = 0;
+
+        for (int i = 0; i < n; i++) {
+            count += countPalindromesAroundCenter(s, i, i); // Odd length
+            count += countPalindromesAroundCenter(s, i, i + 1); // Even length
+        }
+
+        return count;
+    }
+
+    private int countPalindromesAroundCenter(String s, int left, int right) {
+        int count = 0;
+
+        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
+            count++;
+            left--;
+            right++;
+        }
+
+        return count;
+    }
+}
+```
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@ImmidiSivani"/>
+
+```python
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        n = len(s)
+        count = 0
+
+        for i in range(n):
+            count += self.countPalindromesAroundCenter(s, i, i) # Odd length
+            count += self.countPalindromesAroundCenter(s, i, i + 1) # Even length
+
+        return count
+
+    def countPalindromesAroundCenter(self, s: str, left: int, right: int) -> int:
+        count = 0
+
+        while left >= 0 and right < len(s) and s[left] == s[right]:
+            count += 1
+            left -= 1
+            right += 1
+
+        return count
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- **Time Complexity**: $O(n^2)$, where `n` is the length of the input string `s`.
+- **Space Complexity**: $O(1)$, as we only use constant extra space.
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['ImmidiSivani'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>