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1930. Unique Length-3 Palindromic Subsequences.py

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+# Medium
+# Topics
+# premium lock icon
+# Companies
+# Hint
+# Given a string s, return the number of unique palindromes of length three that are a subsequence of s.
+
+# Note that even if there are multiple ways to obtain the same subsequence, it is still only counted once.
+
+# A palindrome is a string that reads the same forwards and backwards.
+
+# A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.
+
+# For example, "ace" is a subsequence of "abcde".
+ 
+
+# Example 1:
+
+# Input: s = "aabca"
+# Output: 3
+# Explanation: The 3 palindromic subsequences of length 3 are:
+# - "aba" (subsequence of "aabca")
+# - "aaa" (subsequence of "aabca")
+# - "aca" (subsequence of "aabca")
+# Example 2:
+
+# Input: s = "adc"
+# Output: 0
+# Explanation: There are no palindromic subsequences of length 3 in "adc".
+# Example 3:
+
+# Input: s = "bbcbaba"
+# Output: 4
+# Explanation: The 4 palindromic subsequences of length 3 are:
+# - "bbb" (subsequence of "bbcbaba")
+# - "bcb" (subsequence of "bbcbaba")
+# - "bab" (subsequence of "bbcbaba")
+
+class Solution:
+    def countPalindromicSubsequence(self, s: str) -> int:
+        first = {}
+        last = {}
+        for i, ch in enumerate(s):
+            if ch not in first:
+                first[ch] = i
+            last[ch] = i
+        
+        result = 0
+        
+        # Step 2: for each character, check possible middle chars
+        for ch in set(s):
+            if first[ch] < last[ch]:  # must have at least 2 occurrences
+                middle_chars = set(s[first[ch] + 1 : last[ch]])
+                result += len(middle_chars)
+        
+        return result
+ 
+