Jan 4

Author: siddydutta

2025-01-January-LeetCoding-Challenge/README.md

@@ -6,7 +6,7 @@
 | January 1  | [1422. Maximum Score After Splitting a String](https://leetcode.com/problems/maximum-score-after-splitting-a-string/) | Easy | Solved |
 | January 2  | [2559. Count Vowel Strings in Ranges](https://leetcode.com/problems/count-vowel-strings-in-ranges/) | Medium | Solved |
 | January 3  | [2270. Number of Ways to Split Array](https://leetcode.com/problems/number-of-ways-to-split-array/) | Medium | Solved |
-| January 4  | []() |  |  |
+| January 4  | [1930. Unique Length-3 Palindromic Subsequences](https://leetcode.com/problems/unique-length-3-palindromic-subsequences/) | Medium | Solved |
 | January 5  | []() |  |  |
 | January 6  | []() |  |  |
 | January 7  | []() |  |  |
@@ -40,5 +40,5 @@
 | Level  | Problems | Solved | Unsolved |
 | ---    | --- | --- | --- |
 | Easy   | 1 | 1 | 0 |
-| Medium | 2 | 2 | 0 |
+| Medium | 3 | 3 | 0 |
 | Hard   | 0 | 0 | 0 |

2025-01-January-LeetCoding-Challenge/Unique Length-3 Palindromic Subsequences.py

@@ -0,0 +1,30 @@
+from collections import Counter
+
+
+class Solution:
+    def countPalindromicSubsequence(self, s: str) -> int:
+        freq = Counter(s)  # frequency of right string
+        seen = set()  # seen characters of left string
+        subs = set()  # set of palindromic subsequences
+        for char in s:
+            freq[char] -= 1
+            for ch in seen:
+                if freq[ch] > 0:
+                    subs.add((ch, char))  # same as ch+char+ch
+            seen.add(char)
+        return len(subs)
+
+
+def main():
+    s = 'aabca'
+    assert Solution().countPalindromicSubsequence(s) == 3
+
+    s = 'adc'
+    assert Solution().countPalindromicSubsequence(s) == 0
+
+    s = 'bbcbaba'
+    assert Solution().countPalindromicSubsequence(s) == 4
+
+
+if __name__ == '__main__':
+    main()