Time: 514 ms (48.59%), Space: 18.4 MB (26.6%) - LeetHub

Author: hogan-tech

2059-unique-length-3-palindromic-subsequences/2059-unique-length-3-palindromic-subsequences.py

@@ -0,0 +1,30 @@
+# time complexity: O(n)
+# space complexity: O(1)
+
+class Solution:
+    def countPalindromicSubsequence(self, s: str) -> int:
+        first = [-1] * 26
+        last = [-1] * 26
+        for i in range(len(s)):
+            curr = ord(s[i]) - ord('a')
+            if first[curr] == -1:
+                first[curr] = i
+            last[curr] = i
+
+        ans = 0
+        for i in range(26):
+            if first[i] == -1:
+                continue
+            between = set()
+            for j in range(first[i] + 1, last[i]):
+                between.add(s[j])
+            ans += len(between)
+        return ans
+
+
+s = "aabca"
+print(Solution().countPalindromicSubsequence(s))
+s = "adc"
+print(Solution().countPalindromicSubsequence(s))
+s = "bbcbaba"
+print(Solution().countPalindromicSubsequence(s))