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| 1 | +# Time: O(n^2) Space: O(n^2) - For all three solutions |
| 2 | +class Solution: |
| 3 | + def longestPalindromeSubseq(self, s: str) -> int: |
| 4 | + # Dynamic Programming |
| 5 | + dp = [ [0] * (len(s) + 1) for i in range(len(s) + 1)] |
| 6 | + res = 0 |
| 7 | + |
| 8 | + for i in range(len(s)): |
| 9 | + for j in range(len(s) - 1, i - 1, -1): |
| 10 | + if s[i] == s[j]: |
| 11 | + dp[i][j] = 1 if i == j else 2 |
| 12 | + if i - 1 >= 0: |
| 13 | + dp[i][j] += dp[i - 1][j + 1] |
| 14 | + else: |
| 15 | + dp[i][j] = dp[i][j + 1] |
| 16 | + if i - 1 >= 0: |
| 17 | + dp[i][j] = max(dp[i][j], dp[i - 1][j]) |
| 18 | + res = max(res, dp[i][j]) |
| 19 | + return res |
| 20 | + |
| 21 | + |
| 22 | + # Memoization |
| 23 | + cache = {} |
| 24 | + |
| 25 | + def dfs(i, j): |
| 26 | + if i < 0 or j == len(s): |
| 27 | + return 0 |
| 28 | + if (i, j) in cache: |
| 29 | + return cache[(i, j)] |
| 30 | + |
| 31 | + if s[i] == s[j]: |
| 32 | + length = 1 if i == j else 2 |
| 33 | + cache[(i, j)] = length + dfs(i - 1, j + 1) |
| 34 | + else: |
| 35 | + cache[(i, j)] = max(dfs(i - 1, j), dfs(i, j + 1)) |
| 36 | + return cache[(i, j)] |
| 37 | + |
| 38 | + for i in range(len(s)): |
| 39 | + dfs(i, i) # odd length |
| 40 | + dfs(i, i + 1) # even length |
| 41 | + |
| 42 | + return max(cache.values()) |
| 43 | + |
| 44 | +# LCS Solution |
| 45 | +class Solution: |
| 46 | + def longestPalindromeSubseq(self, s: str) -> int: |
| 47 | + return self.longestCommonSubsequence(s, s[::-1]) |
| 48 | + |
| 49 | + |
| 50 | + def longestCommonSubsequence(self, s1: str, s2: str) -> int: |
| 51 | + N, M = len(s1), len(s2) |
| 52 | + dp = [[0] * (M+1) for _ in range(N+1)] |
| 53 | + |
| 54 | + for i in range(N): |
| 55 | + for j in range(M): |
| 56 | + if s1[i] == s2[j]: |
| 57 | + dp[i+1][j+1] = 1 + dp[i][j] |
| 58 | + else: |
| 59 | + dp[i+1][j+1] = max(dp[i][j+1], dp[i+1][j]) |
| 60 | + return dp[N][M] |
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