Merge pull request #410 from nandanabhishek/patch-1

added LongestPalindromicSubsequence.py

21. Dynamic Programming/LongestPalindromicSubsequence.cpp

@@ -0,0 +1,41 @@
+class Solution {
+public:
+    // Bottom-up approach
+    // TC: O(N*N), SC: O(N*N)
+    int longestPalindromeSubseqTabular(string& s) {
+        if(s.empty())
+            return 0;
+
+        const int N = s.size();
+        // dp(i, j): length of longest palindromic substring in s[i:j]
+        vector<vector<int> > dp(N, vector<int>(N, 0));
+
+        // all single chars are palindromic
+        for(int i = 0; i < N; i++)
+            dp[i][i] = 1;
+
+        for(int l = 1; l < N; l++) {
+            // starting index of window
+            for(int i = 0; i < N - l; i++) {
+                // ending index of window
+                int j = i + l;
+
+                // if there are only two chars
+                if((j - i + 1) == 2) {
+                    // if the chars are same, then that contributes 2, otherwise since
+                    // individually they are palindromic so max length 1
+                    dp[i][j] = 1 + (s[i] == s[j]);
+                }
+                else {
+                    // for s[i:j]:
+                    // longest palindrome length in s[i+1 : j-1] and +1 if s[i] == s[j]
+                    // if s[i] != s[j], longest length palindrome = Max of longest palin in s[i:j-1] and s[i+1:j]
+                    if(s[i] == s[j])
+                        dp[i][j] = dp[i+1][j-1] + 2;
+                    else
+                        dp[i][j] = max(dp[i+1][j], dp[i][j-1]);
+                }
+            }
+        }
+        return dp[0][N-1];
+    }