Length of Longest Palindromic Subsequence

Length of Longest Palindromic Subsequence (Dynamic Programming Approach).cpp

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+#include<bits/stdc++.h>
+using namespace std;
+
+int findLPS(string s) {
+
+    int n = s.length();
+    int dp[n][n];
+    memset(dp, 0, sizeof(dp));
+
+    // Every single character is a palindrome itself, of length 1
+    for (int i = 0; i < n; ++i)
+        dp[i][i] = 1;
+
+    // Check every subsequence of length 2 and greater, upto length of the string
+    for (int len = 2; len <= n; ++len) {
+
+        for (int i = 0; i <= n - len; ++i) {
+
+            int j = i + len - 1;
+
+            if (s[i] == s[j] && len == 2)
+                dp[i][j] = 2;
+            else if (s[i] == s[j])
+                dp[i][j] = dp[i+1][j-1] + 2;
+            else
+                dp[i][j] = max(dp[i][j-1], dp[i+1][j]);
+        }
+    }
+
+    // Top-right corner cell will contain the length of LPS
+    return dp[0][n-1];
+}
+
+int main() {
+
+    string s;
+    cout << "Enter the string : ";
+    cin >> s;
+
+    int res = findLPS(s);
+    cout << "Length of the Longest Palindromic Subsequence is : " << res << '\n';
+
+    return 0;
+}
+
+/*
+ ~ Time Complexity   : O(n^2)
+ ~ Space Complexity  : O(n^2)
+*/