create 2002 maximum-product-of-the-length-....cpp

cpp/2002-maximum-product-of-the-length-of-two-palindromic-subsequences.cpp

@@ -0,0 +1,81 @@
+ /*
+    Approach: 
+    Need to create all the disjoin subsequence and check if they are palindrome.
+    keep track of maximum product 
+
+    Time complexity : O(N*N^3)
+    Space complexity: O(N)
+
+    N is length of the string
+*/
+
+
+class Solution {
+public:
+
+    int answer = INT_MIN;
+
+    // function to check if the string is a palindrome
+    bool isPalindrome(string &s){
+        int start = 0;
+        int end = s.length() - 1;
+
+        while(start<end){
+            if(s[start]!=s[end]){
+                return false;
+            }
+            start++;
+            end--;
+        }
+
+        return true;
+    }
+
+    // function to generate all the disjoint subsequence
+    void generateAll(int idx, string &s1, string &s2, string& s){
+
+        if(idx >= s.length())
+        {
+            if(isPalindrome(s1)&&isPalindrome(s2)){
+                int l = s1.length()*s2.length();
+                answer = max(answer,l);
+            }
+            return;
+        }
+
+        char c = s[idx];
+
+        /* 
+        we have three options
+        1. Add the char to the first string
+        2. Add the char to the second string
+        3. Add the char to none of the string
+        */
+
+        // add the character in the first string
+        s1.push_back(c);
+        generateAll(idx+1,s1,s2,s);
+        s1.pop_back();
+
+        // add the character in the second string
+        s2.push_back(c);
+        generateAll(idx+1,s1,s2,s);
+        s2.pop_back();
+
+        // add character in no string
+        generateAll(idx+1,s1,s2,s);
+    }
+
+    int maxProduct(string s) {
+
+        string s1 = "";
+        string s2 = "";
+        int idx = 0;
+
+        generateAll(idx,s1,s2,s);
+
+        return answer;
+    }
+
+
+};