📘 Longest Palindromic Subsequence (LPS)

🔹 Key Idea (VERY IMPORTANT 🔥)

👉 Convert LPS → LCS

LPS(s) = LCS(s, reverse(s))

---

🔹 Why does this work?

• Palindrome = same forward & backward
• So if we reverse string and find LCS → we get longest palindrome subsequence

---

🔹 Your Code Explained

string t = s;
reverse(t.begin(), t.end());
return lcs(s, t);

👉 Steps:

1. Copy string → t
2. Reverse it
3. Find LCS between s and t

---

🔹 Example

s = "bbabcbcab"

Reverse:

t = "bacbcbabb"

👉 LCS = "babcbab" (length = 7)

---

🔥 Full Code (Clean)

int lcs(string s1, string s2) {
    int n = s1.size(), m = s2.size();
    vector<vector<int>> dp(n+1, vector<int>(m+1, 0));

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(s1[i-1] == s2[j-1])
                dp[i][j] = 1 + dp[i-1][j-1];
            else
                dp[i][j] = max(dp[i-1][j], dp[i][j-1]);
        }
    }
    return dp[n][m];
}

int longestPalindromeSubsequence(string s) {
    string t = s;
    reverse(t.begin(), t.end());
    return lcs(s, t);
}

---

🔹 Space Optimized Version 🚀

int lcs(string s1, string s2) {
    int n = s1.size(), m = s2.size();
    vector<int> prev(m+1, 0), curr(m+1, 0);

    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= m; j++){
            if(s1[i-1] == s2[j-1])
                curr[j] = 1 + prev[j-1];
            else
                curr[j] = max(prev[j], curr[j-1]);
        }
        prev = curr;
    }
    return prev[m];
}

---

🔥 Print Longest Palindromic Subsequence

👉 Just print LCS of s and reverse(s)

string printLPS(string s) {
    string t = s;
    reverse(t.begin(), t.end());

    int n = s.size();
    vector<vector<int>> dp(n+1, vector<int>(n+1, 0));

    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            if(s[i-1]==t[j-1])
                dp[i][j]=1+dp[i-1][j-1];
            else
                dp[i][j]=max(dp[i-1][j],dp[i][j-1]);
        }
    }

    // backtrack
    string ans="";
    int i=n,j=n;
    while(i>0 && j>0){
        if(s[i-1]==t[j-1]){
            ans+=s[i-1];
            i--; j--;
        }
        else if(dp[i-1][j]>dp[i][j-1]) i--;
        else j--;
    }

    reverse(ans.begin(), ans.end());
    return ans;
}

---

🔹 Complexity

Metric	Value
Time	O(n²)
Space	O(n²) → optimized to O(n)

---

🧠 Important Observations

• LPS is a subsequence, not substring
• Works even if characters are not continuous
• Always think:

“Reverse + LCS”

---

🔥 Related Problems

• Minimum insertions to make palindrome:

n - LPS

• Minimum deletions:

n - LPS

---

⚡ Interview Tip

If asked:

“Longest palindromic subsequence?”

👉 Say instantly:

“We can solve it using LCS between string and its reverse.”

---

If you want next 🚀
I can show:
✅ Direct DP without LCS trick
✅ Palindromic substring (different problem ⚠️)
✅ Hard variations

Just tell me 👍