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Longest Palindromic Subsequence in C++
Suppose we have a string s, we have to find the longest palindromic subsequence's length in s. We can assume that the maximum length of s is 1000. So if the input is like “bbbab”, then output will be 4. One possible palindromic subsequence is “bbbb”.
To solve this, we will follow these steps −
- x := s, then reverse x, n := size of s
- if n is 0, then return 0
- update s by adding one blank space before s, and update x by adding one blank space before x
- ret := 0
- make one matrix dp of size (n + 1) x (n + 1)
- for i in range 1 to n
- for j in range n to n
- dp[i, j] := max of dp[i, j – 1], dp[i – 1, j]
- if x[i] = s[j], then dp[i, j] := max of dp[i, j] and 1 + dp[i – 1, j – 1]
- for j in range n to n
- return dp[n, n]
Let us see the following implementation to get better understanding −
Example
#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
int longestPalindromeSubseq(string s) {
string x = s;
reverse(x.begin(), x.end());
int n = s.size();
if(!n) return 0;
s = " " + s;
x = " " + x;
int ret = 0;
vector < vector <int> > dp(n + 1, vector <int>(n + 1));
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n ; j++){
dp[i][j] = max(dp[i][j - 1], dp[i - 1][j]);
if(x[i] == s[j]) {
dp[i][j] = max(dp[i][j], 1 + dp[i - 1][j - 1]);
}
}
}
return dp[n][n];
}
};
main(){
Solution ob;
cout << (ob.longestPalindromeSubseq("bbbab"));
}Input
"bbbab"
Output
4
Updated on: 02-May-2020
582 Views
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