Is there a way to optimize space in this DP program? - c++

Question

Problem Statement:
A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a string, determines the minimal number of characters to be inserted into the string in order to obtain a palindrome. As an example, by inserting 2 characters, the string "Ab3bd" can be transformed into a palindrome ("dAb3bAd" or "Adb3bdA"). However, inserting fewer than 2 characters does not produce a palindrome.
Input
The first line contains one integer: the length of the input string N, 3≤N≤5000. The second line contains one string with length N. The string is formed from uppercase letters from ‘A’ to ‘Z’, lowercase letters from ‘a’ to ‘z’ and digits from ‘0’ to ‘9’. Uppercase and lowercase letters are to be considered distinct.
Output
The first line contains one integer, which is the desired minimal number.
Link to problem => http://www.spoj.com/problems/IOIPALIN/
My solution:
#include <iostream>
#include <memory.h>
#include <cstdio>
using namespace std;
long memo[5010][5010];
string s;
long int n;
long solve(long i,long j){
if(memo[i][j]!=-1){
return memo[i][j];
}
if(i>=j)
return 0;
if(s[i]==s[j])
return solve(i+1,j-1);
return memo[i][j]= min(solve(i,j-1)+1,solve(i+1,j)+1);
}
int main()
{
ios_base::sync_with_stdio(false);
long int n ;
cin>>n;
cin>>s;
memset(memo,-1,sizeof(memo));
long int a = solve(0,n-1);
cout << a << endl;
return 0;
}
I am getting "time limit exceeded" for this code. How can I fix this ?

Answer 1

Well being a programmer its a bad practice to ask directly the solution to the question. I won't provide you with the exact code, but I will help you solve the problem.
To solve this an n^2 approach is needed. Dynamic programming is what you need. To achieve that create a string rev_str that is reverse of str. Now see the following pseudo code.
for(i = 0 to n)
for(j = 0 to n)
if(i = 0 or j = 0) dp[j][1] = 0;
else if(str[i - 1] = rev_str[j - 1]) dp[j][1] = dp[j - 1][0] + 1;
else dp[j][1] = max(dp[j][0], dp[j - 1][1];
for (j = 0 to n) dp[j][0] = dp[j][1];
your answer will be n - dp[n][0].
Basically what we are trying to do is at every step we are finding the maximum number of characters that match from str to rev_str. Those remaining will be needed to be added. Thus giving an answer.