C. Python Indentation — Conversion from O(N^3) to O(N^2)

Revision en1, by lax_99, 2019-01-19 21:57:33

Hello Codeforces community!

I have a memoized solution for this problem. The issue here is that my complexity is : O(N^3), which leads me to TLE. All the editorials and blogs I have seen have used Bottom up approach and suffix sum to get it to O(N^2). But I'm not sure how to translate this idea to the memoized version.

int dpSol(int i, int level)
{
    if(i == n - 1)
        return 1;
    
    if(cache[i][level] != -1)
        return cache[i][level];
    
    if(arr[i] == 's')
    {
        int sum = 0;
        for(int j = 0; j <= level; ++j)
            sum = (sum + dpSol(i + 1, j)) % MOD;
        return cache[i][level] = sum;
    }
    return cache[i][level] = dpSol(i + 1, level + 1);
}

Any help will be appreciated! Thanks in advance.

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  Rev. Lang. By When Δ Comment
en2 English lax_99 2019-01-19 22:08:56 76 Tiny change: 'm/909/C)\nI have a' -> 'm/909/C)\n\nI have a'
en1 English lax_99 2019-01-19 21:57:33 849 Initial revision (published)