$\pi(n) = \text{ number of prime number smaller than or equal to n}$ . $rub(n) = \text{ number of palindromic number smaller than or equal to n}$ . For a given $p$ and $q$ find maximum such $n$ so that, $\pi(n)\le \frac{p}{q}\times rub(n)$ Inside Math: $\pi(n) \approx \frac{n}{ln(n)}$ [Prime number approximation] $rub(n) \approx 2\sqrt{n}$ maximum value of $\frac{p}{q} = 42$ Hence , \begin{align} \pi(n) &\le \frac{p}{q}\times rub(n) \newline \Longrightarrow \frac{n}{ln(n)} &\le 42 \times 2\sqrt{n} \newline \Longrightarrow \frac{\sqrt{n}}{ln(n)} &\le 84 \newline \Longrightarrow n_{max} &\approx \boxed{1415344} \newline \end{align}

insert minimum number of character in a given string after which the resulting string contain a subsequence abcdefgh...xyz

knapsack dp with ntt