Valid Palindrome III - Solution & Explanation

The problem requires us to remove at most $k$ characters to make the remaining string a palindrome. This can be transformed into finding the longest palindromic subsequence.

We define $f[i][j]$ as the length of the longest palindromic subsequence in the substring $s[i..j]$. Initially, we have $f[i][i] = 1$ for all $i$, since each single character is a palindrome.

If $s[i] = s[j]$, then we have $f[i][j] = f[i+1][j-1] + 2$, since we can add both $s[i]$ and $s[j]$ to the longest palindromic subsequence of $s[i+1..j-1]$.

If $s[i] \neq s[j]$, then we have $f[i][j] = \max(f[i+1][j], f[i][j-1])$, since we need to remove either $s[i]$ or $s[j]$ to make the remaining substring a palindrome.

Finally, we check whether there exists $f[i][j] + k \geq n$, where $n$ is the length of the string $s$. If so, it means that we can remove at most $k$ characters to make the remaining string a palindrome.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string $s$.