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2430. Maximum Deletions on a String

Description

You are given a string s consisting of only lowercase English letters. In one operation, you can:

  • Delete the entire string s, or
  • Delete the first i letters of s if the first i letters of s are equal to the following i letters in s, for any i in the range 1 <= i <= s.length / 2.

For example, if s = "ababc", then in one operation, you could delete the first two letters of s to get "abc", since the first two letters of s and the following two letters of s are both equal to "ab".

Return the maximum number of operations needed to delete all of s.

 

Example 1:

Input: s = "abcabcdabc"
Output: 2
Explanation:
- Delete the first 3 letters ("abc") since the next 3 letters are equal. Now, s = "abcdabc".
- Delete all the letters.
We used 2 operations so return 2. It can be proven that 2 is the maximum number of operations needed.
Note that in the second operation we cannot delete "abc" again because the next occurrence of "abc" does not happen in the next 3 letters.

Example 2:

Input: s = "aaabaab"
Output: 4
Explanation:
- Delete the first letter ("a") since the next letter is equal. Now, s = "aabaab".
- Delete the first 3 letters ("aab") since the next 3 letters are equal. Now, s = "aab".
- Delete the first letter ("a") since the next letter is equal. Now, s = "ab".
- Delete all the letters.
We used 4 operations so return 4. It can be proven that 4 is the maximum number of operations needed.

Example 3:

Input: s = "aaaaa"
Output: 5
Explanation: In each operation, we can delete the first letter of s.

 

Constraints:

  • 1 <= s.length <= 4000
  • s consists only of lowercase English letters.

Solutions

We design a function $dfs(i)$, which represents the maximum number of operations needed to delete all characters from $s[i..]$. The answer is $dfs(0)$.

The calculation process of the function $dfs(i)$ is as follows:

  • If $i \geq n$, then $dfs(i) = 0$, return directly.
  • Otherwise, we enumerate the length of the string $j$, where $1 \leq j \leq (n-1)/2$. If $s[i..i+j] = s[i+j..i+j+j]$, we can delete $s[i..i+j]$, then $dfs(i)=max(dfs(i), dfs(i+j)+1)$. We need to enumerate all $j$ to find the maximum value of $dfs(i)$.

Here we need to quickly determine whether $s[i..i+j]$ is equal to $s[i+j..i+j+j]$. We can preprocess all the longest common prefixes of string $s$, that is, $g[i][j]$ represents the length of the longest common prefix of $s[i..]$ and $s[j..]$. In this way, we can quickly determine whether $s[i..i+j]$ is equal to $s[i+j..i+j+j]$, that is, $g[i][i+j] \geq j$.

To avoid repeated calculations, we can use memoization search and use an array $f$ to record the value of the function $dfs(i)$.

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

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class Solution:
    def deleteString(self, s: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i == n:
                return 0
            ans = 1
            for j in range(1, (n - i) // 2 + 1):
                if s[i : i + j] == s[i + j : i + j + j]:
                    ans = max(ans, 1 + dfs(i + j))
            return ans

        n = len(s)
        return dfs(0)

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class Solution {
    private int n;
    private Integer[] f;
    private int[][] g;

    public int deleteString(String s) {
        n = s.length();
        f = new Integer[n];
        g = new int[n + 1][n + 1];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j)) {
                    g[i][j] = g[i + 1][j + 1] + 1;
                }
            }
        }
        return dfs(0);
    }

    private int dfs(int i) {
        if (i == n) {
            return 0;
        }
        if (f[i] != null) {
            return f[i];
        }
        f[i] = 1;
        for (int j = 1; j <= (n - i) / 2; ++j) {
            if (g[i][i + j] >= j) {
                f[i] = Math.max(f[i], 1 + dfs(i + j));
            }
        }
        return f[i];
    }
}

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class Solution {
public:
    int deleteString(string s) {
        int n = s.size();
        int g[n + 1][n + 1];
        memset(g, 0, sizeof(g));
        for (int i = n - 1; ~i; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j]) {
                    g[i][j] = g[i + 1][j + 1] + 1;
                }
            }
        }
        int f[n];
        memset(f, 0, sizeof(f));
        function<int(int)> dfs = [&](int i) -> int {