2430. Maximum Deletions on a String

Problem

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.

Solution (Java)

class Solution {
    Integer[] dp = new Integer[4001];
    public int deleteString(String s) {
        return helper(s, 0);
    }

    public int helper(String s, int idx) {
        if (idx == s.length()) return 0;
        if (dp[idx] != null) {
            return dp[idx];
        }
        int answer = 1;
        for (int i = idx; 2 * i <= s.length() - 2 + idx; i++) {
            if (dp[i + 1] != null) continue;
            int len = i - idx + 1;
            if (s.substring(idx, idx + len).equals(s.substring(idx + len, idx + 2 * len))) {
                answer = Math.max(answer, 1 + helper(s, i + 1));
            }
        }
        return dp[idx] = answer;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).