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

Solution: Dynamic Programming

• Time complexity: O(n²)
• Space complexity: O(n²)

 

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
const deleteString = function (s) {
  const n = s.length;
  const lcp = Array.from({ length: n + 1 }, () => new Array(n + 1).fill(0));
  const dp = Array.from({ length: n }, () => 1);

  for (let a = n - 1; a >= 0; a--) {
    for (let b = n - 1; b > a; b--) {
      if (s[a] === s[b]) {
        lcp[a][b] = lcp[a + 1][b + 1] + 1;
      }

      if (lcp[a][b] >= b - a) {
        dp[a] = Math.max(dp[b] + 1, dp[a]);
      }
    }
  }

  return dp[0];
};