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Formatted question description: https://leetcode.ca/all/2430.html

2430. Maximum Deletions on a String

  • Difficulty: Hard.
  • Related Topics: .
  • Similar Questions: Shortest Palindrome, Longest Happy Prefix, Remove All Occurrences of a Substring.

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, C++, Python)

  • class Solution {
    public int deleteString(String s) {
        int n = s.length();
        int[][] lcp = new int[n + 1][n + 1];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = n - 1; j >= 0; --j) {
                if (s.charAt(i) == s.charAt(j)) {
                    lcp[i][j] = 1 + lcp[i + 1][j + 1];
                }
            }
        }
        int[] dp = new int[n];
        Arrays.fill(dp, 1);
        for (int i = n - 1; i >= 0; --i) {
            for (int j = 1; j <= (n - i) / 2; ++j) {
                if (lcp[i][i + j] >= j) {
                    dp[i] = Math.max(dp[i], dp[i + j] + 1);
                }
            }
        }
        return dp[0];
    }
}

  • class Solution {
public:
    int deleteString(string s) {
        int n = s.size();
        int lcp[n + 1][n + 1];
        memset(lcp, 0, sizeof lcp);
        for (int i = n - 1; i >= 0; --i) {
            for (int j = n - 1; j >= 0; --j) {
                if (s[i] == s[j]) {
                    lcp[i][j] = 1 + lcp[i + 1][j + 1];
                }
            }
        }
        int dp[n];
        for (int i = n - 1; i >= 0; --i) {
            dp[i] = 1;
            for (int j = 1; j <= (n - i) / 2; ++j) {
                if (lcp[i][i + j] >= j) {
                    dp[i] = max(dp[i], dp[i + j] + 1);
                }
            }
        }
        return dp[0];
    }
};

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


  • func deleteString(s string) int {
	n := len(s)
	lcp := make([][]int, n+1)
	for i := range lcp {
		lcp[i] = make([]int, n+1)
	}
	for i := n - 1; i >= 0; i-- {
		for j := n - 1; j >= 0; j-- {
			if s[i] == s[j] {
				lcp[i][j] = 1 + lcp[i+1][j+1]
			}
		}
	}
	dp := make([]int, n)
	for i := n - 1; i >= 0; i-- {
		dp[i] = 1
		for j := 1; j <= (n-i)/2; j++ {
			if lcp[i][i+j] >= j {
				dp[i] = max(dp[i], dp[i+j]+1)
			}
		}
	}
	return dp[0]
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • function deleteString(s: string): number {
    const n = s.length;
    const f: number[] = new Array(n).fill(1);
    for (let i = n - 1; i >= 0; --i) {
        for (let j = 1; j <= (n - i) >> 1; ++j) {
            if (s.slice(i, i + j) === s.slice(i + j, i + j + j)) {
                f[i] = Math.max(f[i], f[i + j] + 1);
            }
        }
    }
    return f[0];
}

Explain:

nope.

Complexity:

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

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