[Solved] Maximum Deletions on a String LeetCode Contest 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.

Maximum Deletions on a String LeetCode Contest

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

lcs[i][j] means the length of the longest common substring.
If lcs[i][j] = k,
then s.substring(i, i + k) == s.substring(j, j + k)
and s.substring(i, i + k + 1) != s.substring(j, j + k + 1).
This can be done in O(n^2).

dp[i] mean the maximum number of operations to delete
the substring starting at s[i].

If lcs[i][j] >= j - i,
s.substring(i, j) == s.substring(j, j + j - i)
this means we can delete the prefix s.substring(i, j) from s.substring(i),
and it changes to s.substring(j).
And we update dp[i] = max(dp[i], dp[j] + 1)

Complexity

Time O(n^2)
Space O(n^2)

Java

    public int deleteString(String s) {
        int n = s.length();
        int[][] lcs = new int[n + 1][n + 1];
        int[] dp = new int[n];
        for (int i = n - 1; i >= 0; --i) {
            lcs[i] = new int[n + 1];
            dp[i] = 1;
            for (int j = i + i; j < n; ++j) {
                if (s.charAt(i) == s.charAt(j))
                    lcs[i][j] = lcs[i + 1][j + 1] + 1;
                if (lcs[i][j] >= j - i)
                    dp[i] = Math.max(dp[i], dp[j] + 1);
            }
        }
        return dp[0];
    }

C++

    int deleteString(string s) {
        int n = s.size();
        vector<vector<int>> lcs(n + 1, vector<int>(n + 1, 0));
        vector<int> dp(n, 1);
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                if (s[i] == s[j])
                    lcs[i][j] = lcs[i + 1][j + 1] + 1;
                if (lcs[i][j] >= j - i)
                    dp[i] = max(dp[i], dp[j] + 1);
            }
        }
        return dp[0];
    }

Python

    def deleteString(self, s):
        n = len(s)
        lcs = [[0] * (n + 1) for i in range(n + 1)]
        dp = [1] * n
        for i in range(n-1, -1, -1):
            for j in range(i + 1, n):
                if s[i] == s[j]:
                    lcs[i][j] = lcs[i + 1][j + 1] + 1
                if lcs[i][j] >= j - i:
                    dp[i] = max(dp[i], dp[j] + 1)
        return dp[0]

Happy Learning – If you require any further information, feel free to contact me.