Leetcode 2430 Maximum Deletions on a String Solution in Java | Hindi Coding Community

 

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.

Java Code :

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 + 1; 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];

}