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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
}


Explain:

nope.

Complexity:

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