# 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 1: Memoization Search

We design a function $dfs(i)$, which represents the maximum number of operations needed to delete all characters from $s[i..]$. The answer is $dfs(0)$.

The calculation process of the function $dfs(i)$ is as follows:

• If $i \geq n$, then $dfs(i) = 0$, return directly.
• Otherwise, we enumerate the length of the string $j$, where $1 \leq j \leq (n-1)/2$. If $s[i..i+j] = s[i+j..i+j+j]$, we can delete $s[i..i+j]$, then $dfs(i)=max(dfs(i), dfs(i+j)+1)$. We need to enumerate all $j$ to find the maximum value of $dfs(i)$.

Here we need to quickly determine whether $s[i..i+j]$ is equal to $s[i+j..i+j+j]$. We can preprocess all the longest common prefixes of string $s$, that is, $g[i][j]$ represents the length of the longest common prefix of $s[i..]$ and $s[j..]$. In this way, we can quickly determine whether $s[i..i+j]$ is equal to $s[i+j..i+j+j]$, that is, $g[i][i+j] \geq j$.

To avoid repeated calculations, we can use memoization search and use an array $f$ to record the value of the function $dfs(i)$.

The time complexity is $O(n^2)$, and the space complexity is $O(n^2)$. Here, $n$ is the length of the string $s$.

Solution 2: Dynamic Programming

We can change the memoization search in Solution 1 to dynamic programming. Define $f[i]$ to represent the maximum number of operations needed to delete all characters from $s[i..]$. Initially, $f[i]=1$, and the answer is $f[0]$.

We can enumerate $i$ from back to front. For each $i$, we enumerate the length of the string $j$, where $1 \leq j \leq (n-1)/2$. If $s[i..i+j] = s[i+j..i+j+j]$, we can delete $s[i..i+j]$, then $f[i]=max(f[i], f[i+j]+1)$. We need to enumerate all $j$ to find the maximum value of $f[i]$.

The time complexity is $O(n^2)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$.

• class Solution {
private int n;
private Integer[] f;
private int[][] g;

public int deleteString(String s) {
n = s.length();
f = new Integer[n];
g = new int[n + 1][n + 1];
for (int i = n - 1; i >= 0; --i) {
for (int j = i + 1; j < n; ++j) {
if (s.charAt(i) == s.charAt(j)) {
g[i][j] = g[i + 1][j + 1] + 1;
}
}
}
return dfs(0);
}

private int dfs(int i) {
if (i == n) {
return 0;
}
if (f[i] != null) {
return f[i];
}
f[i] = 1;
for (int j = 1; j <= (n - i) / 2; ++j) {
if (g[i][i + j] >= j) {
f[i] = Math.max(f[i], 1 + dfs(i + j));
}
}
return f[i];
}
}

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

• class Solution:
def deleteString(self, s: str) -> int:
@cache
def dfs(i: int) -> int:
if i == n:
return 0
ans = 1
for j in range(1, (n - i) // 2 + 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)
g := make([][]int, n+1)
for i := range g {
g[i] = make([]int, n+1)
}
for i := n - 1; i >= 0; i-- {
for j := i + 1; j < n; j++ {
if s[i] == s[j] {
g[i][j] = g[i+1][j+1] + 1
}
}
}
f := make([]int, n)
var dfs func(int) int
dfs = func(i int) int {
if i == n {
return 0
}
if f[i] > 0 {
return f[i]
}
f[i] = 1
for j := 1; j <= (n-i)/2; j++ {
if g[i][i+j] >= j {
f[i] = max(f[i], dfs(i+j)+1)
}
}
return f[i]
}
return dfs(0)
}

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