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 \(s\).

Python3JavaC++GoTypeScript

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)

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);
    }
};

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);
}

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 \(s\).