Question

2430. Maximum Deletions on a String

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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 $s$.

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 $s$.

class Solution:
  def deleteString(self, s: str) -> int:
    n = len(s)
    g = [[0] * (n + 1) for _ in range(n + 1)]
    for i in range(n - 1, -1, -1):
      for j in range(i + 1, n):
        if s[i] == s[j]:
          g[i][j] = g[i + 1][j + 1] + 1

    f = [1] * n
    for i in range(n - 1, -1, -1):
      for j in range(1, (n - i) // 2 + 1):
        if g[i][i + j] >= j:
          f[i] = max(f[i], f[i + j] + 1)
    return f[0]

class Solution {
  public int deleteString(String s) {
    int n = s.length();
    int[][] 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;
        }
      }
    }
    int[] f = new int[n];
    for (int i = n - 1; i >= 0; --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], f[i + j] + 1);
        }
      }
    }
    return f[0];
  }
}

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];
    for (int i = n - 1; ~i; --i) {
      f[i] = 1;
      for (int j = 1; j <= (n - i) / 2; ++j) {
        if (g[i][i + j] >= j) {
          f[i] = max(f[i], f[i + j] + 1);
        }
      }
    }
    return f[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)
  for i := n - 1; i >= 0; i-- {
    f[i] = 1
    for j := 1; j <= (n-i)/2; j++ {
      if g[i][i+j] >= j {
        f[i] = max(f[i], f[i+j]+1)
      }
    }
  }
  return f[0]
}

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