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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 $i\ge n$, then $dfs(i)=0$, return directly.
• Otherwise, we enumerate the length of the string $j$, where $1\le j\le (n-1)/2$. 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, $g[i][i+j]\ge j$.

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 $1\le j\le (n-1)/2$. 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 $s$.

• Java
• C++
• Python
• Go
• TypeScript

• 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);
  }
  
  

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