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Formatted question description: https://leetcode.ca/all/2472.html

2472. Maximum Number of Non-overlapping Palindrome Substrings

• Difficulty: Hard.
• Related Topics: String, Dynamic Programming.
• Similar Questions: Longest Palindromic Substring, Palindrome Partitioning, Palindrome Partitioning II, Palindrome Partitioning III, Maximum Number of Non-Overlapping Substrings, Palindrome Partitioning IV.

Problem

You are given a string s and a positive integer k.

Select a set of non-overlapping substrings from the string s that satisfy the following conditions:

• The length of each substring is at least k.

• Each substring is a palindrome.

Return the **maximum number of substrings in an optimal selection**.

A substring is a contiguous sequence of characters within a string.

  Example 1:

Input: s = "abaccdbbd", k = 3
Output: 2
Explanation: We can select the substrings underlined in s = "abaccdbbd". Both "aba" and "dbbd" are palindromes and have a length of at least k = 3.
It can be shown that we cannot find a selection with more than two valid substrings.

Example 2:

Input: s = "adbcda", k = 2
Output: 0
Explanation: There is no palindrome substring of length at least 2 in the string.

  Constraints:

• 1 <= k <= s.length <= 2000

• s consists of lowercase English letters.

Solution (Java, C++, Python)

• Java
• C++
• Python
• Go

• class Solution {
      private boolean[][] dp;
      private int[] f;
      private String s;
      private int n;
      private int k;
  
      public int maxPalindromes(String s, int k) {
          n = s.length();
          f = new int[n];
          this.s = s;
          this.k = k;
          dp = new boolean[n][n];
          for (int i = 0; i < n; ++i) {
              Arrays.fill(dp[i], true);
              f[i] = -1;
          }
          for (int i = n - 1; i >= 0; --i) {
              for (int j = i + 1; j < n; ++j) {
                  dp[i][j] = s.charAt(i) == s.charAt(j) && dp[i + 1][j - 1];
              }
          }
          return dfs(0);
      }
  
      private int dfs(int i) {
          if (i >= n) {
              return 0;
          }
          if (f[i] != -1) {
              return f[i];
          }
          int ans = dfs(i + 1);
          for (int j = i + k - 1; j < n; ++j) {
              if (dp[i][j]) {
                  ans = Math.max(ans, 1 + dfs(j + 1));
              }
          }
          f[i] = ans;
          return ans;
      }
  }
  

• class Solution {
  public:
      int maxPalindromes(string s, int k) {
          int n = s.size();
          vector<vector<bool>> dp(n, vector<bool>(n, true));
          vector<int> f(n, -1);
          for (int i = n - 1; i >= 0; --i) {
              for (int j = i + 1; j < n; ++j) {
                  dp[i][j] = s[i] == s[j] && dp[i + 1][j - 1];
              }
          }
          function<int(int)> dfs = [&](int i) -> int {
              if (i >= n) return 0;
              if (f[i] != -1) return f[i];
              int ans = dfs(i + 1);
              for (int j = i + k - 1; j < n; ++j) {
                  if (dp[i][j]) {
                      ans = max(ans, 1 + dfs(j + 1));
                  }
              }
              f[i] = ans;
              return ans;
          };
          return dfs(0);
      }
  };
  

• class Solution:
      def maxPalindromes(self, s: str, k: int) -> int:
          @cache
          def dfs(i):
              if i >= n:
                  return 0
              ans = dfs(i + 1)
              for j in range(i + k - 1, n):
                  if dp[i][j]:
                      ans = max(ans, 1 + dfs(j + 1))
              return ans
  
          n = len(s)
          dp = [[True] * n for _ in range(n)]
          for i in range(n - 1, -1, -1):
              for j in range(i + 1, n):
                  dp[i][j] = s[i] == s[j] and dp[i + 1][j - 1]
          ans = dfs(0)
          dfs.cache_clear()
          return ans
  
  

• func maxPalindromes(s string, k int) int {
  	n := len(s)
  	dp := make([][]bool, n)
  	f := make([]int, n)
  	for i := 0; i < n; i++ {
  		dp[i] = make([]bool, n)
  		f[i] = -1
  		for j := 0; j < n; j++ {
  			dp[i][j] = true
  		}
  	}
  	for i := n - 1; i >= 0; i-- {
  		for j := i + 1; j < n; j++ {
  			dp[i][j] = s[i] == s[j] && dp[i+1][j-1]
  		}
  	}
  	var dfs func(int) int
  	dfs = func(i int) int {
  		if i >= n {
  			return 0
  		}
  		if f[i] != -1 {
  			return f[i]
  		}
  		ans := dfs(i + 1)
  		for j := i + k - 1; j < n; j++ {
  			if dp[i][j] {
  				ans = max(ans, 1+dfs(j+1))
  			}
  		}
  		f[i] = ans
  		return ans
  	}
  	return dfs(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).

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