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2472. Maximum Number of Non-overlapping Palindrome Substrings

Description

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.

Solutions

Solution 1: Preprocessing + Memoization Search

First, preprocess the string $s$ to get $dp[i][j]$, which represents whether the substring $s[i,..j]$ is a palindrome.

Then, define a function $dfs(i)$ to represent the maximum number of non-overlapping palindrome substrings that can be selected from the substring $s[i,..]$, i.e.,

\[\begin{aligned} dfs(i) &= \begin{cases} 0, & i \geq n \\ \max\{dfs(i + 1), \max_{j \geq i + k - 1} \{dfs(j + 1) + 1\}\}, & i < n \end{cases} \end{aligned}\]

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 {
        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)
    }
    

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