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1745. Palindrome Partitioning IV

Description

Given a string s, return true if it is possible to split the string s into three non-empty palindromic substrings. Otherwise, return false.​​​​​

A string is said to be palindrome if it the same string when reversed.

 

Example 1:

Input: s = "abcbdd"
Output: true
Explanation: "abcbdd" = "a" + "bcb" + "dd", and all three substrings are palindromes.

Example 2:

Input: s = "bcbddxy"
Output: false
Explanation: s cannot be split into 3 palindromes.

 

Constraints:

  • 3 <= s.length <= 2000
  • s​​​​​​ consists only of lowercase English letters.

Solutions

Solution 1

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class Solution:
    def checkPartitioning(self, s: str) -> bool:
        n = len(s)
        g = [[True] * n for _ in range(n)]
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n):
                g[i][j] = s[i] == s[j] and (i + 1 == j or g[i + 1][j - 1])
        for i in range(n - 2):
            for j in range(i + 1, n - 1):
                if g[0][i] and g[i + 1][j] and g[j + 1][-1]:
                    return True
        return False

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class Solution {
    public boolean checkPartitioning(String s) {
        int n = s.length();
        boolean[][] g = new boolean[n][n];
        for (var e : g) {
            Arrays.fill(e, true);
        }
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = s.charAt(i) == s.charAt(j) && (i + 1 == j || g[i + 1][j - 1]);
            }
        }
        for (int i = 0; i < n - 2; ++i) {
            for (int j = i + 1; j < n - 1; ++j) {
                if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
                    return true;
                }
            }
        }
        return false;
    }
}

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class Solution {
public:
    bool checkPartitioning(string s) {
        int n = s.size();
        vector<vector<bool>> g(n, vector<bool>(n, true));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i + 1; j < n; ++j) {
                g[i][j] = s[i] == s[j] && (i + 1 == j || g[i + 1][j - 1]);
            }
        }
        for (int i = 0; i < n - 2; ++i) {
            for (int j = i + 1; j < n - 1; ++j) {
                if (g[0][i] && g[i + 1][j] && g[j + 1][n - 1]) {
                    return true;
                }
            }
        }
        return false;
    }
};

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func checkPartitioning(s string) bool {
    n := len(s)
    g := make([][]bool, n)
    for i := range g {
        g[i] = make([]bool, n)
        for j := range g[i] {
            g[i][j] = true
        }
    }
    for i := n - 1; i >= 0; i-- {
        for j := i + 1; j < n; j++ {
            g[i][j] = s[i] == s[j] && (i+1 == j || g[i+1][j-1])
        }
    }
    for i := 0; i < n-2; i++ {
        for j := i + 1; j < n-1; j++ {
            if g[0][i] && g[i+1][j] && g[j+1][n-1] {
                return true
            }
        }
    }
    return false
}

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