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1278. Palindrome Partitioning III

Description

You are given a string s containing lowercase letters and an integer k. You need to :

  • First, change some characters of s to other lowercase English letters.
  • Then divide s into k non-empty disjoint substrings such that each substring is a palindrome.

Return the minimal number of characters that you need to change to divide the string.

 

Example 1:

Input: s = "abc", k = 2
Output: 1
Explanation: You can split the string into "ab" and "c", and change 1 character in "ab" to make it palindrome.

Example 2:

Input: s = "aabbc", k = 3
Output: 0
Explanation: You can split the string into "aa", "bb" and "c", all of them are palindrome.

Example 3:

Input: s = "leetcode", k = 8
Output: 0

 

Constraints:

  • 1 <= k <= s.length <= 100.
  • s only contains lowercase English letters.

Solutions

  • class Solution {
    public int palindromePartition(String s, int k) {
        int n = s.length();
        int[][] g = new int[n][n];
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                g[i][j] = s.charAt(i) != s.charAt(j) ? 1 : 0;
                if (i + 1 < j) {
                    g[i][j] += g[i + 1][j - 1];
                }
            }
        }
        int[][] f = new int[n + 1][k + 1];
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= Math.min(i, k); ++j) {
                if (j == 1) {
                    f[i][j] = g[0][i - 1];
                } else {
                    f[i][j] = 10000;
                    for (int h = j - 1; h < i; ++h) {
                        f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h][i - 1]);
                    }
                }
            }
        }
        return f[n][k];
    }
}

  • class Solution {
public:
    int palindromePartition(string s, int k) {
        int n = s.size();
        vector<vector<int>> g(n, vector<int>(n));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                g[i][j] = s[i] != s[j] ? 1 : 0;
                if (i + 1 < j) g[i][j] += g[i + 1][j - 1];
            }
        }
        vector<vector<int>> f(n + 1, vector<int>(k + 1));
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= min(i, k); ++j) {
                if (j == 1) {
                    f[i][j] = g[0][i - 1];
                } else {
                    f[i][j] = 10000;
                    for (int h = j - 1; h < i; ++h) {
                        f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1]);
                    }
                }
            }
        }
        return f[n][k];
    }
};

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

        f = [[0] * (k + 1) for _ in range(n + 1)]
        for i in range(1, n + 1):
            for j in range(1, min(i, k) + 1):
                if j == 1:
                    f[i][j] = g[0][i - 1]
                else:
                    f[i][j] = inf
                    for h in range(j - 1, i):
                        f[i][j] = min(f[i][j], f[h][j - 1] + g[h][i - 1])
        return f[n][k]


  • func palindromePartition(s string, k int) int {
	n := len(s)
	g := make([][]int, n)
	for i := range g {
		g[i] = make([]int, n)
	}
	for i := n - 1; i >= 0; i-- {
		for j := 1; j < n; j++ {
			if s[i] != s[j] {
				g[i][j] = 1
			}
			if i+1 < j {
				g[i][j] += g[i+1][j-1]
			}
		}
	}
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, k+1)
	}
	for i := 1; i <= n; i++ {
		for j := 1; j <= min(i, k); j++ {
			if j == 1 {
				f[i][j] = g[0][i-1]
			} else {
				f[i][j] = 100000
				for h := j - 1; h < i; h++ {
					f[i][j] = min(f[i][j], f[h][j-1]+g[h][i-1])
				}
			}
		}
	}
	return f[n][k]
}

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