2911. Minimum Changes to Make K Semi-palindromes

Description

Given a string s and an integer k, partition s into k substrings such that the sum of the number of letter changes required to turn each substring into a semi-palindrome is minimized.

Return an integer denoting the minimum number of letter changes required.

Notes

A string is a palindrome if it can be read the same way from left to right and right to left.
A string with a length of len is considered a semi-palindrome if there exists a positive integer d such that 1 <= d < len and len % d == 0, and if we take indices that have the same modulo by d, they form a palindrome. For example, "aa", "aba", "adbgad", and, "abab" are semi-palindrome and "a", "ab", and, "abca" are not.
A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "abcac", k = 2
Output: 1
Explanation: We can divide s into substrings "ab" and "cac". The string "cac" is already a semi-palindrome. If we change "ab" to "aa", it becomes a semi-palindrome with d = 1.
It can be shown that there is no way to divide the string "abcac" into two semi-palindrome substrings. Therefore, the answer would be at least 1.

Example 2:

Input: s = "abcdef", k = 2
Output: 2
Explanation: We can divide it into substrings "abc" and "def". Each of the substrings "abc" and "def" requires one change to become a semi-palindrome, so we need 2 changes in total to make all substrings semi-palindrome.
It can be shown that we cannot divide the given string into two substrings in a way that it would require less than 2 changes.

Example 3:

Input: s = "aabbaa", k = 3
Output: 0
Explanation: We can divide it into substrings "aa", "bb" and "aa".
The strings "aa" and "bb" are already semi-palindromes. Thus, the answer is zero.

Constraints:

2 <= s.length <= 200
1 <= k <= s.length / 2
s consists only of lowercase English letters.

Solutions

Optimization: Preprocessing Divisors List

We can calculate the divisors list for each length, which can save a lot of division operations in the inner loop.

class Solution {
    public int minimumChanges(String s, int k) {
        int n = s.length();
        int[][] g = new int[n + 1][n + 1];
        int[][] f = new int[n + 1][k + 1];
        final int inf = 1 << 30;
        for (int i = 0; i <= n; ++i) {
            Arrays.fill(g[i], inf);
            Arrays.fill(f[i], inf);
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                int m = j - i + 1;
                for (int d = 1; d < m; ++d) {
                    if (m % d == 0) {
                        int cnt = 0;
                        for (int l = 0; l < m; ++l) {
                            int r = (m / d - 1 - l / d) * d + l % d;
                            if (l >= r) {
                                break;
                            }
                            if (s.charAt(i - 1 + l) != s.charAt(i - 1 + r)) {
                                ++cnt;
                            }
                        }
                        g[i][j] = Math.min(g[i][j], cnt);
                    }
                }
            }
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                for (int h = 0; h < i - 1; ++h) {
                    f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h + 1][i]);
                }
            }
        }
        return f[n][k];
    }
}

class Solution {
public:
    int minimumChanges(string s, int k) {
        int n = s.size();
        int g[n + 1][n + 1];
        int f[n + 1][k + 1];
        memset(g, 0x3f, sizeof(g));
        memset(f, 0x3f, sizeof(f));
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = i; j <= n; ++j) {
                int m = j - i + 1;
                for (int d = 1; d < m; ++d) {
                    if (m % d == 0) {
                        int cnt = 0;
                        for (int l = 0; l < m; ++l) {
                            int r = (m / d - 1 - l / d) * d + l % d;
                            if (l >= r) {
                                break;
                            }
                            if (s[i - 1 + l] != s[i - 1 + r]) {
                                ++cnt;
                            }
                        }
                        g[i][j] = min(g[i][j], cnt);
                    }
                }
            }
        }
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= k; ++j) {
                for (int h = 0; h < i - 1; ++h) {
                    f[i][j] = min(f[i][j], f[h][j - 1] + g[h + 1][i]);
                }
            }
        }
        return f[n][k];
    }
};

class Solution:
    def minimumChanges(self, s: str, k: int) -> int:
        n = len(s)
        g = [[inf] * (n + 1) for _ in range(n + 1)]
        for i in range(1, n + 1):
            for j in range(i, n + 1):
                m = j - i + 1
                for d in range(1, m):
                    if m % d == 0:
                        cnt = 0
                        for l in range(m):
                            r = (m // d - 1 - l // d) * d + l % d
                            if l >= r:
                                break
                            if s[i - 1 + l] != s[i - 1 + r]:
                                cnt += 1
                        g[i][j] = min(g[i][j], cnt)

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

func minimumChanges(s string, k int) int {
	n := len(s)
	g := make([][]int, n+1)
	f := make([][]int, n+1)
	const inf int = 1 << 30
	for i := range g {
		g[i] = make([]int, n+1)
		f[i] = make([]int, k+1)
		for j := range g[i] {
			g[i][j] = inf
		}
		for j := range f[i] {
			f[i][j] = inf
		}
	}
	f[0][0] = 0
	for i := 1; i <= n; i++ {
		for j := i; j <= n; j++ {
			m := j - i + 1
			for d := 1; d < m; d++ {
				if m%d == 0 {
					cnt := 0
					for l := 0; l < m; l++ {
						r := (m/d-1-l/d)*d + l%d
						if l >= r {
							break
						}
						if s[i-1+l] != s[i-1+r] {
							cnt++
						}
					}
					g[i][j] = min(g[i][j], cnt)
				}
			}
		}
	}
	for i := 1; i <= n; i++ {
		for j := 1; j <= k; j++ {
			for h := 0; h < i-1; h++ {
				f[i][j] = min(f[i][j], f[h][j-1]+g[h+1][i])
			}
		}
	}
	return f[n][k]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

function minimumChanges(s: string, k: number): number {
    const n = s.length;
    const g = Array.from({ length: n + 1 }, () => Array.from({ length: n + 1 }, () => Infinity));
    const f = Array.from({ length: n + 1 }, () => Array.from({ length: k + 1 }, () => Infinity));
    f[0][0] = 0;
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= n; ++j) {
            const m = j - i + 1;
            for (let d = 1; d < m; ++d) {
                if (m % d === 0) {
                    let cnt = 0;
                    for (let l = 0; l < m; ++l) {
                        const r = (((m / d) | 0) - 1 - ((l / d) | 0)) * d + (l % d);
                        if (l >= r) {
                            break;
                        }
                        if (s[i - 1 + l] !== s[i - 1 + r]) {
                            ++cnt;
                        }
                    }
                    g[i][j] = Math.min(g[i][j], cnt);
                }
            }
        }
    }
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= k; ++j) {
            for (let h = 0; h < i - 1; ++h) {
                f[i][j] = Math.min(f[i][j], f[h][j - 1] + g[h + 1][i]);
            }
        }
    }
    return f[n][k];
}

2911 - Minimum Changes to Make K Semi-palindromes

2911. Minimum Changes to Make K Semi-palindromes

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2911. Minimum Changes to Make K Semi-palindromes

Description

Solutions

All Problems

All Solutions