README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Hard	https://github.com/doocs/leetcode/edit/main/solution/2100-2199/2193.Minimum%20Number%20of%20Moves%20to%20Make%20Palindrome/README_EN.md	2090	Biweekly Contest 73 Q4	Greedy Binary Indexed Tree Two Pointers String

2193. Minimum Number of Moves to Make Palindrome

中文文档

Description

You are given a string s consisting only of lowercase English letters.

In one move, you can select any two adjacent characters of s and swap them.

Return the minimum number of moves needed to make s a palindrome.

Note that the input will be generated such that s can always be converted to a palindrome.

 

Example 1:

Input: s = "aabb"
Output: 2
Explanation:
We can obtain two palindromes from s, "abba" and "baab". 
- We can obtain "abba" from s in 2 moves: "aabb" -> "abab" -> "abba".
- We can obtain "baab" from s in 2 moves: "aabb" -> "abab" -> "baab".
Thus, the minimum number of moves needed to make s a palindrome is 2.

Example 2:

Input: s = "letelt"
Output: 2
Explanation:
One of the palindromes we can obtain from s in 2 moves is "lettel".
One of the ways we can obtain it is "letelt" -> "letetl" -> "lettel".
Other palindromes such as "tleelt" can also be obtained in 2 moves.
It can be shown that it is not possible to obtain a palindrome in less than 2 moves.

 

Constraints:

• 1 <= s.length <= 2000
• s consists only of lowercase English letters.
• s can be converted to a palindrome using a finite number of moves.

Solutions

Solution 1

Python3

class Solution:
    def minMovesToMakePalindrome(self, s: str) -> int:
        cs = list(s)
        ans, n = 0, len(s)
        i, j = 0, n - 1
        while i < j:
            even = False
            for k in range(j, i, -1):
                if cs[i] == cs[k]:
                    even = True
                    while k < j:
                        cs[k], cs[k + 1] = cs[k + 1], cs[k]
                        k += 1
                        ans += 1
                    j -= 1
                    break
            if not even:
                ans += n // 2 - i
            i += 1
        return ans

Java

class Solution {
    public int minMovesToMakePalindrome(String s) {
        int n = s.length();
        int ans = 0;
        char[] cs = s.toCharArray();
        for (int i = 0, j = n - 1; i < j; ++i) {
            boolean even = false;
            for (int k = j; k != i; --k) {
                if (cs[i] == cs[k]) {
                    even = true;
                    for (; k < j; ++k) {
                        char t = cs[k];
                        cs[k] = cs[k + 1];
                        cs[k + 1] = t;
                        ++ans;
                    }
                    --j;
                    break;
                }
            }
            if (!even) {
                ans += n / 2 - i;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minMovesToMakePalindrome(string s) {
        int n = s.size();
        int ans = 0;
        for (int i = 0, j = n - 1; i < j; ++i) {
            bool even = false;
            for (int k = j; k != i; --k) {
                if (s[i] == s[k]) {
                    even = true;
                    for (; k < j; ++k) {
                        swap(s[k], s[k + 1]);
                        ++ans;
                    }
                    --j;
                    break;
                }
            }
            if (!even) ans += n / 2 - i;
        }
        return ans;
    }
};

Go

func minMovesToMakePalindrome(s string) int {
	cs := []byte(s)
	ans, n := 0, len(s)
	for i, j := 0, n-1; i < j; i++ {
		even := false
		for k := j; k != i; k-- {
			if cs[i] == cs[k] {
				even = true
				for ; k < j; k++ {
					cs[k], cs[k+1] = cs[k+1], cs[k]
					ans++
				}
				j--
				break
			}
		}
		if !even {
			ans += n/2 - i
		}
	}
	return ans
}