LeetCode 2193. Minimum Number of Moves to Make Palindrome Solution in Java, C++, Python & Go | Explanation + Code

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2193. Minimum Number of Moves to Make Palindrome

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

PythonJavaC++Go
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(code-box)

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