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Leetcode #2193: Minimum Number of Moves to Make Palindrome

In this guide, we solve Leetcode #2193 Minimum Number of Moves to Make Palindrome in Python and focus on the core idea that makes the solution efficient.

You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.

Leetcode

Problem Statement

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.

Quick Facts

  • Difficulty: Hard
  • Premium: No
  • Tags: Greedy, Binary Indexed Tree, Two Pointers, String

Intuition

The constraints hint that we can reason about two ends of the data at once, which is perfect for a two-pointer scan.

Moving one pointer at a time keeps the invariant intact and avoids nested loops.

Approach

Place pointers at the left and right ends and move them based on the comparison or target condition.

This yields a clean linear pass after any required sorting.

Steps:

  • Set left and right pointers.
  • Move a pointer based on the condition.
  • Update the best answer while scanning.

Example

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.

Python Solution

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

Complexity

The time complexity is O(n) (after optional sort O(n log n)). The space complexity is O(1).

Edge Cases and Pitfalls

Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.

Summary

This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.


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