How do you solve Minimum Number of Moves to Make Palindrome on LeetCode?

Minimum Number of Moves to Make Palindrome (LeetCode #2193) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Greedy strategies often yield optimal solutions for problems involving order and arrangement.

What is the brute force solution for Minimum Number of Moves to Make Palindrome?

The brute force approach for Minimum Number of Moves to Make Palindrome is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible way to swap adjacent characters to form a palindrome. This method is straightforward but inefficient, as it may require many swaps to achieve the desired result. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Moves to Make Palindrome?

Minimum Number of Moves to Make Palindrome uses the following concepts: Two Pointers, String, Greedy, Binary Indexed Tree. The recommended patterns to study are: Two Pointers, Greedy, String Manipulation.

What are the interview tips for Minimum Number of Moves to Make Palindrome?

Always clarify the problem constraints and ensure you understand the input/output requirements. Discuss your thought process and approach with the interviewer; they may provide hints or guidance. Practice explaining your code as you write it, as this simulates the interview environment.

What are common mistakes when solving Minimum Number of Moves to Make Palindrome?

Failing to account for adjacent swaps and their implications. Not recognizing when to stop searching for matches, leading to unnecessary iterations.

#2193

Minimum Number of Moves to Make Palindrome

Hard

Two PointersStringGreedyBinary Indexed TreeTwo PointersGreedyString Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a greedy approach with two pointers to minimize the number of swaps needed to form a palindrome. By always trying to match characters from the outer ends towards the center, we can efficiently determine the minimum moves.

⚙️

Algorithm

6 steps

1Step 1: Initialize two pointers, left at the start and right at the end of the string.
2Step 2: While left is less than right, check if characters at both pointers match.
3Step 3: If they match, move both pointers inward.
4Step 4: If they do not match, find the nearest matching character for the left pointer by moving the right pointer leftward.
5Step 5: Count the number of swaps needed to bring the matching character next to the left pointer.
6Step 6: Increment the move counter and continue until the pointers meet.

solution.py29 lines

1# Full working Python code
2
3def min_moves_to_palindrome(s):
4    s = list(s)
5    moves = 0
6    left, right = 0, len(s) - 1
7    while left < right:
8        if s[left] == s[right]:
9            left += 1
10            right -= 1
11        else:
12            r = right
13            while r > left and s[left] != s[r]:
14                r -= 1
15            if r == left:
16                # No match found, swap left with left + 1
17                s[left], s[left + 1] = s[left + 1], s[left]
18                moves += 1
19            else:
20                # Move the matching character to the right
21                for i in range(r, right):
22                    s[i], s[i + 1] = s[i + 1], s[i]
23                    moves += 1
24                left += 1
25                right -= 1
26    return moves
27
28# Example usage
29print(min_moves_to_palindrome('letelt'))  # Output: 2

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string a limited number of times, making it efficient. The space complexity is O(n) due to the character array used to manipulate the string.

1Greedy strategies often yield optimal solutions for problems involving order and arrangement.
2Two pointers can effectively reduce the complexity of problems involving pairs or sequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.