How do you solve Minimum Adjacent Swaps to Alternate Parity on LeetCode?

Minimum Adjacent Swaps to Alternate Parity (LeetCode #3587) 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: Count even and odd elements.

What is the time complexity of Minimum Adjacent Swaps to Alternate Parity?

The optimal solution for Minimum Adjacent Swaps to Alternate Parity runs in O(n) time and uses O(n) space. The optimal approach runs in linear time by counting and calculating swaps based on the counts, avoiding permutations.

What is the brute force solution for Minimum Adjacent Swaps to Alternate Parity?

The brute force approach for Minimum Adjacent Swaps to Alternate Parity is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach tries all possible swaps to find a valid arrangement. It checks every possible combination, which can be inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Adjacent Swaps to Alternate Parity?

Minimum Adjacent Swaps to Alternate Parity uses the following concepts: Array, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Adjacent Swaps to Alternate Parity?

Explain your thought process clearly. Discuss edge cases upfront. Practice similar problems for familiarity.

What are common mistakes when solving Minimum Adjacent Swaps to Alternate Parity?

Forgetting to check the parity difference. Not considering both valid patterns.

#3587

Minimum Adjacent Swaps to Alternate Parity

Medium

ArrayGreedyHash MapArray

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)

Count the even and odd numbers. If their counts differ by more than one, return -1. Otherwise, compute the minimum swaps needed for two valid patterns: starting with even or odd.

⚙️

Algorithm

3 steps

1Step 1: Count even and odd numbers in nums.
2Step 2: If abs(evenCnt - oddCnt) > 1, return -1.
3Step 3: Calculate swaps needed for both patterns (even-odd and odd-even) and return the minimum.

solution.py14 lines

1def min_swaps_optimal(nums):
2    evens = [x for x in nums if x % 2 == 0]
3    odds = [x for x in nums if x % 2 != 0]
4    if abs(len(evens) - len(odds)) > 1:
5        return -1
6    return min(count_swaps(nums, evens, odds, start=0), count_swaps(nums, odds, evens, start=1))
7
8def count_swaps(nums, first, second, start):
9    swaps = 0
10    for i in range(len(nums)):
11        expected = first[i // 2] if i % 2 == start else second[i // 2]
12        if nums[i] != expected:
13            swaps += 1
14    return swaps

ℹ

Complexity note: The optimal approach runs in linear time by counting and calculating swaps based on the counts, avoiding permutations.

1Count even and odd elements.
2Check parity difference before processing.

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