How do you solve Minimum Operations to Make the Array Alternating on LeetCode?

Minimum Operations to Make the Array Alternating (LeetCode #2170) 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: Maximize the frequency of elements at even and odd indices to minimize changes.

What is the brute force solution for Minimum Operations to Make the Array Alternating?

The brute force approach for Minimum Operations to Make the Array Alternating is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible way to change elements in the array to achieve the alternating pattern. This is simple but inefficient as it may require checking all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make the Array Alternating?

Minimum Operations to Make the Array Alternating uses the following concepts: Array, Hash Table, Greedy, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Operations to Make the Array Alternating?

Always discuss your thought process and approach before jumping into coding. Consider edge cases and constraints early in the discussion. Practice explaining your code and the reasoning behind your choices.

What are common mistakes when solving Minimum Operations to Make the Array Alternating?

Not considering the case where the most frequent elements at even and odd indices are the same. Failing to account for edge cases where the array length is very small.

#2170

Minimum Operations to Make the Array Alternating

Medium

ArrayHash TableGreedyCountingHash 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)

The optimal approach leverages the frequency of numbers at even and odd indices to minimize changes. By maximizing the count of the most frequent elements at these indices, we can reduce the number of operations needed.

⚙️

Algorithm

3 steps

1Step 1: Create two frequency maps: one for elements at even indices and one for odd indices.
2Step 2: Identify the most frequent elements in both maps and their counts.
3Step 3: Calculate the minimum operations needed by considering the most frequent elements at even and odd indices, ensuring they are different.

solution.py18 lines

1# Full working Python code
2from collections import Counter
3
4def min_operations_optimal(nums):
5    even_count = Counter(nums[i] for i in range(0, len(nums), 2))
6    odd_count = Counter(nums[i] for i in range(1, len(nums), 2))
7    even_most_common = even_count.most_common(2)
8    odd_most_common = odd_count.most_common(2)
9    even1, even1_count = even_most_common[0] if even_most_common else (None, 0)
10    even2, even2_count = even_most_common[1] if len(even_most_common) > 1 else (None, 0)
11    odd1, odd1_count = odd_most_common[0] if odd_most_common else (None, 0)
12    odd2, odd2_count = odd_most_common[1] if len(odd_most_common) > 1 else (None, 0)
13
14    if even1 != odd1:
15        return len(nums) - (even1_count + odd1_count)
16    else:
17        return min(len(nums) - (even1_count + odd2_count), len(nums) - (even2_count + odd1_count))
18

ℹ

Complexity note: This complexity is linear because we only traverse the array a couple of times to build frequency maps and then sort them, which is efficient given the constraints.

1Maximize the frequency of elements at even and odd indices to minimize changes.
2Ensure that the most frequent elements at even and odd indices are different.

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