What is the time complexity of Minimum Operations to Make Array Parity Alternating?

The optimal solution for Minimum Operations to Make Array Parity Alternating runs in O(n) time and uses O(1) space. The complexity is O(n) because we only traverse the array once to calculate operations and find min/max values.

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

The brute force approach for Minimum Operations to Make Array Parity Alternating is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible configurations of the array and count operations needed to achieve parity alternating. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

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

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

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

Understand parity and how it affects array structure. Practice similar array manipulation problems. Explain your thought process clearly during coding.

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

Forgetting to check both parity patterns. Not accounting for max - min correctly.

#3854

Minimum Operations to Make Array Parity Alternating

Medium

ArrayGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

Instead of generating patterns, directly calculate the operations needed to adjust each element to fit either of the two required patterns. This reduces unnecessary computations.

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Algorithm

3 steps

1Step 1: Count operations needed to convert nums to fit the 'even, odd, even...' pattern.
2Step 2: Count operations needed to convert nums to fit the 'odd, even, odd...' pattern.
3Step 3: Return the minimum operations and calculate max(nums) - min(nums) for the resulting array.

solution.py10 lines

1def min_operations(nums):
2    even_ops, odd_ops = 0, 0
3    for i, num in enumerate(nums):
4        if i % 2 == 0:
5            even_ops += num % 2
6            odd_ops += (num + 1) % 2
7        else:
8            even_ops += (num + 1) % 2
9            odd_ops += num % 2
10    return [min(even_ops, odd_ops), max(nums) - min(nums)]

ℹ

Complexity note: The complexity is O(n) because we only traverse the array once to calculate operations and find min/max values.

1Parity must alternate between even and odd.
2Each element's parity can be fixed with a single operation.

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