How do you solve Construct Uniform Parity Array II on LeetCode?

Construct Uniform Parity Array II (LeetCode #3876) 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: Identify smallest odd/even for parity adjustments.

What is the time complexity of Construct Uniform Parity Array II?

The optimal solution for Construct Uniform Parity Array II runs in O(n) time and uses O(n) space. Single pass to classify numbers into odds and evens, leading to linear time complexity.

What is the brute force solution for Construct Uniform Parity Array II?

The brute force approach for Construct Uniform Parity Array II is Brute Force, with O(n²) time complexity and O(1) space complexity. Check all combinations of elements from nums1 to see if we can form an array nums2 with uniform parity. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Construct Uniform Parity Array II?

Construct Uniform Parity Array II uses the following concepts: Array, Math. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Construct Uniform Parity Array II?

Clarify the problem requirements. Discuss edge cases early. Think about time complexity before coding.

What are common mistakes when solving Construct Uniform Parity Array II?

Not checking both odd and even cases. Overlooking the condition for valid subtraction.

#3876

Construct Uniform Parity Array II

Medium

ArrayMathHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

To construct nums2, we can fix the parity to either all odd or all even. By using the smallest odd/even element for subtraction, we can ensure all elements match the desired parity.

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Algorithm

3 steps

1Step 1: Identify the smallest odd and even numbers in nums1.
2Step 2: Check if we can create an all-odd array using the smallest even number for subtraction or vice versa.
3Step 3: Return true if either condition is satisfied.

solution.py6 lines

1def canConstruct(nums1):
2    odds = [x for x in nums1 if x % 2 != 0]
3    evens = [x for x in nums1 if x % 2 == 0]
4    if odds and (min(odds) - min(evens) >= 1): return True
5    if evens and (min(evens) - min(odds) >= 1): return True
6    return False

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Complexity note: Single pass to classify numbers into odds and evens, leading to linear time complexity.

1Identify smallest odd/even for parity adjustments.
2Uniform parity can be achieved through careful selection.

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