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

Construct Uniform Parity Array I (LeetCode #3875) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Parity of numbers determines the possibility of uniformity.

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

The optimal solution for Construct Uniform Parity Array I runs in O(n) time and uses O(1) space. The complexity is O(n) because we only need to traverse the array once to count evens and odds.

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

The brute force approach for Construct Uniform Parity Array I is Brute Force, with O(n²) time complexity and O(1) space complexity. We can try every possible combination of selecting elements from nums1 to form nums2. This involves checking all pairs of indices to see if we can create a uniform parity array. The optimal Optimal Solution approach improves this to O(n).

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

Construct Uniform Parity Array I 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 I?

Clarify the problem constraints. Think about edge cases like minimum and maximum values. Discuss your thought process and reasoning.

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

Assuming all elements must be used in nums2. Overlooking the parity of differences.

#3875

Construct Uniform Parity Array I

Easy

ArrayMathHash 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)

To construct a uniform parity array, we can leverage the fact that the difference between any two distinct integers will always yield an odd or even result based on their parities. If we have at least one even and one odd number, we can always create a uniform array.

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Algorithm

3 steps

1Step 1: Count the number of even and odd integers in nums1.
2Step 2: If there are at least one even and one odd integer, return true.
3Step 3: If all integers are even or all are odd, return true.

solution.py4 lines

1def canConstruct(nums1):
2    evens = sum(1 for x in nums1 if x % 2 == 0)
3    odds = len(nums1) - evens
4    return evens == 0 or odds == 0

ℹ

Complexity note: The complexity is O(n) because we only need to traverse the array once to count evens and odds.

1Parity of numbers determines the possibility of uniformity.
2The difference between two numbers affects the resulting parity.

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