3876. Construct Uniform Parity Array II

Difficulty: Medium

Topics: Senior, Array, Math, Weekly Contest 494

You are given an array nums1 of n distinct integers.

You want to construct another array nums2 of length n such that the elements in nums2 are either all odd or all even.

For each index i, you must choose exactly one of the following (in any order):

nums2[i] = nums1[i]
nums2[i] = nums1[i] - nums1[j], for an index j != i, such that nums1[i] - nums1[j] >= 1

Return true if it is possible to construct such an array, otherwise return false.

Example 1:

Input: nums1 = [1,4,7]
Output: true
Explanation:
- Set nums2[0] = nums1[0] = 1.
- Set nums2[1] = nums1[1] - nums1[0] = 4 - 1 = 3.
- Set nums2[2] = nums1[2] = 7.
- nums2 = [1, 3, 7], and all elements are odd. Thus, the answer is true.

Example 2:

Input: nums1 = [2,3]
Output: false
Explanation: It is not possible to construct nums2 such that all elements have the same parity. Thus, the answer is false.

Example 3:

Input: nums1 = [4,6]
Output: true
Explanation:
- Set nums2[0] = nums1[0] = 4.
- Set nums2[1] = nums1[1] = 6.
- nums2 = [4, 6], and all elements are even. Thus, the answer is true.

Example 4:

Input: nums1 = [11,16]
Output: true

Example 5:

Input: nums1 = [2,4,6]
Output: true

Example 6:

Input: nums1 = [1,3,5]
Output: true

Example 7:

Input: nums1 = [1,2]
Output: true

Example 8:

Input: nums1 = [5,10]
Output: true

Example 9:

Input: nums1 = [8,9]
Output: false

Example 10:

Input: nums1 = [3,6,9]
Output: true

Constraints:

1 <= n == nums1.length <= 10⁵
1 <= nums1[i] <= 10⁹
nums1 consists of distinct integers.

Hint:

Try fixing the parity to either all even or all odd.
Use the smallest odd/even element if a subtraction is needed to match the chosen parity.

Solution:

We solve the problem by checking both possible target parities (all odd or all even) for nums2. For each target parity, we determine whether every element in nums1 can either stay as is (if it already matches the target parity) or be transformed by subtracting the smallest number of the opposite parity, as long as the result remains positive. If either parity works, we return true; otherwise, false.

Approach

Fix target parity: Try to make all elements in nums2 odd, then try to make all even.
Key observation: To change the parity of an element, we must subtract an odd number (since odd – odd = even, even – odd = odd). The best candidate for subtraction is the smallest odd number in nums1, because using a larger odd number would make the result smaller and could go below 1, making it invalid.
Check validity for each parity:
- For all odd: Any even number must be subtracted by the smallest odd number minOdd, and the result must be at least 1.
- For all even: Any odd number must be subtracted by minOdd (an odd number) to become even, and the result must be at least 1.
Special case: If there are no odd numbers, all elements are already even, so all-even is trivially possible.

Let's implement this solution in PHP: 3876. Construct Uniform Parity Array II

<?php
/**
 * @param Integer[] $nums1
 * @return Boolean
 */
function uniformArray(array $nums1): bool
{
    ...
    ...
    ...
    /**
     * go to ./solution.php
     */
}

// Test cases
echo uniformArray([1,4,7]) .  "\n";     // Output: true
echo uniformArray([2,3]) .  "\n";       // Output: false
echo uniformArray([4,6]) .  "\n";       // Output: true
echo uniformArray([11,16]) .  "\n";     // Output: true
echo uniformArray([2,4,6]) .  "\n";     // Output: true
echo uniformArray([1,3,5]) .  "\n";     // Output: true
echo uniformArray([1,2]) .  "\n";       // Output: true
echo uniformArray([5,10]) .  "\n";      // Output: true
echo uniformArray([8,9]) .  "\n";       // Output: false
echo uniformArray([3,6,9]) .  "\n";     // Output: true
?>

Explanation:

We first find the minimum odd number in nums1. This is crucial because subtracting any odd number preserves the parity change but the smallest one gives the largest possible result, avoiding negative or zero values.
To check all odd:
- Iterate through nums1.
- If a number is already odd, we can keep it as is.
- If it’s even, we must subtract minOdd. If minOdd doesn’t exist or num - minOdd < 1, this parity fails.
To check all even:
- If no odd numbers exist, all are even → success.
- Otherwise, for each odd number, subtract minOdd; if the result is less than 1, this parity fails.
If either check passes, return true; else false.

Complexity Analysis

Time Complexity: O(n) — We scan the array a constant number of times (to find min odd and to validate each parity).
Space Complexity: O(1) — We use only a few extra variables.

Contact Links

If you found this series helpful, please consider giving the repository a star on GitHub or sharing the post on your favorite social networks 😍. Your support would mean a lot to me!