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3875. Construct Uniform Parity Array I

Description

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

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

 

Example 1:

Input: nums1 = [2,3]

Output: true

Explanation:

  • Choose nums2[0] = nums1[0] - nums1[1] = 2 - 3 = -1.
  • Choose nums2[1] = nums1[1] = 3.
  • nums2 = [-1, 3], and both elements are odd. Thus, the answer is true​​​​​​​.

Example 2:

Input: nums1 = [4,6]

Output: true

Explanation:​​​​​​​

  • Choose nums2[0] = nums1[0] = 4.
  • Choose nums2[1] = nums1[1] = 6.
  • nums2 = [4, 6], and all elements are even. Thus, the answer is true.

 

Constraints:

  • 1 <= n == nums1.length <= 100
  • 1 <= nums1[i] <= 100
  • nums1 consists of distinct integers.

Solutions

Solution 1: Brain Teaser

If all elements in $\textit{nums1}$ are either all odd or all even, we can directly set $\textit{nums2}$ equal to $\textit{nums1}$, which satisfies the condition.

If $\textit{nums1}$ contains both odd and even numbers, we can set each element of $\textit{nums2}$ to the current element of $\textit{nums1}$ minus some element in $\textit{nums1}$ with different parity. Since odd minus even and even minus odd both yield an odd number, all elements of $\textit{nums2}$ will be odd, satisfying the condition.

Therefore, regardless of whether the elements in $\textit{nums1}$ are all odd, all even, or a mix of both, we can always construct a valid $\textit{nums2}$. Thus the answer is always $\text{true}$.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

  • class Solution {
    public boolean uniformArray(int[] nums1) {
        return true;
    }
}


  • class Solution {
public:
    bool uniformArray(vector<int>& nums1) {
        return true;
    }
};


  • class Solution:
    def uniformArray(self, nums1: list[int]) -> bool:
        return True


  • func uniformArray(nums1 []int) bool {
	return true
}


  • function uniformArray(nums1: number[]): boolean {
    return true;
}

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