Special Array I

Problem Statement

In this problem, we define a "special" array based on the parity (evenness or oddness) of its adjacent elements. Specifically, an array is termed as special if each adjacent pair of elements consists of one even and one odd number. The task is to assess whether a given array of integers, nums, fits this definition of special. The function should return true if the array is special, and false otherwise.

Examples

Example 1

Input:

nums = [1]

Output:

true

Explanation:

There is only one element. So the answer is `true`.

Example 2

Input:

nums = [2,1,4]

Output:

true

Explanation:

There is only two pairs: `(2,1)` and `(1,4)`, and both of them contain numbers with different parity. So the answer is `true`.

Example 3

Input:

nums = [4,3,1,6]

Output:

false

Explanation:

`nums[1]` and `nums[2]` are both odd. So the answer is `false`.

Constraints

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100

Approach and Intuition

To determine if the provided array nums is special, we need to evaluate the parity of each pair of adjacent elements. Here is a step-by-step plan to achieve this:

1. If the length of the array is 1, we immediately return true since there are no adjacent pairs to compare, as seen in example 1.
2. Traverse the array element by element starting from the first till the second last element.
3. For each element nums[i], check the parity with nums[i + 1]:
   • If both elements have the same parity (both even or both odd), then the array can immediately be determined as not special, and we return false.
4. If no such pair (with the same parity) is found throughout the array, this implies every adjacent pair has different parity, and hence, the array is special, and we return true.

Each of these steps ensures that we efficiently determine the special nature of the array based solely on the requirement that each pair of adjacent numbers must consist of one odd and one even number. The given constraints make this approach feasible for every possible input.

Solutions

class Solution {
public:
    bool checkSpecialArray(vector<int>& values) {
        for (int i = 0; i < values.size() - 1; i++) {
            if (((values[i] & 1) ^ (values[i + 1] & 1)) == 0) {
                return false;
            }
        }
        return true;
    }
};