3151. Special Array I

[mah-shamim]

Topics: Array

An array is considered special if every pair of its adjacent elements contains two numbers with different parity.

You are given an array of integers nums. Return true if nums is a special array, otherwise, return false.

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

Hint:

1. Try to check the parity of each element and its previous element.

[mah-shamim]

We need to determine if the given array is "special" by checking if every pair of adjacent elements contains two numbers with different parity (one even and one odd). Here's how you can implement this:

1. Check the length of the array: If the array has only one element, it is automatically considered special.
2. Iterate through the array: For each pair of adjacent elements, check if they have different parity.
3. Return the result: If all adjacent pairs have different parity, return true; otherwise, return false.

Let's implement this solution in PHP: 3151. Special Array I

<?php
/**
 * @param Integer[] $nums
 * @return Boolean
 */
function isArraySpecial($nums) {
    $length = count($nums);

    // If there is only one element, it's automatically a special array.
    if ($length === 1) {
        return true;
    }

    // Check each adjacent pair for different parity
    for ($i = 1; $i < $length; $i++) {
        if (($nums[$i] % 2) === ($nums[$i - 1] % 2)) {
            // If two consecutive elements have the same parity, return false
            return false;
        }
    }

    // If all pairs have different parity, return true
    return true;
}

// Test cases
$nums1 = [1];
$nums2 = [2, 1, 4];
$nums3 = [4, 3, 1, 6];

echo isArraySpecial($nums1) ? 'true' : 'false'; // Output: true
echo "\n";
echo isArraySpecial($nums2) ? 'true' : 'false'; // Output: true
echo "\n";
echo isArraySpecial($nums3) ? 'true' : 'false'; // Output: false
echo "\n";
?>

Explanation:

1. Edge Case:

   • If the array has only one element, it's automatically considered a special array since there are no adjacent pairs to check.

2. Checking Adjacent Elements:

   • Iterate through the array starting from the second element.
   • Use the modulo operator % to determine the parity of adjacent elements:
     • nums[i] % 2 == 0 indicates the number is even.
     • nums[i] % 2 == 1 indicates the number is odd.
   • Compare the parity of the current element nums[i] with the previous element nums[i-1]. If they have the same parity, return false.

3. Return true:

   • If all adjacent pairs have different parities, return true.

This solution runs in O(n), where n is the length of the array, making it efficient for the given constraints.

[topugit]

Thanks for solving the problem of "Special Array I"

[mah-shamim]

Thanks