LeetCode 3151: Special Array I Solution

Problem Explanation

Explanation

To determine if an array is special, we need to check if the parity of every pair of adjacent elements is different. We can iterate through the array and compare the parity of each pair of adjacent elements. If at any point we find two adjacent elements with the same parity, we return false. If we iterate through the entire array without finding any such pair, we return true.

Algorithm:

Iterate through the array from index 1 to n-1.
Check the parity of the current element and the previous element.
If the parities are the same, return false.
If we complete the iteration without finding any same parities, return true.

Time Complexity

The time complexity of this solution is O(n), where n is the number of elements in the input array.

Space Complexity

The space complexity of this solution is O(1) as we are using a constant amount of extra space regardless of the input size.

Solution Code

class Solution {
    public boolean isSpecialArray(int[] nums) {
        for (int i = 1; i < nums.length; i++) {
            if ((nums[i] % 2 == 0 && nums[i - 1] % 2 == 0) || (nums[i] % 2 != 0 && nums[i - 1] % 2 != 0)) {
                return false;
            }
        }
        return true;
    }
}

Frequently Asked Questions

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The time complexity for LeetCode 3151 (Special Array I) varies by solution. Check the detailed explanation section for specific complexities in Java, C++, and Python implementations.

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3151. Special Array I