3151. Special Array I - Explanation

Description

An array is considered special if the parity of every pair of adjacent elements is different. In other words, one element in each pair must be even, and the other must be odd.

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

Topics

Array

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Prerequisites

Before attempting this problem, you should be comfortable with:

Array Traversal - Iterating through adjacent pairs in an array
Modulo Operator - Using % 2 to determine if a number is even or odd
Bitwise AND - Understanding that n & 1 extracts the least significant bit to check parity

1. Modulo Comparison

Intuition

An array is special if every pair of adjacent elements has different parity (one is even, one is odd). To check this, we compare the parity of each element with its neighbor. Using the modulo operator, we can determine if a number is even (remainder 0 when divided by 2) or odd (remainder 1). If any two adjacent elements have the same parity, the array is not special.

Algorithm

Iterate through the array starting from index 1.
For each index, compare the parity of the current element with the previous element using modulo 2.
If both elements have the same parity (both even or both odd), return false.
If the loop completes without finding adjacent elements of the same parity, return true.

class Solution:
    def isArraySpecial(self, nums: List[int]) -> bool:
        for i in range(1, len(nums)):
            if nums[i - 1] % 2 == nums[i] % 2:
                return False
        return True

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

class Solution {
public:
    bool isArraySpecial(vector<int>& nums) {
        for (int i = 1; i < nums.size(); i++) {
            if (nums[i - 1] % 2 == nums[i] % 2) {
                return false;
            }
        }
        return true;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @return {boolean}
     */
    isArraySpecial(nums) {
        for (let i = 1; i < nums.length; i++) {
            if (nums[i - 1] % 2 === nums[i] % 2) {
                return false;
            }
        }
        return true;
    }
}

public class Solution {
    public bool IsArraySpecial(int[] nums) {
        for (int i = 1; i < nums.Length; i++) {
            if (nums[i - 1] % 2 == nums[i] % 2) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
    fun isArraySpecial(nums: IntArray): Boolean {
        for (i in 1 until nums.size) {
            if (nums[i - 1] % 2 == nums[i] % 2) {
                return false
            }
        }
        return true
    }
}

class Solution {
    func isArraySpecial(_ nums: [Int]) -> Bool {
        for i in 1..<nums.count {
            if nums[i - 1] % 2 == nums[i] % 2 {
                return false
            }
        }
        return true
    }
}

impl Solution {
    pub fn is_array_special(nums: Vec<i32>) -> bool {
        for i in 1..nums.len() {
            if nums[i - 1] % 2 == nums[i] % 2 {
                return false;
            }
        }
        true
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

2. Bitwise Comparison

Intuition

Instead of using the modulo operator to check parity, we can use bitwise AND with 1. The least significant bit of a number determines its parity: if the bit is 1, the number is odd; if the bit is 0, the number is even. Bitwise operations are typically faster than modulo operations, making this approach slightly more efficient.

Algorithm

Iterate through the array starting from index 1.
For each index, extract the least significant bit of the current and previous elements using bitwise AND with 1.
If both bits are the same, the elements have the same parity, so return false.
If all adjacent pairs have different parities, return true.

class Solution:
    def isArraySpecial(self, nums: List[int]) -> bool:
        for i in range(1, len(nums)):
            if nums[i - 1] & 1 == nums[i] & 1:
                return False
        return True

public class Solution {
    public boolean isArraySpecial(int[] nums) {
        for (int i = 1; i < nums.length; i++) {
            if ((nums[i - 1] & 1) == (nums[i] & 1)) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
    /**
     * @param {number[]} nums
     * @return {boolean}
     */
    isArraySpecial(nums) {
        for (let i = 1; i < nums.length; i++) {
            if ((nums[i - 1] & 1) === (nums[i] & 1)) {
                return false;
            }
        }
        return true;
    }
}

public class Solution {
    public bool IsArraySpecial(int[] nums) {
        for (int i = 1; i < nums.Length; i++) {
            if ((nums[i - 1] & 1) == (nums[i] & 1)) {
                return false;
            }
        }
        return true;
    }
}

class Solution {
    fun isArraySpecial(nums: IntArray): Boolean {
        for (i in 1 until nums.size) {
            if ((nums[i - 1] and 1) == (nums[i] and 1)) {
                return false
            }
        }
        return true
    }
}

class Solution {
    func isArraySpecial(_ nums: [Int]) -> Bool {
        for i in 1..<nums.count {
            if (nums[i - 1] & 1) == (nums[i] & 1) {
                return false
            }
        }
        return true
    }
}

impl Solution {
    pub fn is_array_special(nums: Vec<i32>) -> bool {
        for i in 1..nums.len() {
            if (nums[i - 1] & 1) == (nums[i] & 1) {
                return false;
            }
        }
        true
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Common Pitfalls

Off-by-One Errors in Loop Bounds

When iterating to compare adjacent elements, starting from index 0 instead of 1 can cause an out-of-bounds access when checking nums[i-1]. Always start from index 1 when comparing each element with its previous neighbor.

Forgetting Single-Element Arrays

An array with only one element has no adjacent pairs to check, so it is trivially special. Forgetting this edge case can lead to incorrect logic or unnecessary iterations. The loop naturally handles this since iterating from 1 to len(nums) produces zero iterations when len(nums) == 1.