Finding special array - JavaScript

An array is a special array if:

- All the elements at odd indices are odd
- All the elements at even indices are even

We need to write a JavaScript function that takes an array and checks if it's a special array or not.

Understanding the Logic

For an array to be special:

• Index 0 (even): element must be even
• Index 1 (odd): element must be odd
• Index 2 (even): element must be even
• Index 3 (odd): element must be odd

The key insight is that arr[i] % 2 should equal i % 2 for all indices.

Example

const arr = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
const isSpecial = (arr = []) => {
    for(let i = 0; i < arr.length; i++){
        if(arr[i] % 2 === i % 2){
            continue;
        }
        return false;
    }
    return true;
};
console.log(isSpecial(arr));

true

Testing with Different Arrays

// Test case 1: Special array
console.log(isSpecial([2, 3, 4, 5])); // true

// Test case 2: Not special array
console.log(isSpecial([1, 2, 3, 4])); // false (1 at even index)

// Test case 3: Empty array
console.log(isSpecial([])); // true (vacuously true)

// Test case 4: Single element arrays
console.log(isSpecial([4])); // true (even at index 0)
console.log(isSpecial([3])); // false (odd at index 0)

true
false
true
true
false

Alternative Implementation

Here's a more concise version using the every() method:

const isSpecialArray = (arr) => {
    return arr.every((element, index) => element % 2 === index % 2);
};

console.log(isSpecialArray([0, 1, 2, 3, 4, 5])); // true
console.log(isSpecialArray([1, 0, 3, 2])); // false

true
false

How It Works

The function checks each element's parity (even/odd) against its index's parity:

• element % 2 returns 0 for even numbers, 1 for odd numbers
• index % 2 returns 0 for even indices, 1 for odd indices
• If they match for all elements, the array is special

Conclusion

A special array requires elements and indices to have matching parity. The solution efficiently checks this condition using modulo operations and early termination for better performance.