Finding the only even or the only odd number in a string of space separated numbers in JavaScript

We are required to write a JavaScript function that takes in a string that contains numbers separated by spaces.

The string either contains all odd numbers and only one even number or all even numbers and only one odd number. Our function should return that one different number from the string.

Problem

Given a string of space-separated numbers where all numbers are either odd or even except for one, we need to find and return that unique number that differs from the rest.

Example

Following is the code:

const str = '2 4 7 8 10';
const findDifferent = (str = '') => {
    const odds = [];
    const evens = [];
    const arr = str
        .split(' ')
        .map(Number);
    arr.forEach(num => {
        if(num % 2 === 0){
            evens.push(num);
        }else{
            odds.push(num);
        };
    });
    return odds.length === 1 ? odds[0] : evens[0];
};
console.log(findDifferent(str));

Output

Following is the console output:

7

How It Works

The function splits the input string into an array of numbers, then separates them into odd and even arrays. Since we know exactly one number will be different, we check which array has only one element and return it.

Alternative Approach

Here's a more efficient solution that doesn't require storing all numbers:

const str1 = '1 3 5 2 7';
const str2 = '2 4 6 8 9';

const findDifferentOptimized = (str = '') => {
    const numbers = str.split(' ').map(Number);
    
    // Check if first three numbers to determine the pattern
    const evenCount = numbers.slice(0, 3).filter(num => num % 2 === 0).length;
    const lookingForEven = evenCount === 1;
    
    return numbers.find(num => (num % 2 === 0) === lookingForEven);
};

console.log(findDifferentOptimized(str1)); // 2 (even among odds)
console.log(findDifferentOptimized(str2)); // 9 (odd among evens)

2
9

Conclusion

Both approaches work effectively, but the optimized version is more memory-efficient as it doesn't store all numbers in separate arrays. The key insight is identifying the pattern early and finding the number that breaks it.