Picking the odd one out in JavaScript

We are required to write a JavaScript function that takes in an array of literals that contains all similar elements but one.

Our function should return the unlike number.

Therefore, let's write the code for this function ?

Example

The code for this will be ?

const arr = [2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4];
// considering that the length of array is at least 3
const findUnlike = arr => {
    for(let i = 1; i 

Output

The output in the console will be ?

2


Alternative Approach Using Frequency Count

A simpler approach is to count the frequency of each element and return the one that appears only once:

const findUnlikeSimple = arr => {
    const count = {};
    
    // Count frequency of each element
    for(let num of arr) {
        count[num] = (count[num] || 0) + 1;
    }
    
    // Find the element that appears only once
    for(let num in count) {
        if(count[num] === 1) {
            return parseInt(num);
        }
    }
};

console.log(findUnlikeSimple([2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4])); // 2
console.log(findUnlikeSimple([1, 1, 1, 3, 1, 1])); // 3
console.log(findUnlikeSimple([5, 7, 7, 7, 7])); // 5


2
3
5


Comparison

Conclusion

Both methods effectively find the odd element. The frequency count approach is more readable and handles various array arrangements, while the comparison method uses less memory space.