Find the Duplicate Number

I was given a challenge by a friend to find the duplicates in an array.

For instance, a function when it receives an array -

[1, 2, 3, 4, 5, 2, 2, 2]

should return the duplicated number 2

He challenged me to write the code without using more than 1 loop.

My first try

const findDuplicate = nums => {

    const uniqueSet = []
    let duplicate = null
    nums.map(x => {
        if(uniqueSet.includes(x)){
           duplicate = x 
        }
        uniqueSet.push(x)
    })
    return duplicate
}

so, findDuplicate([1, 2, 3, 4, 5, 2, 2, 2]) returns 2

He was not satisfied but wanted to try differently rather than looping it through.

After hours of breaking my head, I came up with this -

const findDuplicate = nums => {

    const uniqueSet = new Set(nums)
    const arrSum = nums.reduce((a, b) => a + b)

    let uniqueArray = Array.from(uniqueSet)
    const uniqueSetSum = uniqueArray.reduce((a, b) => a + b)
    const duplicate = Math.abs(arrSum - uniqueSetSum) / (nums.length - uniqueArray.length)

    return duplicate
}

He did not understand what is going on at the first look. After a while, he understood the code.

Which one do you prefer? Simpler code or Complex code? Or is there any other way you would solve this?

Share your thoughts. Thanks.

Comments

[Melvin Carvalho]

1loc.dev/

const countOccurrences = arr => arr.reduce((prev, curr) => (prev[curr] = ++prev[curr] || 1, prev), {});

// Examples
countOccurrences([2, 1, 3, 3, 2, 3]); // { '1': 1, '2': 2, '3': 3 }
countOccurrences(['a', 'b', 'a', 'c', 'a', 'b']); // { 'a': 3, 'b': 2, 'c': 1 }

[Ashwin]

This is good!

[Melvin Carvalho]

Do look at the other functions in 1loc.dev

I liked this trick:

++prev[curr] || 1