firstDuplicate.js

firstDuplicate = a => {
    r = new Set()
    for (e of a)
        if (r.has(e))
            return e
        else
            r.add(e)
    return -1
}

firstDuplicate.md

Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.

Example

For a = [2, 1, 3, 5, 3, 2], the output should be firstDuplicate(a) = 3.

There are 2 duplicates: numbers 2 and 3. The second occurrence of 3 has a smaller index than the second occurrence of 2 does, so the answer is 3.

For a = [2, 4, 3, 5, 1], the output should be firstDuplicate(a) = -1.

Input/Output

[execution time limit] 4 seconds (js)

[input] array.integer a

Guaranteed constraints: 1 ≤ a.length ≤ 105, 1 ≤ a[i] ≤ a.length.

[output] integer

The element in a that occurs in the array more than once and has the minimal index for its second occurrence. If there are no such elements, return -1.

[JavaScript (ES6)] Syntax Tips

// Prints help message to the console // Returns a string function helloWorld(name) { console.log("This prints to the console when you Run Tests"); return "Hello, " + name; }