/**
 * 
 * Ashish Patel
 * e: ashishsushilPatel@gmail.com
 * w: https://ashish.me
 *
 */

/* 
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.
*/

function firstDuplicate(array) {
  const dupDict = {}
  for (let index = 0; index < array.length; index++) {
    const num = array[index]
    if(dupDict.hasOwnProperty(num)){
      return num
    }
    dupDict[num] = num
    
  }
  return -1
}

console.log(firstDuplicate([2, 1, 3, 5, 3, 2]))

// test('first Duplicate', () => {
//   expect(firstDuplicate([2, 1, 3, 5, 3, 2])).toEqual(3)
// });