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JavaScript

Posted by Brian Hernandez
Brian Hernandez
20,285 Points

Trying to create an array of duplicate values from another array, why doesn't my code work?

I am working on this piece of code here where the goal is to create an array of all the values that are duplicated from another array. The resulting array I'd like to have should only enter the duplicated values from the first array once. The catch is I can't use any array functions or methods. This is what I have so far:

var numbers = [8,24,20,5,13,3,1,12,11,24,8,24,20,4,5,23,24,23,21,2,19,3,21,2,14,17,21,5,7,10,20,11,0,5,18,2,13,11,14,3,20,1,23,6,21,10,14,0,15,20];
var results = [];
var tempArr = [];

for (var i = 0; i <= numbers.length; i++) {
    if (tempArr[numbers[i]] === undefined) {
    tempArr[numbers[i]] = 1;
  } else if (results[numbers[i]] === undefined)  {
    results.push(numbers[i]);
  }
}

console.log(results);

I am getting closer to me desired output... but for some reason the results array continues to contain multiple entries of the values that are duplicated in the numbers array. Where am I going wrong here?

4 Answers

Brian Hernandez
Brian Hernandez
20,285 Points

After doing some searching around, I found this cool solution to my problem I posted earlier. It uses an object to test weather a number is duplicated in any position in the array:

var numbers = [8,24,20,5,13,3,1,12,11,24,8,24,20,4,5,23,24,23,21,2,19,3,21,2,14,17,21,5,7,10,20,11,0,5,18,2,13,11,14,3,20,1,23,6,21,10,14,0,15,20];

var results = [];
var seenIt = {};

for (var i = 0; i < numbers.length; i++) {
    var number = numbers[i];

  seenIt[number] = 1 + (seenIt[number] || 0);

  if (seenIt[number] === 2) {
    results.push(number);
  }

}

console.log(results);

So, as the loop progresses through the array, the value at the current index position of the numbers array is used to create a key for the seenIt object. That key is given the sum value of 1 plus either what the current value of the seenIt[number] key is, if it exists, or 0. So, if the value of the current numbers array index has not been seen before, its corresponding seenIt[number] key is set to the value of 1. And if it was seen before, its prior value would have been set to 1 so the current value then gets evaluated to 2 (1 + 1 = 2). And if it is 2, then that index value gets pushed into the results array. And when we print the results array, we will see all of the values, given only once, that have at least one or more duplicates in the numbers array.

Seth Kroger
Seth Kroger
56,416 Points
Seth Kroger
Seth Kroger
56,416 Points

results[numbers[i]] doesn't correlate to where the number is in the results because the duplicate number is always pushed to the end of the array, not put into the index equal to the number. I think you could count the times you encounter a number by incrementing the spot in tempArr and only pushing the number to results on the 2nd encounter.

Antonio De Rose
Antonio De Rose
20,886 Points
Antonio De Rose
Antonio De Rose
20,886 Points

Very good question :)

var array = [24,20,5,13,3,1,12,11,24,8,24,20,4,5,23,24,23,21,2,19,3,21,2,14,17,21,5,7,10,20,11,0,5,18,2,13,11,14,3,20,1,23,6,21,10,14,0,15,20];
    var a = [], prev;

    array.sort();
    for ( var i = 0; i < array.length; i++ ) {
        if ( array[i] !== prev ) {
            a.push(array[i]);
        } 
        prev = array[i];
    }

    console.log(a)
Brian Hernandez
Brian Hernandez
20,285 Points

Thank you but I can't use a method like sort() for this solution.

Antonio De Rose
Antonio De Rose
20,886 Points
Antonio De Rose
Antonio De Rose
20,886 Points

wiil this help you

var array = [24,20,5,13,3,1,12,11,24,8,24,20,4,5,23,24,23,21,2,19,3,21,2,14,17,21,5,7,10,20,11,0,5,18,2,13,11,14,3,20,1,23,6,21,10,14,0,15,20];
    var a = [];var b=[];


    for ( var i = 0; i < array.length; i++ ) {
        if ( b.indexOf(array[i]) > -1  && a.indexOf(array[i]) < 0) {
            a.push(array[i]);
        } 
        b.push(array[i])
    }

    console.log(a)