Javascript getting the oldest element in the array

Question

Lets say that I am to remove the oldest element of an array, because I wanted to replace it with a new value. (first-in-first-out).

For example, I have this array of values

var arr = [1,2,3,4,5,6,7]

And I wanted to only get the first three, values, and then on the 4th, replace the element who came first

[1,2,3], [4,2,3], [4,5,3] and so on..

I came up with a solution

var arr = [1,1,2,3,4,5,6,7];
var newArr = [];
for(i=0; i<arr.length; i++){
    if(newArr.length == 3 && newArr.indexOf(arr[i]) < 0) {
       newArr[i%3] = arr[i];
    } else if(newArr.indexOf(arr[i]) < 0) {
       newArr.push(arr[i])
    }
    console.log(newArr)
}

Which will render:

1
1
1,2
1,2,3
1,4,3
1,4,5
6,4,5
6,7,5

Instead of

1
1,2
1,2,3
4,2,3
4,5,3
4,5,6
7,5,6

What am I missing out.

Any reason why you're also testing for unique values? if(newArr.indexOf(arr[i]) < 0) — godfrzero
– godfrzero, Commented Oct 13, 2013 at 8:38
@godfrzero, when you are assigning jobs, you won't do the same job for right? — Joey Hipolito
– Joey Hipolito, Commented Oct 13, 2013 at 9:10
@JoeySalacHipolito No idea what the context for the code is, just asked for clarification :D — godfrzero
– godfrzero, Commented Oct 13, 2013 at 13:29

thefourtheye · Accepted Answer · 2013-10-13 08:38:04Z

2

var arr = [1, 1, 2, 3, 4, 5, 6, 7];
var newArr = [], currentIndex = 0;
for (i = 0; i < arr.length; i++) {
    if (newArr.length === 3 && newArr.indexOf(arr[i]) < 0) {
        newArr[currentIndex % 3] = arr[i];
        currentIndex += 1;
    } else if (newArr.indexOf(arr[i]) < 0) {
        newArr.push(arr[i]);
    }
    console.log(newArr)
}

Output

[ 1 ]
[ 1 ]
[ 1, 2 ]
[ 1, 2, 3 ]
[ 4, 2, 3 ]
[ 4, 5, 3 ]
[ 4, 5, 6 ]
[ 7, 5, 6 ]

You just need to track the current index where you need to place the number using a separate variable.

answered Oct 13, 2013 at 8:38

thefourtheye

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