javascript get the last duplicate value index javascript

Question

I have an array to be like this
var arr = [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5].

so it contains duplicate values. In here last index value of
'0' is '3',
'1' is '4',
'2' is '5',
'3' is '13',
and so on.
And i counted total duplicate values

var counts = {};
arr.forEach(function(x) { counts[x] = (counts[x] || 0)+1; })

but i want to know last duplicate value index only. please help me.
thanks in advance

Seeing as counts is an object, there is no "last", as there is no order — adeneo
– adeneo, Commented Feb 2, 2016 at 19:10
Will the array always have values grouped together (ie. [0,0,1,1,2,2]) or can they be scattered (ie. [0,1,2,0,1,2])? — william.taylor.09
– william.taylor.09, Commented Feb 2, 2016 at 19:16

Shaun · Accepted Answer · 2016-02-02 19:15:46Z

7

var arr = [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5];
var existingItems = {};

arr.forEach(function(value, index) {
    existingItems[value] = index;
});

console.log(existingItems);

answered Feb 2, 2016 at 19:15

Shaun

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3 Comments

nem035 Over a year ago

@adeneo, nope. It uses the second parameter of forEach, which is the index, instead of the first one, which is the value. It also overwrites each previous index with the new index, thus at the end it will contain the last index. +1 @Shaun, we think alike :)

adeneo Over a year ago

@nem - I finally got the point, it just seemed so strange I got confused

TimoStaudinger Over a year ago

I like this solution. Also works for unsorted arrays! +1

epascarello · Accepted Answer · 2016-02-02 19:13:44Z

1

Simple loop and check to see if next index is different

var arr = [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5]
var indexes = arr.reduce(function(result, cur, ind, arr){
   if (ind+1===arr.length || cur != arr[ind+1]) {  //check to see if last or different
       result.push(ind);  //if different, store the index
   }
   return result;
},[]);
console.log(indexes);

answered Feb 2, 2016 at 19:13

epascarello

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1 Comment

nem035 Over a year ago

This is probably the only correct answer, since OP said he wants an array. Specially since he doesn't specify anything about how the array should look.

TimoStaudinger · Accepted Answer · 2016-02-02 19:20:31Z

0

Array.prototype.lastIndexOf() comes in handy here:

var arr = [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5];

var out = {};

arr.forEach(function(item) {
  if(!out.hasOwnProperty(item)) {
    out[item] = arr.lastIndexOf(item);
  }
});

console.log(out); // Object {0: 3, 1: 4, 2: 5, 3: 13, 4: 15, 5: 22}

This will also work for unsorted input arrays.

edited Feb 2, 2016 at 19:20

answered Feb 2, 2016 at 19:19

TimoStaudinger

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4 Comments

nem035 Over a year ago

This might be an overkill considering that lastIndexOf performs a loop over the array for each element, thus O(N^2), while this can be done in O(N).

TimoStaudinger Over a year ago

@nem I agree. However, depending on the situation, the more readable and maintainable code may be worth the (minor) performance penalty. Also, this way we do not rely on the premise that the array is sorted.

nem035 Over a year ago

Going from O(N) to O(N^2) is certainly not a minor performance penalty. Also, you already have the index passed into forEach, so yo can just use it instead of calling lastIndexOf, hence readability as well ^_^.

TimoStaudinger Over a year ago

I just saw Shaun's solution and I like it better than mine. Note though that lastIndexOf will only be called once per unique item, not once per element of arr, which makes it a little less bad... :-)

nem035 · Accepted Answer · 2016-02-02 19:24:42Z

Well, what you can do is check if a value is a duplicate (count > 1) and then use the second parameter of forEach which is the index to remember it. Something like:

var arr = [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5];
var counts = {};
var indexes = {};
arr.forEach(function(x, idx) {
  counts[x] = (counts[x] || 0) + 1;
  if(counts[x] > 1) { indexes[x] = idx; } // check if a value is a duplicate and update its index
});

// logs the last index of all duplicate values
console.log(indexes); // {0: 3, 3: 13, 4: 15, 5: 22}

If you want the last index of all values, not just duplicates, you can omit the count > 1 check:

var arr = [0, 0, 0, 0, 1, 2, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 5, 5, 5, 5, 5, 5, 5];
var counts = {};
var indexes = {};
arr.forEach(function(x, idx) {
  counts[x] = (counts[x] || 0) + 1;
  indexes[x] = idx; // update value index
});

// logs the last index of all values
console.log(indexes); // {0: 3, 1: 4, 2: 5, 3: 13, 4: 15, 5: 22}

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