Javascript: How to find first duplicate value and return its index?

Question

I have to find first duplicate value in array and then return its index in variable firstIndex. This has to be done with for loop which should stop after having found first duplicate. I know this is probably pretty simple but I got stuck. So far I have this but it doesn't seem to be working:

var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 0; a < numbers4.length; a++) {
    for (var b = a+1; b < numbers4.length; b++) {
        if (numbers4[a] === numbers4[b])
            firstIndex = numbers4.indexOf(numbers4[a]);
            break;
    }
}
console.log(firstIndex);

Console prints out 1 which is fine because 2 is first duplicate, but when I change numbers in array, the loop doesn't work. Can you advise what can be changed here?

Thanks in advance!

Answer 1

If I correctly understand your question, that's should help you... Basically, you need for a double iteration.

const firstDupeIndex = list => list.findIndex(
  (item, index) => list.lastIndexOf(item) !== index
);

console.log(
  "First Dupe at index:", 
  firstDupeIndex([5, 2, 3, 4, 4, 6, 7, 1, 2, 3])
);

---

Thee above implementation comes with the drawback of being O(n2), due to nesting the lastIndexOf within the findIndex function.

A better solution would be to index your occurrences by building a dictionary, therefore keeping time complexity to just O(n) in the worst case. Probably a little bit less neat, but surely more efficient with big inputs.

const firstDupeIndex = (list) => {
  const dict = {};
  
  for (const [index, value] of list.entries()) {
    if (dict.hasOwnProperty(value)) {
      return dict[value];
    }
  
    dict[value] = index;
  }
  
  return -1;
};

console.log(
  "First Dupe at index:", 
  firstDupeIndex(['a', 'b', 'c', 'd', 'e', 'b', 'z', 't', 'c'])
);

Answer 2

Change your code with the following

    var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
    var firstIndex = "";
   var isMatch=false;
    for (var a = 0; a < numbers4.length; a++) {
        for (var b = a+1; b < numbers4.length; b++) {
            if (numbers4[a] === numbers4[b]){
                firstIndex = numbers4.indexOf(numbers4[a]);
                isMatch=true;
                break;
          }
        }
           if (isMatch) {break;}
    }
    console.log(firstIndex);

Answer 3

I would use an object remembering the values already found... Something like that should work ;)

var numbers4 = [5, 2, 3, 4, 4, 6, 7, 1, 2, 3];

function findFirstDuplicateIndex(arr){
  var found = {};
  
  for (var a = 0, aa = arr.length; a < aa ; a++) {
    if (found[arr[a]])
      return found[arr[a]];
  
    found[numbers4[a]] = a
  }
}

console.log(findFirstDuplicateIndex(numbers4));

It's quite fast because you just loop one time through the array. The rest of the time you just access an object property or you set the object property... Let me know if you have questions ;)

However maybe there something faster... It's just an idea ^^

PS: It also works with words, not just numbers

Answer 4

You don't even need nested for loops.

var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 1; a < numbers4.length; a++) { //start from second elem since first is never a duplicate
        if (numbers4.lastIndexOf(numbers4[a])> a) {
            firstIndex = a;
            break;
        }
}
console.log(firstIndex); //1

All you have to do during iterating is to check if current value exists somewhere further in array. That is done by checking last index of this value's occurrence using lastIndexOf().

Answer 5

Your problem is - you have two loops and only one break, you need to break out of both.

Why not simply return the index as soon as values match?

var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
function getDuplicate(numbers4)
{
    for (var a = 0; a < numbers4.length; a++) {
        for (var b = a+1; b < numbers4.length; b++) {
            if (numbers4[a] === numbers4[b]){
                return a;
            }
        }
    }
}
console.log(getDuplicate(numbers4 ));

However, you can optimize your code further by keeping a map

function getDuplicate( numbers )
{
    var map = {};
    for (var a = 0; a < numbers.length; a++) 
    {
        if( map[ numbers[ a ] ] )
        {
          return a;
        }
        map[ numbers[ a ] ] = true;
    }
    return -1;
}

Answer 6

You can check if indexOf() is not equal to lastIndexOf() and return value and break loop

var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var i = 0; i < numbers4.length; i++) {
  if (numbers4.indexOf(numbers4[i]) != numbers4.lastIndexOf(numbers4[i])) {
    firstIndex = i;
    break;
  }
}

console.log(firstIndex)

Answer 7

Your break; terminates b loop, because if() is not using parenthesis {}.

Additionally, firstIndex should be simply set to either a or b depending on whether you need to return the duplicate's first occurance or first repetition.

It should be:

if (numbers4[a] === numbers4[b])
{
            firstIndex = a;
            break;
}

Answer 8

var numbers = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex;
var len = numbers.length;

for (var i = 0; i < len; i++) {
  var tmpArray = numbers.slice(i + 1, len);
  var index = tmpArray.indexOf(numbers[i]);
  if (index !== -1) {
    firstIndex = index + i + 1;
    break;
  }
}
console.log(firstIndex);

Update:

Actually your logic is right, but you missed braces for if condition and also if the condition satisfies then it means firstIndex is the same as a

This is your code with braces,

var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3];
var firstIndex = "";
for (var a = 0; a < numbers4.length; a++) {
  for (var b = a + 1; b < numbers4.length; b++) {
    if (numbers4[a] === numbers4[b]) {
      firstIndex = a
      break;
    }
  }
}
console.log(firstIndex);

Answer 9

The question states the first dupe in the array has to be found along with it's index. So I return an object where the i property is the index and the e property is the first duplicate element itself. One way of performing this task would be

var numbers4 = [5, 2, 3, 4, 2, 6, 7, 1, 2, 3],
    headDupe = (a,r={}) => (r.e = a.find((n,i) => (r.i = a.indexOf(n), r.i < i)),r);

console.log(headDupe(numbers4));