Printing index of duplicate numbers in array javascript

Question

I need to find the first duplicate number in array and then print index of this number in console. In my code below I reach the moment when code finds the duplicate but prints its index. In simple words, 2 is duplictaing in array but it's stands in first place so I need to print index "0".

var sameNum = [2, 4, 5, 2, 3, 5, 1, 2, 4];
var firstIndex = [];

for (var i = 0; i < sameNum.length; i++) {
  for (var j = i; j < sameNum.length; j++) {
    if (i != j && sameNum[i] == sameNum[j]) {
      firstIndex = [i];
    }
  }
}
console.log(firstIndex);

Possible duplicate of Get the array index of duplicates

Hassan Imam
– Hassan Imam

2017-07-05 17:48:43 +00:00
Commented Jul 5, 2017 at 17:48 — Hassan Imam
– Hassan Imam, Commented Jul 5, 2017 at 17:48

Jeff Hechler · Accepted Answer · 2017-07-05 18:05:57Z

2

If I'm understanding the question correctly (you're looking for the first index of a number that occurs more than once), then the value of i is what you're looking for as that represents the index.

if(i != j && sameNum[i] == sameNum[j]) {
  firstIndex.push(i);
  // or print it //
  console.log(i)
}

edited Jul 5, 2017 at 18:05

answered Jul 5, 2017 at 17:49

Jeff Hechler

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2 Comments

Jakub Karki Over a year ago

yes, that's what i'm looking for but i need to print index not number behind index

Jeff Hechler Over a year ago

I added an example to your answer, does that make sense? You can also refactor it to collect values of j if you need all of the index locations of duplicates.

Hassan Imam · Accepted Answer · 2017-07-05 17:50:35Z

0

I have used this solution

Array.prototype.getDuplicates = function () {
    var duplicates = {};
    for (var i = 0; i < this.length; i++) {
        if(duplicates.hasOwnProperty(this[i])) {
            duplicates[this[i]].push(i);
        } else if (this.lastIndexOf(this[i]) !== i) {
            duplicates[this[i]] = [i];
        }
    }

    return duplicates;
};

console.log([1, 2 ,3 ,4 ,5 ,6 ,6 ,7 ,8 ,9].getDuplicates());

answered Jul 5, 2017 at 17:50

Hassan Imam

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Comments

EricG · Accepted Answer · 2017-07-05 17:52:26Z

0

I would associative array for (fast) lookup/storing encountered values.

var inputArray = [1, 2 ,3 ,4 ,5 ,6 ,6 ,7 ,8 ,9];
var encounteredIndices = {};

for(var i = 0; i < inputArray.length; i++)
  if (encounteredIndices[inputArray[i]])
    console.log(i); // Or add to some array if you wish
  else
    encounteredIndices[inputArray[i]] = 1;

answered Jul 5, 2017 at 17:52

EricG

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Comments

brk · Accepted Answer · 2017-07-05 17:59:54Z

0

You can first sort the array if required.On doing this it will in ascending order.Then use forEach to loop over the array, and get the index

var sameNum = [1, 3, 2, 4, 5, 6, 6, 6, 7, 7, 7, 8, 9];
var firstIndex = [];
// sorting the array and iterating over it
// forEach array method accepts three parameter of which index is one of the parameter
sameNum.sort().forEach(function(item, index) {
  if (sameNum[index + 1] == sameNum[index]) {
    console.log(index + 1);
  }
})

answered Jul 5, 2017 at 17:59

brk

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Comments

Sachin Kumar · Accepted Answer · 2017-07-05 18:00:50Z

Try this one:-

<!doctype html>
<html lang="en">
<head>
    <meta charset="UTF-8">
    <meta name="viewport"
          content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
    <meta http-equiv="X-UA-Compatible" content="ie=edge">
    <title>Document</title>
</head>
<body>

<input id="input" type="time">
<button onclick="live()">run</button>
<!---->
<p id="output">Output</p>
<script>
       function live() {


  var sameNum = [1, 2 ,3 ,4 ,6 ,6 ,7 ,8 ,9];
  var firstIndex;

  for(var i = 0; i < sameNum.length; i++) {
      for(var j = i; j < sameNum.length; j++) {
        if(i != j && sameNum[i] == sameNum[j]) {
        firstIndex = i;
        }
        if (firstIndex != null){break;}
      }
  }

if (firstIndex == null){
  alert("No repeating element");
}
else{
  alert(firstIndex);
}
       }
</script>
</body>
</html>

Alberto Trindade Tavares · Accepted Answer · 2017-07-05 18:08:45Z

0

Nested loops, in this case, are a bad solution. It gives O(n^2) complexity, where we can solve linearly with a map.

We can use a map to track letters that showed up so far in a single loop:

var sameNum = [1, 2 ,3 ,4 ,5 ,6 ,6 ,7 ,8 ,9];
var lettersMap = {};
var firstIndex = -1;

for (var i = 0; i < sameNum.length; i++) {
  if (lettersMap[sameNum[i]]) {
    firstIndex = i;
    break;
  }
  else {
    lettersMap[sameNum[i]] = true;
  }
};

console.log(firstIndex);

edited Jul 5, 2017 at 18:08

answered Jul 5, 2017 at 18:03

Alberto Trindade Tavares

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Comments

Matheus de Quadros · Accepted Answer · 2017-08-29 18:46:09Z

0

You can use the "j" to get the index of duplicate values, like this

var sameNum = [1, 3, 2 ,3 ,4 ,5 ,6 ,6 ,7 ,8 ,9];
var firstIndex = [];

for(var i = 0; i < sameNum.length; i++) {
  for(var j = i; j < sameNum.length; j++) {
    if(i != j && sameNum[i] == sameNum[j]) {
      console.log("--------"+sameNum[i]+"--------");
      console.log("Index: "+ j);
    }
  }
}

edited Aug 29, 2017 at 18:46

answered Jul 5, 2017 at 18:03

Matheus de Quadros

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