Question 54204 in Javascript Question posted in
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Finding recurring elements in an array of strings using ES6 concepts – Javascript

I came across a basic question of array in javascript of list out the recurring elements or duplicate elements in an array of strings.

So, lets say we have an array, arr = ['Ash', 'Bob', 'Jarvis', 'Ash', 'Ash', 'Dylan', 'Bob', 'Dylan','Tom', 'John']

So now I tried to solve this by below approach:

var duplicate = [];

function duplicateValues (arr) {
   let l = arr.length;
   
   for (let i=0; i<l; i++){
      if(arr[i+1] == arr[i]) {
     duplicate.push(arr[i]);
   }
 }
   return duplicate;
}

duplicateValues(['Ash', 'Bob', 'Jarvis', 'Ash', 'Ash', 'Dylan', 'Bob', 'Dylan','Tom', 'John']
);

Expected output:

['Ash', 'Bob', 'Dylan']

But I am getting an empty array in the console. What should be the best approach to solve this problem?

Tags:

4

Answers


  1. 0

    This should be very simple, look below.

    const d = ['Ash', 'Bob', 'Jarvis', 'Ash', 'Ash', 'Dylan', 'Bob', 'Dylan','Tom', 'John'];

const dub = d.filter(x=> d.filter(f=> f ==x).length>1).reduce((a,v)=> {
  if(!a.find(x=> x == v))
      a.push(v)
    return a
},[])
console.log(dub)
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  2. 0

    Wouldn’t that be easier?

    const array = ['Ash', 'Bob', 'Jarvis', 'Ash', 'Ash', 'Dylan', 'Bob', 'Dylan', 'Tom', 'John']

const result = [...new Set(array.filter((e, i, a) => a.indexOf(e, i + 1) !== -1))]

console.log(result)
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  3. 0

    All you need to do is to just check if a map has that key.

    const arr = ['Ash', 'Bob', 'Jarvis', 'Ash', 'Ash', 'Dylan', 'Bob', 'Dylan', 'Tom', 'John']

let map = new Map();

const dups = arr.reduce((acc, curr) => {
  if (map.has(curr.toLowerCase()) && acc.indexOf(curr) == -1) {
    acc.push(curr)
  } else {
    map.set(curr.toLowerCase(), curr)
  }

  return acc;
}, []);

console.log(dups)
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  4. 0

    You could use reduce to get the frequency of each name in the array, then filter only those entries that appear with a frequency of 2 or more.

    This approach avoids the use of find and indexOf, which would scan over the array an unnecessary number of times.

    const arr = ['Ash', 'Bob', 'Jarvis', 'Ash', 'Ash', 'Dylan', 'Bob', 'Dylan','Tom', 'John']

console.log(Object.entries(arr.reduce((a,c)=>
  (a[c]??=0,a[c]++,a),{})).filter(([k,v])=>v>1).map(([k])=>k))
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