js find duplicates in array

let arr =["a","b","c"];
// ES6 way
const duplicate = [...arr];

// older method
const duplicate = Array.from(arr);const names = ['Mike', 'Matt', 'Nancy', 'Adam', 'Jenny', 'Nancy', 'Carl']

const count = names =>
  names.reduce((a, b) => ({ ...a,
    [b]: (a[b] || 0) + 1
  }), {}) // don't forget to initialize the accumulator

const duplicates = dict =>
  Object.keys(dict).filter((a) => dict[a] > 1)

console.log(count(names)) // { Mike: 1, Matt: 1, Nancy: 2, Adam: 1, Jenny: 1, Carl: 1 }
console.log(duplicates(count(names))) // [ 'Nancy' ]function getDuplicateArrayElements(arr){
    var sorted_arr = arr.slice().sort();
    var results = [];
    for (var i = 0; i < sorted_arr.length - 1; i++) {
        if (sorted_arr[i + 1] === sorted_arr[i]) {
            results.push(sorted_arr[i]);
        }
    }
    return results;
}

var colors = ["red","orange","blue","green","red","blue"];
var duplicateColors= getDuplicateArrayElements(colors);//["blue", "red"]// JavaScript - finds if there is duplicate in an array. 
// Returns True or False.

const isThereADuplicate = function(arrayOfNumbers) {
    // Create an empty associative array or hash. 
    // This is preferred,
    let counts = {};
    // // but this also works. Comment in below and comment out above if you want to try.
    // let counts = [];

    for(var i = 0; i <= arrayOfNumbers.length; i++) {
        // As the arrayOfNumbers is being iterated through,
        // the counts hash is being populated.
        // Each value in the array becomes a key in the hash. 
        // The value assignment of 1, is there to complete the hash structure.
        // Once the key exists, meaning there is a duplicate, return true.
        // If there are no duplicates, the if block completes and returns false.
        if(counts[arrayOfNumbers[i]] === undefined) {
            counts[arrayOfNumbers[i]] = 1;
        } else {
            return true;
        }
    }
    return false;
}arr.reduce((b,c)=>((b[b.findIndex(d=>d.el===c)]||b[b.push({el:c,count:0})-1]).count++,b),[]);[1, 2, 2, 4, 3, 4].filter((e, i, a) => a.indexOf(e) !== i) // [2, 4]