let arr = [1, 2, 3, 4, 5, 5];
const seen = new Set();
const duplicates = arr.filter(n => seen.size === seen.add(n).size);
console.log(duplicates); // [ 5 ]
console.log(duplicates.length); // 1function hasDuplicates(array) {
return (new Set(array)).size !== array.length;
}// This code was pasted by ProgrammerRimon, 2022-02-18
// Find dublicates numbers from array;
const a = [1,1,1,3,3,4,5,5,6,7,8,8,8,8,8,9,2];
function findDublicatesNumbersFromArray(a) {
var d = [];
for (let i = 0; i < a.length; i++) {
var ct = a[i];
var cmt = 0;
for(var x = 0; x<a.length;++x) {
if(ct === a[x]) {
cmt++
if(cmt > 1) {
d.push(a[i])
}
}
}
}
return d.filter((i, ix)=> d.indexOf(i) === ix);
}
console.log(findDublicatesNumbersFromArray(a)) [1, 2, 3].every((e, i, a) => a.indexOf(e) === i) // true
[1, 2, 1].every((e, i, a) => a.indexOf(e) === i) // false
Array(5).fill(2)
//=> [2, 2, 2, 2, 2]
const counts = {};
const sampleArray = ["1", "5", "9", "14", "5", "22", "48", "25", "22", "20", "9" ,"13"]
sampleArray.forEach(function (x) { counts[x] = (counts[x] || 0) + 1; });
console.log(counts)
// output: {
'1': 1,
'5': 2,
'9': 2,
'13': 1,
'14': 1,
'20': 1,
'22': 2,
'25': 1,
'48': 1
}