I have to find a unique number in unsorted array, but my function returns wrong number, I can't understand why. Here is my code:
function findUniq(arr) {
let sorted = [...arr].sort()
if(sorted.length === 0) return 0
// do magic
let num = 0
for(let i = 1; i < sorted.length; i++){
if(sorted[num] !== sorted[i]){
num++;
sorted[num] = sorted[i]
}
}
return num + 1
}
if I invoke findUniq([9,7,7,6,6,5,5,5]) it gives 4. What do I do wrong? Thanks in advance. I forgot to mention I have to have just one for loop to implement O(n) time complexity
This should work:
It returns only one unique number in array or undefined
If you want all unique numbers replace return arr[0] with return arr. It will then return array of all unique numbers (or empty array if there are not any)
function findUniq(arr) {
arr.filter((item, index) => {
arr.splice(index, 1)
const unique = !arr.includes(item)
arr.splice(index, 0, item)
return unique
})
return arr[0]
}
ES6 aproach:
function findUniq(arr) {
return arr
.map((c) => arr.filter((b) => c == b))
.filter((e) => e.length < 2)
.reduce((total, cur) => total.concat(cur), [])
}
You can use reduce and can find the item that repeats once in the last index.
var arr = [9,7,7,6,6,5,5,5]
var uniqueNumber;
arr.reduce((obj,val,index) =>
{
obj[val] ? ++obj[val] : obj[val] = 1;
if(index == (arr.length - 1))
{
uniqueNumber = Object.keys(obj).find(key => obj[key] === 1)
}
return obj
},{})
console.log(uniqueNumber)To do it in a single O(n) loop, reduce, keeping track of counts as well as a set of singular items.
function findUniq(arr) {
let [single] = arr.reduce((acc, el) => {
if (!acc[el]) acc[el] = 0;
if (++acc[el] === 1) acc.singular.add(el);
else acc.singular.delete(el);
return acc;
}, { singular: new Set() }).singular;
return single;
}
const input = [2, 2, 9, 7, 7, 6, 6, 5, 5, 5];
const result = findUniq(input);
console.log(result);