I'm doing this codewars I want return the object property is possible do it when I know the object value ?
function findOdd(A) {
let counter = {};
let finalResult;
const NumInt = A.sort((a, b)=> a - b)
NumInt.forEach((e)=>{counter[e] = (counter[e] || 0) +1})
console.log(counter)
const values = Object.values(counter).filter((value)=>{
return value % 2 !== 0
})
Object.values(counter).filter((value)=>{
if(value % 2 !== 0){
return console.log(value)
}
})
}
findOdd([1,1,2,-2,5,2,4,4,-1,-2,5])
Using the Object.values() method returns an array of a given object's own enumerable property values so it's not what you want in your case.
You can iterate through the for...in statement which iterates over all enumerable properties of an object that are keyed by strings to achieve that.
function findOdd(A) {
let counter = {};
let finalResult;
const NumInt = A.sort((a, b)=> a - b)
NumInt.forEach((e)=>{counter[e] = (counter[e] || 0) +1})
console.log(counter)
const values = Object.values(counter).filter((value)=>{
return value % 2 !== 0
})
for (const el in counter){
if(counter[el] % 2 !== 0){
return console.log(el, counter[el]); //logs property name and it's value
}
}
}
findOdd([1,1,2,-2,5,2,4,4,-1,-2,5])You can create an object storing the key,occurrences couples and then filter the only one key with an odd occurrence with reduce and Object#keys:
function findOdd(arr) {
const occurrences = arr.reduce(function(acc, curr) {
acc[curr] = acc[curr] ? acc[curr] + 1 : 1;
return acc;
}, {});
const result = Object.keys(occurrences).find(key => (occurrences[key] % 2) === 1);
return Number(result);
}
console.log(findOdd([7]) === 7);
console.log(findOdd([0]) === 0);
console.log(findOdd([1, 1, 2]) === 2);
console.log(findOdd([0, 1, 0, 1, 0]) === 0);
console.log(findOdd([1, 2, 2, 3, 3, 3, 4, 3, 3, 3, 2, 2, 1]) === 4);You can't loop an object using array built-in functions like map, forEach, for...of
I'm posting one solution here.
const data = {
2: 0,
1: 2,
3: 1,
4: 7,
5: 9
}
for (const key in data) {
if (data[key] % 2 === 1)
console.log(key, data[key])
}
Other possible ways are using
Object.keys() and Object.values()
Or
Object.entries()