Hi i seem to be struggling with this. i want to find the odd number
if ((storedNumbers[storedNumber] % 2) != 0) {
return storedNumber;
it returns undefined. How can i loop through each object value and return the key once i find an odd number. I thought != would work,
Any idea? thanks
function findOdd(A) {
//create an empty object that stores count of numbers
storedNumbers = {};
//iterate through the array and add 1 to the counter (and create the object if it has not already been counted)
for (i = 0; i < A.length; i++) {
currentValue = A[i];
if (storedNumbers[currentValue] >= 1) {
storedNumbers[currentValue] = storedNumbers[currentValue] + 1;
console.log(storedNumbers);
} else {
storedNumbers[currentValue] = 1;
console.log(storedNumbers);
}
}
// //for each object, run a loop that %2 = 0. then return that number
// const valuesList = Object.values(storedNumbers);
// console.log(valuesList);
// for (i = 0; i < valuesList.length; i++) {
// currentValue = valuesList[i];
// if (valuesList[currentValue] % 2 == 0) {
// console.log("no problem");
// }
// if (valuesList[currentValue] % 2 != 0) {
// return valuesList[currentValue];
// console.log("found");
const run = Object.keys(storedNumbers).forEach(storedNumber => {
console.log(storedNumber); // the name of the current key.
console.log(storedNumbers[storedNumber]); // the value of the current key.
if ((storedNumbers[storedNumber] % 2) != 0) {
return storedNumber;
}
});
}
findOdd([20, 1, 1, 2, 2, 3, 3, 5, 5, 4, 20, 4, 5]);
function findOdd(array){
let result = [ ] ;
for (let i = 0 ; i < array.length ; i++){
if ( array[i] % 2 === 1){
result.push(array[i] );
}
}
return result;
}
Have you commented out all of the code you have above the findOdd() function? The code Jlassi gave is all the code you need to return an array of odd numbers.
function findOdd(A) {
//create an empty object that stores count of numbers
storedNumbers = {};
//iterate through the array and add 1 to the counter (and create the object if it has not already been counted)
for (i = 0; i < A.length; i++) {
currentValue = A[i];
if (storedNumbers[currentValue] >= 1) {
storedNumbers[currentValue] = storedNumbers[currentValue] + 1;
console.log(storedNumbers);
} else {
storedNumbers[currentValue] = 1;
console.log(storedNumbers);
}
}
// Object.keys(storedNumbers).forEach(storedNumber => {
// console.log(storedNumber); // the name of the current key.
// console.log(storedNumbers[storedNumber]); // the value of the current key.
// let currentVal = storedNumbers[storedNumber];
// if (currentVal % 2 !== 0) {
// return storedNumber;
// }
// });
// return storedNumber;
function findOdd(array) {
let result = [];
for (let i = 0; i < array.length; i++) {
if (array[i] % 2 === 1) {
result.push(array[i]);
}
}
return result;
}
findOdd(storedNumbers);
}
findOdd([20, 1, 1, 2, 2, 3, 3, 5, 5, 4, 20, 4, 5]);
its required i return the key from the key value pair of the odd number
Could you give an example of what you want your object to look like? What is the output you want, exactly as it should be displayed? Because I am unsure why you are using an object.
Ah…That makes more sense. The challenge is asking for an integer from an array that is repeated an odd number of times.
Well, this is how I did it using arrays, not objects:
function findOdd(A) {
const arr = A.sort();
let single = [[0,0]];
a = 0;
let i = 1;
for (let int of arr) {
if (single[a][0] != int) {
let newarr = [int,1]
single.push(newarr);
i = 1;
a++;
}
else {
i++
single[a][1] = i;
}
}
let odd;
for (let ele of single) {
if (ele[1] % 2 == 1) {
odd = ele[0];
}
}
return odd;
}
Hi thanks, actually i am confused specifically as to why the ForEach solution doesn’t work? What makes it return undefined?
Works (For in loop)
function findOdd(A) {
//create an empty object that stores count of numbers
storedNumbers = {};
//iterate through the array and add 1 to the counter (and create the object if it has not already been counted)
for (i = 0; i < A.length; i++) {
currentValue = A[i];
if (storedNumbers[currentValue] >= 1) {
storedNumbers[currentValue] = storedNumbers[currentValue] + 1;
console.log(storedNumbers);
} else {
storedNumbers[currentValue] = 1;
console.log(storedNumbers);
}
}
for (const number in storedNumbers) {
const count = storedNumbers[number];
// if the count is odd
if (count % 2 !== 0) {
// return that number
return +number;
}
}
}
findOdd([20, 1, 1, 2, 2, 3, 3, 5, 5, 4, 20, 4, 5]);
doesn’t work (for each loop):
function findOdd(A) {
//create an empty object that stores count of numbers
storedNumbers = {};
//iterate through the array and add 1 to the counter (and create the object if it has not already been counted)
for (i = 0; i < A.length; i++) {
currentValue = A[i];
if (storedNumbers[currentValue] >= 1) {
storedNumbers[currentValue] = storedNumbers[currentValue] + 1;
console.log(storedNumbers);
} else {
storedNumbers[currentValue] = 1;
console.log(storedNumbers);
}
}
Object.keys(storedNumbers).forEach(storedNumber => {
console.log(storedNumber); // the name of the current key.
console.log(storedNumbers[storedNumber]); // the value of the current key.
let currentVal = storedNumbers[storedNumber];
if (storedNumbers[storedNumber] % 2 !== 0) {
return +storedNumber;
}
});
}
findOdd([20, 1, 1, 2, 2, 3, 3, 5, 5, 4, 20, 4, 5]);
Because a return statement in the forEach method does not return back to the calling function. You use the forEach method to iterate through an array without stopping to return something.
