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undefined elements inside an array which has randomized numbers inside it

so I created this script to generate an array of 5 random numbers where the numbers must be unique, but I get +2 undefined elements. There are 5 numbers which are unique but the length of the array isn't anymore 5
let arr = [];
while (arr.length <= 5) {
let x = Math.floor(Math.random() * 11);
if (arr.includes(x) === false) {
arr.push(x);
}
arr.length += 1;
}
console.log(arr);
Thanks in advance guys.

You should remove the arr.length += 1 line. The array length will be automatically incremented each time a new element is pushed, no need to modify it by yourself: this is a bad practice.

the length is not 5 because this loop runs 6 times, arr.length = 0,1,2,3,4,5. and you have a 3 undefined elements because of ( arr.length += 1; ) this skips step 1,3,5.

let arr = [];
while (arr.length<5){
let x = Math.floor(Math.random()*11);
if (arr.includes(x) === false){
arr.push(x);
}
}
console.log(arr);

creating a function that only returns odd numbers [duplicate]

This question already has answers here:
How do I extract even elements of an Array?
(8 answers)
How to do a script for odd and even numbers from 1 to 1000 in Javascript?
(8 answers)
Closed 2 years ago.
I've spent an embarrassing amount of time on this question only to realize my function is only right 50% of the time. So the goal here is to return only the odd numbers of all the numbers in between the two arguments. (for instance if the arguments are 1 and 5 i'd need to return 2 & 3) the function I wrote is completely dependent on the first argument. if it's even my function will return odds, but if the first number is odd it'll return evens. does anyone know how i can fix this?
function oddNumbers(l, r) {
const arr = [];
const theEvens = [];
for (let i= l; i<r; i++) {
arr.push(i)
}
console.log(arr)
for (let i= 0; i < arr.length; i+= 2 ) {
const evens = arr[0] + i;
theEvens.push(evens);
}
theEvens.forEach(item => arr.splice(arr.indexOf(item), 1));
console.log(arr)
}
oddNumbers(2, 20);

I modified the code a bit to return only odd numbers
We use the % operator that behaves like the remainder operator in math:
so when we say i % 2 if the number is even the result of the operation will be 0
but when the "i" is an odd number the result will be 1
so now we can filter the even from the odd numbers using this operation
function oddNumbers(l, r) {
const arr = [];
for (let i= l; i<r; i++) {
if(i % 2 !== 0) arr.push(i);
}
console.log(arr);
}
oddNumbers(2, 20);

You can loop from initial to end parameters and get odd numbers using modulo, try this:
let result = [];
let returnOdd = (n1, n2) => {
for(i = n1; i < n2; i++){
if(i % 2 != 0){
result.push(i)
}
}
return result;
}
console.log(returnOdd(2, 20));

You could use the filter method.
This method creates a new array based on the condition it has. In this case it will to go through all the numbers in the array, and check if the number is odd (though the remainder operator).
For example:
1 % 2 = 1 ( true, keep in the new array )
2 % 2 = 0 ( false ignore in the new array )
function OddNumbers(start, end) {
// Create an array from the given range
const nums = Array(end - start + 1).fill().map((_, idx) => start + idx);
// Use filter to return the odd numbers via the % operator
return nums.filter(num => num % 2);
}
console.log(OddNumbers(2,20))

Print the value of the first element that occurs N number of times

So I'm making this ReactJS application.
Part of a filtering system I have the following problem:
Simplified...
I have an array, let's say its simple one like let arr = [1,7,4,3,4,7];
and I also input an N variable from the user that is a simple integer input.
I need to return or console.log() the integers from the array that is repeated N times. If there is nonsuch repeating number log err msg or return -1;
For instance,
let arr = [1,7,4,3,4,7]; and let n = 2; i get 7- cos 7 repeats 2 times
let arr = [7,4,5,3,5,5,3,4,3,2,1]; and let n = 3; i get 5 - cos 5 repeats 3 times
let arr = [1,6,4,6,4,6]; and let n = 4; i get -1 or cl("err") - cos nothing repeats 4 times
Code from comments:
const getRepeatingNumber = (arr, n) => {
for (unit of arr) {
if (unit === maj_unit) count++;
else {
count--;
if (count === 0) {
maj_unit = unit;
count = 1;
}
}
}
return maj_unit;
}

You can use array#every, create an accumulator and place the number as key and its frequency as value, once the value reaches the specified number of occurrence, break from the loop using return false, then return the result.
const getRepeatingNumber = (arr, count) => {
let result = -1, hash = {};
arr.every(n => {
hash[n] = (hash[n] || 0) + 1;
if(hash[n] === count) {
result = n;
return false;
}
return true;
});
return result;
}
console.log(getRepeatingNumber([1,7,4,3,4,7],2));
console.log(getRepeatingNumber([7,4,5,3,5,5,3,4,3,2,1], 3));
console.log(getRepeatingNumber([1,6,4,6,4,6], 4));

Below is the code which will fix your problem I hope.
You need to loop over all array values and find how many time each value occurred and save number of occurrence in a result array because there may be multiple values occurred n number of times. i.e. in your provided array let arr = [1,7,4,3,4,7]; 7 and 4 occurred twice but i am returning result[0] since you might only need first occurred value for n times.
let arr = [1,7,4,3,4,7];
let getRepeatingNumber = function (arr, n) {
let result = [];
arr.forEach((value) => {
if (arr.filter(val => val === value).length === n && !result.includes(value)) {
result.push(value);
}
});
return result.length ? result[0] : -1
}
console.log(getRepeatingNumber(arr, 2)); // 7

Addition with carryover in arrays in pure Javascript

I want to create simple addition of array with carryover. Also need carryover and result value for display.
Something like this:-
e.g var input = [[0,0,9],[0,9,9]];
var carryover = [];
var result = [];
Thanks...

The two parts you might have been struggling with, I assume would be how you get the carry, and how you get the result..
result [diget] = t % 10;
The % 10 part is what is called modulus, here I'm doing a modulus by 10, so that gets you the 10's unit value.
carryover [diget] = Math.trunc(t / 10);
For the carryover, you just then divide by 10, and then we strip the decimals,. That's what Math.trunc does.
var input = [[0,0,0,9],[0,9,9]];
var carryover = [];
var result = [];
var digetlength = Math.max(input[0].length, input[1].length);
//lets padd inputs to be same size
input[0].unshift(
...new Array(digetlength - input[0].length).fill(0));
input[1].unshift(
...new Array(digetlength - input[1].length).fill(0));
for (var diget = digetlength - 1; diget >= 0; diget -= 1) {
var t = input[0][diget] + input[1][diget];
if (diget < digetlength - 1)
t += carryover[diget + 1];
result [diget] = t % 10;
carryover [diget] = Math.trunc(t / 10);
}
result.unshift(carryover[0]);
console.log('result: ' + result.join(', '));
console.log('carry: ' + carryover.join(', '));

1.turn both numbers into array of digits, reverse them.
2.determine the end index of the for-loop with max length of above 2 arrays.
3.create the 3rd carryover digits array of zeros (don't forget the extra digit).
4.Add the respective digits from step1 and step3,
as you iterate through each of digits from right to left,
4.1 if the sum is greater than 9 then add 1 into next carryover slot.
5. you should have array of carried over digits when the for-loop is done
count the number of 1s you have in them.
function numberOfCarryOperations(num1, num2) {
const dd1 = [...num1.toString()].reverse()
const dd2 = [...num2.toString()].reverse()
const end = Math.max(dd1.length, dd2.length)
const carry = Array(end+1).fill(0)
for (let i = 0; i < end; i++) {
//console.log(i,(Number(dd1[i]?dd1[i]:0)),Number(dd2[i]?dd2[i]:0),carry)
if (((Number(dd1[i]?dd1[i]:0)) + Number(dd2[i]?dd2[i]:0) + carry[i]) > 9) {
carry[i+1] = 1
}
//console.log('-----',carry)
}
//console.log(num1, num2,carry)
return carry.reduce((sum,curr)=>sum+curr)
}

Here is my attempt. It will accept the following as input:
Any number of input arrays
The input arrays don't all need to have the same number of items
I've added code comments to explain what goes on, I hope they're informative enough to explain the answer.
const
input = [
[0,0,9],
[0,9,9],
[1,0,9,9]
];
function getMaxArrayLength(values) {
// Determine the number of items in the longest array. Initialize the reduce with 0.
return values.reduce((maxLength, array) => {
// Return the largets number between the last largest number and the
// length of the current array.
return Math.max(maxLength, array.length);
}, 0);
}
function sumValues(values) {
const
// Determine the number of items in the longest array.
maxLength = getMaxArrayLength(values),
result = [],
carry = [];
// Loop over the length of the longest array. The value of index will be substracted from
// the length of the input arrays. Therefore it is easier to start at 1 as this will
// return a proper array index without needing to correct it.
for (let index = 1; index <= maxLength; index++) {
const
// Get the carryover value from the last sum or 0 in case there is no previous value.
carryValue = (carry.length === 0) ? 0 : carry[carry.length-1],
// Sum up all the values at the current index of all the input arrays. After summing up
// all the values, also add the carry over from the last sum.
sum = values.reduce((sum, array) => {
// Determine the index for the current array. Start at the end and substract the
// current index. This way the values in the array are processed last to first.
const
arrayIndex = array.length - index;
// It could be the current array doesn't have as many items as the longest array,
// when the arrayIndex is less than 0 just return the current result.
if (arrayIndex < 0) {
return sum;
}
// Return the accumulated value plus the value at the current index of the
// current source array.
return sum + array[arrayIndex];
}, 0) + carryValue;
// The carry over value is the number of times 10 fits into the sum. This should be rounded
// down so for instance 5/10=.5 becomes 0.
carry.push(Math.floor(sum / 10));
// Push the remainder of the sum divided by 10 into the result so 15 % 10 = 5.
result.push(sum % 10);
}
// Return the carry over and the result, reverse the arrays before returning them.
return {
carryOver: carry.reverse(),
result: result.reverse()
};
}
const
result = sumValues(input);
console.log(`Carry over: ${result.carryOver}`);
console.log(`Result: ${result.result}`);

javascript hackerranks sherlock and array performance issue

Watson gives Sherlock an array A of length N. Then he asks him to
determine if there exists an element in the array such that the sum of
the elements on its left is equal to the sum of the elements on its
right. If there are no elements to the left/right, then the sum is
considered to be zero. Formally, find an i, such that,
Input Format
The first line contains T, the number of test cases. For each test
case, the first line contains N, the number of elements in the array
A. The second line for each test case contains N space-separated
integers, denoting the array A.
Constraints
1<=T<=10
1<=N<=10^5
1<=Ai<=2*10^4
1<=i<=N
Output Format
For each test case print YES if there exists an element in the array,
such that the sum of the elements on its left is equal to the sum of
the elements on its right; otherwise print NO.
Sample Input
2
3
1 2 3
4
1 2 3 3
Sample Output
NO
YES
Explanation
For the first test case, no such index exists. For the second test
case,
therefore index 3 satisfies the given conditions.
I'm having timeout issues on 3 of the test cases
function check(input) {
var result = "NO";
var sum=0;
input.map(function(data){
sum=sum+(+data);
})
sumLeft=0;
sumRight=sum-(+input[0]);
for(var i=1;i<input.length;i++){
sumLeft=sumLeft+(+input[i-1]);
sumRight=sumRight-(+input[i])
if(sumLeft==sumRight)
{
console.log("YES");
return;
}
}
console.log("NO");
}
function processData(input) {
//Enter your code here
var lines = input.split("\r\n");
for (var m = 2; m < lines.length; m = m + 2) {
check(lines[m].split(" "));
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function(input) {
_input += input;
});
process.stdin.on("end", function() {
processData(_input);
});

Loop over the array once to find the sum. Declare two variables: sumLeft and sumRight. sumLeft should have an initial value of 0 and sumRight should be totalSum-arr[0].
Iterate over the array again and increment sumLeft by the (n-1) element and decrement sumRight by the nth element. Keep comparing the two variables to check if they equal each other. You cut your time complexity down to O(n)
The below code passed the test on https://www.hackerrank.com/challenges/sherlock-and-array . The tricky part was setting up default responses for when the array length was 1. I will admit that #trincot 's answer was more efficient (n as opposed to 2n) for arrays containing only positive integers.
function check(input) {
var result = "NO";
var sum=0;
if(input.length == 1){
console.log("YES");
return;
}
input.map(function(data){
sum=sum+(+data);
})
sumLeft=0;
sumRight=sum-(+input[0]);
for(var i=1;i<input.length-1;i++){
sumLeft=sumLeft+(+input[i-1]);
sumRight=sumRight-(+input[i])
if(sumLeft==sumRight)
{
console.log("YES");
return;
}else if (sumLeft>sumRight) { ///worked both with and without this optimization
console.log("NO");
return;
}
}
console.log("NO");
}
function processData(input) {
//var lines = input.split("\r\n");
var lines = input.split(/\r|\n/)
for (var m = 2; m < lines.length; m = m + 2) {
check(lines[m].split(" "));
}
}
process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function(input) {
_input += input;
});
process.stdin.on("end", function() {
processData(_input);
});

You could go through the array from both ends in inwards direction using two pointers (indices). Keep a balance, starting with 0, as follows:
When the balance is negative move the left pointer one step to the right while increasing the balance with the value you leave behind. When the balance is positive, move the right pointer one step to the left while decreasing the balance with the value you leave behind.
When the two pointers meet each other, check the balance. If it is zero, you have success.
Here is the algorithm in ES6 code, together with a text area where you can adapt the input according to the required input format:
function hasMiddle(a) {
var balance = 0, i = 0, j = a.length-1;
while (i < j) balance += balance > 0 ? -a[j--] : a[i++];
return !balance;
}
// I/O: event handling, parsing input, formatting output
var input = document.querySelector('textarea');
var output = document.querySelector('pre');
input.oninput = function() {
var lines = this.value.trim().split(/[\r\n]+/).filter(s => s.trim().length);
// Strip the case count and array element counts:
lines = lines.slice(1).filter( (s, i) => i % 2 );
// Call function for each test case, returning array of booleans:
var results = lines.map( line => hasMiddle(line.match(/\d+/g).map(Number)) );
// Output results
output.textContent = results.map( pos => pos ? 'YES' : 'NO' ).join('\n');
}
// Evaluate input immediately
input.oninput();
Input:<br>
<textarea style="width:100%; height:120px">2
3
1 2 3
4
1 2 3 3
</textarea>
<pre></pre>
This algorithm requires your input array to consist of non-negative numbers.
If you need to support negative numbers in your array, then the algorithm needs to go through the array first to calculate the sum, and then go through the array again to find the point where the balance reaches 0:
function hasMiddle(a) {
var balance = a.reduce( (sum, v) => sum + v );
return !a.every ( (v, i) => balance -= v + (i ? a[i-1] : 0) );
}
// I/O for snippet
var input = document.querySelector('textarea');
var output = document.querySelector('pre');
input.oninput = function() {
var lines = this.value.trim().split(/[\r\n]+/).filter(s => s.trim().length);
// Strip the case count and array element counts:
lines = lines.slice(1).filter( (s, i) => i % 2 );
// Call function for each test case, returning array of booleans:
var results = lines.map( line => hasMiddle(line.match(/[\d-]+/g).map(Number)));
// Output results
output.textContent = results.map( pos => pos ? 'YES' : 'NO' ).join('\n');
}
// Evaluate input immediately
input.oninput();
Input:<br>
<textarea style="width:100%; height:120px">2
3
1 2 3
4
1 2 3 3
</textarea>
<pre></pre>

Given that we have a proper array one might do as follows
var arr = [...Array(35)].map(_ => ~~(Math.random()*10)+1),
sum = arr.reduce((p,c) => p+c),
half = Math.floor(sum/2),
ix;
console.log(JSON.stringify(arr));
midix = arr.reduce((p,c,i,a) => { (p+=c) < half ? p : !ix && (ix = i);
return i < a.length - 1 ? p : ix;
},0);
console.log("best possible item in the middle # index", midix,": with value:",arr[midix]);
console.log("sums around midix:",
arr.slice(0,midix)
.reduce((p,c) => p+c),
":",
arr.slice(midix+1)
.reduce((p,c) => p+c));
Of course for randomly populated arrays as above, we can not always get a perfect middle index.