Create multidimensionnal array with for loop - javascript

I want to create a multidimensionnale array using a for loop.
The result i need is :
[
[1, 2, 3],
[4,5,6],
[7,8,9]
]
So how can i loop from 1 to 10 and each 3 numbers i create a new array ?
I don't find a solution.. Thanks a lot for your help

This should be what you're looking for:
var outer = [];
var inner = [];
for (var i = 1; i < 10; i++) {
inner.push(i);
if (inner.length == 3) {
outer.push(inner);
inner = [];
}
}
if (inner.length > 0) outer.push(inner);
console.log(outer);
Remove the if after the loop, if you don't want any elements with less than 3 inner elements.

This might be the simplest solution:
const result = [];
for(let i = 1; i < 10; i = i + 3) {
result.push([i, i+1, i+2]);
}
console.log(result);
If you're sure you want 3 consecutive numbers, you can loop from 1 to 10 skipping 3 every time, so your i inside for loop would be 1, then 4 and then 7 in last iteration. With every iteration you create [i, i+1, i+2] array on the spot and push it to result array. However this solution is based on above conditions. It works well for 10, for any number it would require additional if statement inside for loop.

You could take a single loop and use a variable for the first index of the 2D array.
var i, r, array = [];
for (i = 0; i < 9; i++) {
r = Math.floor(i / 3);
array[r] = array[r] || [];
array[r].push(i + 1);
}
console.log(array);

var c = 1;
for(var i = 0; i < 3; i++)
for(var j = 0; j < 3; j++)
array[i][j] = ++c;

Related

find all numbers disappeared in array

Please help me to solve this leetcode problem using javascript as I am a beginner and dont know why this code is not working
Ques: Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
var findDisappearedNumbers = function (nums) {
var numLength = nums.length;
nums.sort(function (a, b) { return a - b });
for (var i = 0; i < nums.length - 1; i++) {
if (nums[i + 1] === nums[i]) {
nums.splice(i, 1);
}
}
for (var k = 0; k < nums.length; k++) {
for (var j = 1; j <= numLength; j++) {
if (nums[k] !== j) {
return j;
}
}
}
};
if there is any error in my code please let me know;
i have done the following thing
first i have sorted the array in ascending order
then i have cut all the duplicate elements
then i have created loop that will check if nums[k] !== j ;
and it will return j which is the missing number;
for example this is the testcase [4,3,2,7,8,2,3,1]
first my code will sort this in ascending order [1,2,2,3,3,4,7,8]
then it will remove all duplicate elements and it will return [1,2,3,4,,7,8]
and then it will check nums[k] is not equal to j and it will print j
I think it'd be easier to create a Set of numbers from 1 to n, then just iterate through the array and delete every found item from the set:
var findDisappearedNumbers = function(nums) {
const set = new Set();
for (let i = 0; i < nums.length; i++) {
set.add(i + 1);
}
for (const num of nums) {
set.delete(num);
}
return [...set];
};
console.log(findDisappearedNumbers([4,3,2,7,8,2,3,1]));
To fix your existing code, I'm not sure what the logic you're trying to implement in the lower section, but you can iterate from 1 to numLength (in the outer loop, not the inner loop) and check to see if the given number is anywhere in the array. Also, since you're mutating the array with splice while iterating over it in the upper loop, make sure to subtract one from i at the same time so you don't skip an element.
var findDisappearedNumbers = function(nums) {
var numLength = nums.length;
nums.sort(function(a, b) {
return a - b
});
for (var i = 0; i < nums.length - 1; i++) {
if (nums[i + 1] === nums[i]) {
nums.splice(i, 1);
i--;
}
}
const notFound = [];
outer:
for (var j = 1; j < numLength; j++) {
for (var k = 0; k < nums.length; k++) {
if (nums[k] === j) {
continue outer;
}
}
notFound.push(j);
}
return notFound;
};
console.log(findDisappearedNumbers([4, 3, 2, 7, 8, 2, 3, 1]));
#CertainPerformance certainly cracked it again using the modern Set class. Here is a slighly more conservative approach using an old fashioned object:
console.log(missingIn([4,3,2,7,8,2,3,1]));
function missingIn(arr){
const o={};
arr.forEach((n,i)=>o[i+1]=1 );
arr.forEach((n) =>delete o[n] );
return Object.keys(o).map(v=>+v);
}
My solution for the problem to find the missing element
var findDisappearedNumbers = function(nums) {
const map={};
const result=[];
for(let a=0;a<nums.length;a++){
map[nums[a]]=a;
}
for(let b=0;b<nums.length;b++){
if(map[b+1]===undefined){
result.push(b+1)
}
}
return result;
};
Example 1:
Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]
Example 2:
Input: nums = [1,1]
Output: [2]

Loop trough array and sum the length

EDITED:
Can someone help me with the problem below further. I have a class and an array inside the class. I want now use a for loop to sum the length of the previous array length, but for each iteration. If i == 1 I want sum the length of arr[0].x.length, If i == 2 I want sum the length of arr[0].x.length+arr[1].x.length, ect. It will be a lot of code to check each iteration.
Is there a simple way to do this? Instead allways use a new line like
if (i == 1) n = n + arr[i-1].x.length;
if (i == 2) n = n + arr[i-1].x.length+arr[i-2].x.length;
if (i == 3) n = n + arr[i-1].x.length+arr[i-2].x.length+arr[i-3].x.length;
function Class() {
var x = [];
}
for (var i = 0; i < 9; i++) {
arr[i] = new Class();
}
I add 4 items to each object.
arr[0].x.push(...)
arr[0].x.push(...)
...
arr[1].x.push(...)
arr[1].x.push(...)
...
var n = 0;
for (var i = 0; i < arr.length; i++) {
if (i == 1) {
n = n + arr[i-1].x.length;
} else if (i == 2) {
n = n + arr[i-1].x.length+arr[i-2].x.length;
} else if (i == 3) {
n = n + arr[i-1].x.length+arr[i-2].x.length+arr[i-3].x.length;
}
// ect.
}
You could use reduce to get a total of all the lengths of your sub-arrays. For example:
const arrs = [[1, 2, 3], [4, 5, 6]];
const sum = arrs.reduce((acc, arr) => acc += arr.length, 0);
console.log(sum);
// 6
Just nest the loop two times: go over the indexes once then go up to that index from 0 in an inner loop:
for (var i = 0; i < arr1.length; i++) {
for(var j = 0; j <= i; j++) {
n = n + arr1[j].length;
}
}
Edit: benvc's answer is what you are looking for if you want to use reduce.
var arr = [[1,2,3], [4,5,6], [7]];
var n = 0;
for (var i = 0; i < arr.length; i++){
n += arr[i].length;
}
console.log(n);

How to get even numbers array to print first instead of odds?

So I have this function where I've need to take out the evens and odds and put them into separate arrays but I need the evens array to print first rather than the odds.
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
function divider(numbersArray) {
var evensOdds = [[], []];
for (var i = 0; i < numbersArray.length; i++) {
evensOdds[i & 1].push(numbersArray[i]);
}
return evensOdds;
}
If you want to split the number by their even and odd values, instead of using the index (i), determine the sub array to push into using the value - numbersArray[i] % 2.
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
function divider(numbersArray) {
var evensOdds = [[], []];
for (var i = 0; i < numbersArray.length; i++) {
evensOdds[numbersArray[i] % 2].push(numbersArray[i]);
}
return evensOdds;
}
console.log(divider(numbersArray));
If you want to split them by even and odd indexes use (i + 1) % 2 to determine the sub array to push into:
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
function divider(numbersArray) {
var evensOdds = [[], []];
for (var i = 0; i < numbersArray.length; i++) {
evensOdds[(i + 1) % 2].push(numbersArray[i]);
}
return evensOdds;
}
console.log(divider(numbersArray));
Just for fun, a forEach version of the accepted answer.
var numbersArray = [1,2,34,54,55,34,32,11,19,17,54,66,13];
var even_odd = [ [], [] ];
numbersArray.forEach( e => even_odd[e%2].push(e) );
console.log(even_odd);

Group items of two in one array

I am trying to push numbers in an array into another array in groups of two.
If I have an array [1,4,3,2]; it should return [[1,4],[3,2]];
var arrayPairSum = function(nums) {
var len = nums.length / 2;
var arr = [];
for(var i = 0; i < len; i ++) {
var newArr = [];
newArr.push(nums[i]);
newArr.push(nums[i + 1]);
arr.push(newArr);
}
console.log(arr); //this should give me [[1,4],[3,2]];
};
arrayPairSum([1,4,3,2]);
can anyone see what I need to do to achieve this? I cannot figure it out.
You can use reduce method to achieve this. reduce method accepts a callback method provided on every item in the array.
In the other words, this method applies a function against an accumulator and each element in the array (from left to right) to reduce it to a single value.
var array=[1,4,3,2,8];
var contor=array.reduce(function(contor,item,i){
if(i%2==0)
contor.push([array[i],array[i+1]].filter(Boolean));
return contor;
},[]);
console.log(contor);
If you really want to iterate over the array, may skip every second index, so i+=2 ( as satpal already pointed out) :
var arrayPairSum = function(nums) {
var len = nums.length - 1;//if nums.length is not even, it would crash as youre doing nums[i+1], so thats why -1
var arr = [];
for (var i = 0; i < len; i += 2) {
var newArr = [];
newArr.push(nums[i]);
newArr.push(nums[i + 1]);
arr.push(newArr);
}
console.log(arr); //this should give me [[1,4],[3,2]];
};
arrayPairSum([1, 4, 3, 2]);
The upper one crops away every non pair at the end. If you want a single [value] at the end, may go with
len=nums.length
And check later before pushing
if(i+1<nums.length) newArr.push(nums[i+1]);
You were pretty close. Simply change the length to nums.length and in the loop increment i by 2.
var arrayPairSum = function(nums) {
var len = nums.length - 1;
var arr = [];
for(var i = 0; i < len; i+=2) {
var newArr = [];
newArr.push(nums[i]);
newArr.push(nums[i + 1]);
arr.push(newArr);
}
console.log(arr); //this should give me [[1,4],[3,2]];
};
arrayPairSum([1,4,3,2]);

Reversing certain number of elements in an array javascript

I am working on a code where I need to reverse certain no of elements in an array and rest should remain same. For example is an array has values of 1,2,3,4,5,6 and I have to reverse 4 elements of it then output should be 4,3,2,1,5,6. I am using below code to achieve this but getting error, please suggest.
function reverseArray(n, a) {
var interimArray1 = [];
//var interimArray2=[];
//var finalArray=[];
for (var i < n; i >= 0; i--) {
interimArray1.push[a[i]];
}
for (var i = n; i < a.length; i++) {
interimArray1.push[a[i]];
}
for (var i = 0; i < interimArray1.length; i++) {
console.log(interimArray1[i]);
}
}
var arr = [1, 2, 3, 4, 5, 6];
var num = 4;
reverseArray(num, arr);
The error in your code is that you intend to call the push method on a[i] like so:
interimArray1.push(a[i]);
but instead you write:
interimArray1.push[a[i]];
You make that mistake twice. To give arguments to the push method, you must use round parenthesis ().
With that fixed, you will see that your code works perfectly.
You can use Array#slice, Array#splice as follow.
function partialReverse(arr, num, from = 0) {
var slicedArr = arr.slice(from, num + from);
arr.splice(from, num); // Remove `num` items from array
arr.splice(from, 0, ...slicedArr.reverse()); // Add `num` reversed items
return arr;
}
var arr = [1, 2, 3, 4, 5, 6];
console.log(partialReverse(arr, 4, 0)); // Reverse four items from `arr` starting from 0th index
console.log(partialReverse(arr, 4, 1)); // Reverse four items from `arr` starting from 1st index
Lots of hints but you seem to be missing them. ;-)
You need to assign an initial value to i, so:
for (var i = n; ... )
===========^
Also, you need to use () to call functions, not [], so:
interimArray1.push(a[i]);
==================^====^
Same in the following for block. Otherwise, the code works though it's more verbose than it needs to be.
This is working :
I'm sure there are faster ways of doing it. Also, it will only work for elements at the beginning of the array but you can adjust the function for what you want to achieve.
var reverseArray = function(arr,elementsToReverse) {
var tempArrayRev = [];
var tempArray = [];
for (var i=0;i<arr.length;i++) {
if (i < elementsToReverse) {
tempArrayRev[i] = arr[i];
} else {
tempArray.push(arr[i]);
}
}
return tempArrayRev.reverse().concat(tempArray);
}
var array = [1,2,3,4,5,6];
document.getElementById('arrayOutput').innerHTML += reverseArray(array,4);
<div id="arrayOutput">Array :<br></div>
This is the answer you can test it.
function reverseArray(n, a) {
var interimArray1 = [];
for (var i = 0; i < a.length; i++) {
interimArray1.push(a[i]);
}
for (var i = num; i >=0; i--) {
interimArray1[i-1] = a[n - i];
}
for (var i = 0; i < interimArray1.length; i++) {
console.log(interimArray1[i]);
}
}
var arr = [1, 2, 3, 4, 5, 6];
var num = 4;
reverseArray(num, arr);
You could use something like this.
function reverseArray(n, arrIn) {
// Splice splits the array in 2 starting at 0 index going n long
var arrOut = arrIn.splice(0,n);
// reverse is pretty straight forward
arrOut = arrOut.reverse();
// Concat joins the two together
return arrOut.concat(arrIn);
}

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