I know various ways to sort an array, but when the thing comes like i will have to manage time and space complexity i just don’t know what to do how to do
Help me to sort an array in better way with less time and space complexity
Assuming like i have 10k elements in an array and i will have to sort that array, which sorting algorithm will be better?
var Arr = [ 3, 4,1,2, 5,7];
Arr.sort((a,b)=> a-b);
console.log('Arr.sort------',Arr);//[ 1, 2, 3, 4, 5, 7 ]
OR
for (var i = 1; i < Arr.length; i++){
console.log('i:',i)
for (var j = 0; j < i; j++){
console.log(j)
if (Arr[i] < Arr[j]) {
var x = Arr[i];
Arr[i] = Arr[j];
Arr[j] = x;
}
}
}
console.log('sort------',Arr);//[ 1, 2, 3, 4, 5, 7 ]
OR
function selectionSort(arr) {
var minIdx, temp,
len = arr.length;
for (var i = 0; i < len; i++) {
minIdx = i;
for (var j = i + 1; j < len; j++) {
if (arr[j] < arr[minIdx]) {
minIdx = j;
}
}
temp = arr[i];
arr[i] = arr[minIdx];
arr[minIdx] = temp;
}
return arr;
}
var selectionSortArr=Arr
var ddd = selectionSort(selectionSortArr);
console.log('sort -------',selectionSortArr);//[ 1, 2, 3, 4, 5, 7 ]
I am practing at hackerrank and I have an exercise with
two-dimensional matrix. I am facing an error in my implementation
11 2 4
4 5 6
10 8 -12
I need to sum across the primary diagonal: 11 + 5 - 12 = 4 after the other diagonal 4 + 5 +10 = 19 finally 19 - 4 = 15
function diagonalDifference(arr) {
var sumRigth = 0;
var sumLeft = 0;
var array = new Array();
for(var i = 0; i < arr.length ; i++ ){
for(var j = 0; j < arr[i].length; j++){
array.push(arr[i][j]);
}
}
for (var i = 0 ; i < array.length; i = i + 4){
sumRigth += array[i];
}
for (var j = 2 ; j < array.length - 1 ; j = j + 2 ){
sumLeft += array[j];
}
return sumLeft - sumRigth;
}
you can try this
function sumDiagonal(matrix) {
let firstSum = 0, secondSum = 0;
for (let row = 0; row < matrix.length; row++) {
firstSum += matrix[row][row];
secondSum += matrix[row][matrix.length - row - 1];
}
console.log(firstSum + ' ' + secondSum);
console.log(firstSum-secondSum);
}
sumDiagonal([[11,2,4],[4,5,6],[10,8,-12]]);
I don't think you're on the right path. A general solution would first sum the elements from top-left to bottom-right (saved here as sumRigth). Then, sum the elements from top-right to bottom-left (saved here as sumLeft). I took it for granted that arrays contain numbers and are of the same size.
function diagonalDifference(array) {
let sumRigth = 0, sumLeft = 0, count = 0;
for (var i = 0 ; i < array.length; i++){
sumRigth += array[i][count++];
}
count = array.length-1;
for (var i = 0; i < array.length; i++){
sumLeft += array[i][count--];
}
return sumLeft - sumRigth;
}
let arr = [
[11, 2, 4],
[4, 5, 6],
[10, 8, -12]
];
console.log(diagonalDifference(arr));
You could take a single loop and get two values dierctly for summing.
function getValue(matrix) {
let sum = 0;
for (let i = 0, l = matrix.length; i < l; i++)
sum += matrix[i][l - i - 1] - matrix[i][i];
return sum;
}
console.log(getValue([[11, 2, 4], [4, 5, 6], [10, 8, -12]]));
I am just starting with JS and need some help with calculating a gcd.
I would like to calculate the gcd for every combination of two elements from two arrays.
I mean: for each element A[i] from array A, for each element B[j] from B, calculate the gcd value of A[i] and B[j] and print it in the console. I do have 16 prints, but they are not correct. I use Euclid's algorithm to calculate it and it looks like the A[i] value is overwritten. I have no idea why. Could someone help me? This is my code:
var n = 4;
var A = [2, 5, 6, 7];
var B = [4, 9, 10, 12];
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
while (A[i] != B[j]) {
if (A[i] < B[j]) {
B[j] = B[j] - A[i];
} else {
A[i] = A[i] - B[j];
}
}
console.log(A[i]);
}
}
You are modifying your array elements while performing euclid's algorithm. I recommend creating a separate function for this algorithm, like:
var n = 4;
var A = [2, 5, 6, 7];
var B = [4, 9, 10, 12];
for (var i = 0; i < n; i++) {
for (var j = 0; j < n; j++) {
console.log(euclid(A[i], B[j]));
}
}
function euclid(a, b) {
while (b != 0) {
var r = a % b;
a = b;
b = r;
}
return a;
}
Edit:
You can make and use a storage array in the following way:
var C = []; // The array that will contain the arrays
for (var i = 0; i < n; i++) {
C[i] = []; // "Inner array"
for (var j = 0; j < n; j++) {
C[i][j] = euclid(A[i], B[j]);
console.log(C[i][j]);
}
}
Without complicationg your algorithm , use forEach as following :
const A = [2, 5, 6, 7];
const B = [4, 9, 10, 12];
const gcd = (x, y) => (!y) ? x : gcd(y, (x % y));
A.forEach((a, i) => {
B.forEach((b, j) => {
console.log(
`GCD(A[${i}]=${a}, B[${j}]=${b}) ＝`, gcd(a, b)
);
});
})
i'm just beginning to learn javascript and this is my first question on stackoverflow, so feel free to criticize me if i'm approaching this the wrong way.
var divisibleByThreePairSum = function(array) {
var pairs = [];
for (var i = 0; i < array.length; i++) {
for (var j = i++; j < array.length; j++) {
var sum = array[i] + array[j];
if (sum % 3 === 0) {
pairs.push([i, j]);
}
}
}
return pairs;
}
console.log(divisibleByThreePairSum([3,1,0,2,1,3,2,0]));
This gives me the answer;
[ [ 1, 3 ], [ 1, 6 ], [ 3, 4 ], [ 5, 5 ], [ 5, 7 ], [ 7, 7 ] ]
[Finished in 0.2s]
For the second "for" loop, I formatted it like so, (j = i++) as to avoid repeats like [1, 3], [3, 1], but I can't seem to get rid of getting pairs like [5, 5], and [7, 7]. Is there any possible ways to format the code differently so that this doesn't happen? Again, I apologize if this was asked improperly; i'll definitely be using this site more often so please let me know if i'm doing anything wrong "question format" wise, Thanks!
Issue is j = i++. This will assign value of i to j and then increment value of i. This will also result in skipping of alternate values of i as it is incremented twice.
for(var i = 0; i< 5; i++){
for(var j = i++; j< 5; j++){
console.log(i,j)
}
}
You should rather use j=i+1. This will sent next value and will not increment value of i
var divisibleByThreePairSum = function(array) {
var pairs = [];
for (var i = 0; i < array.length; i++) {
for (var j = i+1; j < array.length; j++) {
var sum = array[i] + array[j];
if (sum % 3 === 0) {
pairs.push([i, j]);
}
}
}
return pairs;
}
console.log(divisibleByThreePairSum([3, 1, 0, 2, 1, 3, 2, 0]));
I am new to JavaScript, and I have an array which contains numbers.
var arr = [2,4,8,1,5,9,3,7,6];
How can I sort it using a native for loop in JavaScript?
I know sort function is available, but I want it through for loop.
The output should be:
var res = [1,2,3,4,5,6,7,8,9];
var Arr = [1, 7, 2, 8, 3, 4, 5, 0, 9];
for (var i = 1; i < Arr.length; i++)
for (var j = 0; j < i; j++)
if (Arr[i] < Arr[j]) {
var x = Arr[i];
Arr[i] = Arr[j];
Arr[j] = x;
}
console.log(Arr);
I would do something like this...
var input = [2,3,8,1,4,5,9,7,6];
var output = [];
var inserted;
for (var i = 0, ii = input.length ; i < ii ; i++){
inserted = false;
for (var j = 0, jj = output.length ; j < jj ; j++){
if (input[i] < output[j]){
inserted = true;
output.splice(j, 0, input[i]);
break;
}
}
if (!inserted)
output.push(input[i])
}
console.log(output);
Maybe there are more efficient ways, but if you want to use the for loop, it's my first idea...
First create an empty array where the sorted numbers will be pushed into.
let sorted = [];
Secondly, create a very large amount of numbers that none of the numbers of the array can match. This number will be used for the very first comparison to determine which number of the array is smaller.
let comparison = 9000000000;
Create a for loop.
This loop will have another loop inside of it. The inner loop will check for the smallest number in a given array, and once the smallest number is gotten, it will be push into the empty array we created. The smallest number will also be removed from the initial array and then the array will run again.
for(a = 0; a < arr.length; a++){
//This inner loop fetches the smallest number.
for(b = 0; b < arr.length; a++){
if(comparison > arr[b]){
comparison = arr[b];
}
}
// The smallest number is assigned to comparison
// Now it being pushed to the empty array
sorted.push(comparison);
// Remove the smallest number from the initial array
let indexOfSmallNumber = arr.indexOf(comparison);
arr.splice(indexOfSmallNumber, 1);
// Set the comparison back to 9000000000;
comparison = 90000000000;
a = -1;
// Here, "a" is our main loop index counter and we are
// setting it to -1 because we don't want it to change
// to 2 by default, doing this will make the loop run
// forever until the initial array is empty.
}
let arr = [4, 2, 5, 1]
let temp;
function converter(arr) {
for(let i=0; i<arr.length; i++) {
for (let j=i+1; j<arr.length; j++) {
if(arr[i] > arr[j]) {
temp = arr[i]
arr[i] = arr[j]
arr[j] = temp
}
}
}
return arr
}
const newArr = converter(arr)
console.log(newArr)
Use:
let s = [4, 6, 3, 1, 2];
for (let i = 0; i < s.length;) {
if (s[i] > s[i + 1]) {
let a = s[i];
s[i] = s[i + 1];
s[i + 1] = a;
i--;
}
else {
i++;
}
}
This is a sorting algorithm which has a best time complexity of O(n) and the worst time of O(n^2).
This code checks for each number, and then compares to all numbers on the left side.
To check the time it takes each code to run, you can also use this code below:
let start = process.hrtime.bigint()
let end = process.hrtime.bigint()
console.log(end - start) // This measures the time used in nano seconds.
Also for microseconds, you can use this performance.now().
Here there is a very simple solution that uses a temporary array to store the values greater than the current one. Then it puts the current value between the lesser and the greater values:
var arr = [2,4,8,1,5,9,3,7,6];
var res = [];
for (const c of arr) {
let tmp = [];
while (c < res[res.length-1]) {
tmp.unshift(res.pop());
}
res = [...res, c, ...tmp];
}
const numberArr = [5, 9, 2, 8, 4, 10, 1, 3, 7, 6];
function sortedFunction(arr) {
let sortedArr = [];
for (let i = 0; i < arr.length; i++) {
for (let j = i + 1; j < arr.length; j++) {
let n = 0;
if (arr[i] > arr[j]) {
n = arr[i];
arr[i] = arr[j];
arr[j] = n;
}
}
sortedArr.push(arr[i]);
}
return sortedArr;
}
sortedFunction(numberArr);
Under the JavaScript array sort section of W3Schools it talks about how to compare a value in an array with the others and then order them based on the values being returned. I updated the code to use a for loop to sort values.
// Ascending points
var points = [5.0, 3.7, 1.0, 2.9, 3.4, 4.5];
var output = [];
var i;
for (i = 0; i < points.length; i++) {
points.sort(function (a, b) {
return a - b
});
output += points[i] + "<br>";
}
console.log(output);
// Descending points
var points = [5.0, 3.7, 1.0, 2.9, 3.4, 4.5];
var output = [];
var i;
for (i = 0; i < points.length; i++) {
points.sort(function (a, b) {
return b - a
});
output += points[i] + "<br>";
}
console.log(output);
const array = [12, 3, 45, 61, 23, 45, 6, 7];
function sortArray(array) {
for (var i = 0; i < array.length; ++i) {
for (var j = 0; j < array.length - 1 - i; ++j) {
if (array[j] > array[j + 1]) {
[array[j], array[j + 1]] = [array[j + 1], array[j]];
}
}
}
return array;
}
console.log(sortArray(array));
Here are the two solutions for the same algorithm:
Solution 1:
We can directly use JavaScript functions:
let arr = [2, 4, 8, 1, 5, 9, 3, 7, 6]
const changeOrder = (arr) => {
return arr.sort((a, b) => a - b)
}
let result = changeOrder(arr);
console.log(result) // [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
Solution 2:
We can use a JavaScript for loop for doing the same
let arr = [2, 4, 8, 1, 5, 9, 3, 7, 6]
const changeOrder = (arr) => {
for(let i=1; i< arr.length; i++) {
for(let j=0; j < i; j++) {
if(arr[i] < arr[j]) {
let x = arr[i]
arr[i] = arr[j]
arr[j] = x
}
}
}
return arr;
}
let result = changeOrder(arr);
console.log(result) // [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
An improvement to previous answer
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j < arr.length - i - 1; j++) {
if (arr[j] > arr[j + 1]) {
let temp = arr[j];
arr[j] = arr[j + 1];
arr[j + 1] = temp;
}
}
}