I have made a bubble sort algorithm (sorta) using JS. It works sometimes, but the problem is that it only iterates through the array once. Here is my code:
function bubble(arr) {
for (var i = 0; i < arr.length; i++) {
if (arr[i] > arr[i + 1]) {
var a = arr[i]
var b = arr[i + 1]
arr[i] = b
arr[i + 1] = a
}
}
return arr;
}
Another bubble sort implementation:
const bubbleSort = array => {
const arr = Array.from(array); // avoid side effects
for (let i = 1; i < arr.length; i++) {
for (let j = 0; j < arr.length - i; j++) {
if (arr[j] > arr[j + 1]) {
[arr[j], arr[j + 1]] = [arr[j + 1], arr[j]];
}
}
}
return arr;
};
You need an inner loop to complete the sort correctly:
function bubble(arr) {
var len = arr.length;
for (var i = 0; i < len ; i++) {
for(var j = 0 ; j < len - i - 1; j++){ // this was missing
if (arr[j] > arr[j + 1]) {
// swap
var temp = arr[j];
arr[j] = arr[j+1];
arr[j + 1] = temp;
}
}
}
return arr;
}
document.write(bubble([1,9,2,3,7,6,4,5,5]));
Please look at the following sequence:
[5, 4, 3, 2, 1]
Now lets say you need to sort this in the ascending order using bubble sort.
So, you iterate the array and swap adjacent elements which are ordered otherwise.
Here is what you will get after the completion of the iteration
[4, 3, 2, 1, 5]
Now if you do this another time, you will get this:
[3, 2, 1, 4, 5]
Likewise, you need to repeat the iteration enough times to get it sorted fully. This means you need 2 nested loops. The inner loop is to iterate the array and the outer loop is to repeat the iteration.
Please see the step-by-step example of this article.
const bubbleSort = (array)=>{
let sorted = false;
let counter =0;
while(!sorted){
sorted = true;
for(let i =0; i < array.length -1 -counter; i++){
if(array[i] > array[i+1]){
helper(i,i+1,array);
sorted = false;
}
}
counter++;
}
return array;
}
//swap function
function helper(i,j, array){
return [array[i],array[j]] = [array[j],array[i]]
}
let array=[8,5,2,9,5,6,3];
console.log(bubbleSort(array))
var array = [6,2,3,7,5,4,1];
function bubbleSort(arr) {
for(let j=0;j<arr.length;j++) {
for(let i = 0; i < arr.length; i++) {
if(arr[i]>arr[i+1] && (i+1 < arr.length)) {
var temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
}
}
}
return arr;
}
console.log(bubbleSort(array));
My bubble sort with just a while loop :
function bubbleSort(arr){
var sorted = false
while (!sorted){
sorted = true;
arr.forEach(function (element, index, array){
if (element > array[index+1]) {
array[index] = array[index+1];
array[index+1] = element;
sorted = false;
}
});
}
}
function bubble(arr) {//You need Two Loops for Bubble sort
for (var i = 0; i < arr.length; i++) {//Outer Loop
for(var j=0; j < arr.length - 1; j++){//Inner Loop
if (arr[j] > arr[j + 1]) {
var a = arr[j]
var b = arr[j + 1]
arr[j] = b
arr[j + 1] = a
}
}
}
return arr;
}
Another form of bubble sort includes starting at the end of the array and placing the smallest element first and going till the largest. This is the code:
function bubbleSort(items) {
var length = items.length;
for (var i = (length - 1); i >= 0; i--) {
//Number of passes
for (var j = (length - i); j > 0; j--) {
//Compare the adjacent positions
if (items[j] < items[j - 1]) {
//Swap the numbers
var tmp = items[j];
items[j] = items[j - 1];
items[j - 1] = tmp;
}
}
}
}
Note Bubble sort is one of the slowest sorting algorithms.
It works for me. I commented the code for more understanding
bubbleSort = (numbersArray) => {
const arrayLenght = numbersArray.length;
for (let i = 0; i < arrayLenght; i++) {
for(let j = 0; j < arrayLenght; j++) {
// Print only to debug
// console.log(`i: ${i} - j: ${j}`);
// console.log(`numbersArray[i]: ${numbersArray[i]} | numbersArray[j]: ${numbersArray[j]}`);
// Check if current number is greater than the next number
if (numbersArray[j] > numbersArray[j + 1]) {
// Store current value to generate swap
const currentNumber = numbersArray[j];
// Now the current position get value of the next position
// And de next position get value of the current position
numbersArray[j] = numbersArray[j + 1];
numbersArray[j + 1] = currentNumber;
}
}
}
// Debug: Print the sorted array
console.log(`sorted array: ${numbersArray.toString()}`);
}
const numbers = [
[3, 10, 5, 7],
[8, 5, 2, 9, 5, 6, 3],
[4, 50, 28, 47, 9, 2097, 30, 41, 11, 3, 68],
[3, 10, 5, 7, 8, 5, 2, 9, 5, 6, 3]
];
numbers.forEach(element => {
bubbleSort(element);
});
Output:
sorted array: 3,5,7,10
sorted array: 2,3,5,5,6,8,9
sorted array: 3,4,9,11,28,30,41,47,50,68,2097
sorted array: 2,3,3,5,5,5,6,7,8,9,10
var arr = [5, 3, 4, 1, 2, 6];
function sort (arr) {
for(let i=0; i < arr.length - 1; i++) {
if(arr[i] > arr[i+1]) {
let b = arr[i+1];
arr[i+1] = arr[i];
arr[i] = b;
i = -1; // Resets the loop
}
}
return arr;
}
console.log(sort(arr));
Try this (performance upgrade):
function bubbleSort(inputArr, reverse = false) {
const len = inputArr.length;
for (let i = 0; i < len; i++) {
for (let j = i + 1; j < len; j++) {
let a = inputArr[i];
let b = inputArr[j];
if (reverse ? a < b : a > b) {
const tmp = inputArr[j];
inputArr[j] = inputArr[i];
inputArr[i] = tmp;
}
}
}
return inputArr;
}
Use:
arr = [234,2,4,100, 1,12,5,23,12];
console.log(bubbleSort(arr)); // or console.log(bubbleSort(arr, true));
You need another loop:
var arr = [2, 1]
for(let i = 0;i<arr.length;i++){
for(let b = 0; b<arr.length;i++){
if(arr[b] > arr[b+1]){
var first = arr[b]
var second = arr[b + 1]
arr[b] = second
arr[b + 1] = first
}
}
}
Hope this helps I would recommend using quick sort if you want a high efficiency though.
const bubbleSort = (inputArr) => {
const len = inputArr.length;
for (let i = 0; i < len; i++) {
for (let j = 0; j < len; j++) {
if (inputArr[j] > inputArr[j + 1]) {
let tmp = inputArr[j];
inputArr[j] = inputArr[j + 1];
inputArr[j + 1] = tmp;
}
}
}
return inputArr;
};
const numbers = [50, 30, 10, 40, 60];
console.log(bubbleSort(numbers));
// Output: [ 10, 30, 40, 50, 60 ]
function bubbleSort(array) {
var done = false;
while (!done) {
//alert(1)
done = true;
for (var i = 1; i < array.length; i += 1) {
if (array[i - 1] > array[i]) {
//alert(2)
done = false;
var tmp = array[i - 1];
array[i - 1] = array[i];
array[i] = tmp;
}
}
}
return array;
}
Another way would be like this:
function bubbleSort(arr) {
let swapped;
do {
swapped = false;
for (var i = 0; i < arr.length; i++) {
if (arr[i] > arr[i + 1]) {
swapped = true;
var tmp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = tmp;
}
}
} while (swapped);
return arr;
}
let myArray = [8, 1, 2, 5, 51, 13, 15, 33, 123, 100, 22];
console.log(bubbleSort(myArray));
Explanation:
In this function we are going to declare a swapped variable that is being set to false inside the DO WHILE loop, this is being done as a fail-safe not to end up with an infinite loop.
Inside the loop, we have another FOR loop which iterates through the given array and checks if the current value is greater than the next (which we don't want, we need ascending order).
When the IF the condition is true, we are going to swap the variables and assign true for the swapped variable, this is done because we want to keep on the DO WHILE loop untill everything is sorted.
package hasan;
public class hssd {
public static void main(String[] args) {
int t=9;
int g=20;
for (t=g;t>19;++t){
System.out.println(7);
int f=12;
int r=15;
for(r=f;r>5;++r)
System.out.println(r+1000000000+"*"+1000000000);
}
}
}
Related
Here is my code to solve leetcode- Two sum
let nums = [2, 7, 11, 15];
let target = 9;
for (let i = 0; i < nums.length; i++) {
let x = target - nums[i];
if (nums[i + 1] === x) {
console.log([i, i+1]) // [0, 1]
return [i, i + 1];
}
} //it ok
nums.forEach((item, index) => {
let x = target - nums[index];
if (nums[index + 1] === x) {
console.log([index, index + 1]);// [0, 1]
return [index,index+1]
}
}); // undefined
I use forEach instead of for(...), both console.log are same, but forEach returns undefined. Why?
You can use JS map
because the forEach always return undefined
let nums = [2, 7, 11, 15],
target = 9;
let m = {};
nums.map((value, index) => {
m[value] = index;
});
for (i = 0; i < nums.length; i++) {
let diff = target - nums[i];
if (m.hasOwnProperty(diff)) {
console.log(i, m[diff]);
}
}
I am trying to generate sprial matrix in javascript.
question
Given an integer A, generate a square matrix filled with elements from 1 to A^2 in spiral order.
input : 3
[ [ 1, 2, 3 ],
[ 8, 9, 4 ],
[ 7, 6, 5 ] ]
when input is 4
[ [1, 2, 3, 4],
[12, 13, 14, 5],
[11, 16, 15, 6],
[10, 9, 8, 7] ]
my approach is to create 2d array with 0 value and after that they will fill values.
let generateMatrix = function(A) {
let arr = [], counter = 1;
for (let i = 0; i < A; i++) {
let items = []
for (let j = 0; j < A; j++) {
items.push(0)
}
arr.push(items)
}
var spiralMatrix = function(arr) {
if (arr.length > 1) {
for (let i = 0; i < arr[0].length; i++) {
arr[0][i] = counter++;
}
}
return arr
}
return spiralMatrix(arr)
}
console.log(generateMatrix(2))
You could take loops for each edges and loop until no more ranges are avaliable.
function spiral(length) {
var upper = 0,
lower = length - 1,
left = 0,
right = length - 1,
i = 0,
j = 0,
result = Array.from({ length }, _ => []),
value = 1;
while (true) {
if (upper++ > lower) break;
for (; j < right; j++) result[i][j] = value++;
if (right-- < left) break;
for (; i < lower; i++) result[i][j] = value++;
if (lower-- < upper) break;
for (; j > left; j--) result[i][j] = value++;
if (left++ > right) break;
for (; i > upper; i--) result[i][j] = value++;
}
result[i][j] = value++;
return result;
}
var target = document.getElementById('out'),
i = 10;
while (--i) target.innerHTML += spiral(i).map(a => a.map(v => v.toString().padStart(2)).join(' ')).join('\n') + '\n\n';
<pre id="out"></pre>
This bit of code should do what you are trying to.
// This is your Editor pane. Write your JavaScript hem and
// use the command line to execute commands
let generateMatrix = function(A) {
let arr = [],
counter = 1;
for (let i = 0; i < A; i++) {
let items = [];
for (let j = 0; j < A; j++) {
items.push(0);
}
arr.push(items);
}
var spiralMatrix = function(arr) {
let count = 1;
let k = 0; // starting row
let m = arr.length; // ending row
let l = 0; // starting column
let n = arr[0].length; //ending column
while (k < m && l < n) {
// top
for (var i = l; i < n; i++) {
arr[k][i] = count;
count++;
}
k++;
// right
for (var i = k; i < m; i++) {
arr[i][n - 1] = count;
count++;
}
n--;
// bottom
if (k < m) {
for (var i = n - 1; i >= l; i--) {
arr[m - 1][i] = count;
count++;
}
m--;
}
// left
if (l < n) {
for (var i = m - 1; i >= k; i--) {
arr[i][l] = count;
count++;
}
l++;
}
}
return arr;
};
return spiralMatrix(arr);
};
console.log(generateMatrix(4));
This is in some ways the reverse of an answer I gave to another question. We can recursively build this up by slicing out the first row and prepending it to the result of rotating the result of a recursive call on the remaining numbers:
const reverse = a =>
[...a] .reverse ();
const transpose = m =>
m [0] .map ((c, i) => m .map (r => r [i]))
const rotate = m =>
transpose (reverse (m))
const makeSpiral = (xs, rows) =>
xs .length < 2
? [[... xs]]
: [
xs .slice (0, xs .length / rows),
... rotate(makeSpiral (xs .slice (xs .length / rows), xs.length / rows))
]
const range = (lo, hi) =>
[...Array (hi - lo + 1)] .map ((_, i) => lo + i)
const generateMatrix = (n) =>
makeSpiral (range (1, n * n), n)
console .log (generateMatrix (4))
A sharp eye will note that rotate is different here from the older question. transpose (reverse (m)) returns a clockwise rotated version of the input matrix. reverse (transpose (m)) returns a counter-clockwise rotated one. Similarly, here we rotate the result of the recursive call before including it; whereas in the other question we recurse on the rotated version of the matrix. Since we're reversing that process, it should be reasonably clear why.
The main function is makeSpiral, which takes an array and the number of rows to spiral it into and returns the spiraled matrix. (If rows is not a factor of the length of the array, the behavior might be crazy.) generateMatrix is just a thin wrapper around that to handle your square case by generating the initial array (using range) and passing it to makeSpiral.
Note how makeSpiral works with rectangles other than squares:
makeSpiral ([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], 2) //=>
// [
// [ 1, 2, 3, 4, 5, 6],
// [12, 11, 10, 9, 8, 7]
// ]
makeSpiral ([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], 3) //=>
// [
// [ 1, 2, 3, 4],
// [10, 11, 12, 5],
// [ 9, 8, 7, 6]
// ]
makeSpiral ([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], 4) //=>
// [
// [ 1, 2, 3],
// [10, 11, 4],
// [ 9, 12, 5],
// [ 8, 7, 6]
// ]
makeSpiral ([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], 6) //=>
// [
// [ 1, 2],
// [12, 3],
// [11, 4],
// [10, 5],
// [ 9, 6],
// [ 8, 7]
// ]
The other functions -- range, reverse, transpose, and rotate -- are general purpose utility functions for working with arrays or matrices.
Here's one solution.
I keep the current "moving direction" in dx and dy, such that the next matrix element indices are given by x+dx and y+dy.
If the next item is already filled or is out of bounds, I change this direction clockwise. Otherwise, I fill it with the next value.
const size = 6;
const matrix = Array(size).fill().map(() => Array(size).fill(0));
let x = -1;
let y = 0;
let dx = 1;
let dy = 0;
function changeDirection() {
if (dx === 1) {
dx = 0;
dy = 1;
} else if (dy === 1) {
dy = 0;
dx = -1;
} else if (dx === -1) {
dx = 0;
dy = -1;
} else {
dx = 1;
dy = 0;
}
}
for (let i = 0; i < size * size; i++) {
const yNext = y + dy;
const xNext = x + dx;
const nextRow = matrix[yNext] || [];
const nextItemContent = nextRow[xNext];
if (nextItemContent === undefined || nextItemContent > 0) {
changeDirection();
i--;
continue;
}
y = yNext;
x = xNext;
matrix[y][x] = i + 1;
}
const result = document.getElementById('result');
matrix.forEach(row => {
row.forEach(value => {
result.innerHTML += value.toString().padStart(3);
});
result.innerHTML += '\n';
});
<pre id="result"></pre>
I'm calculating the index, each number should go in a linear array
console.clear();
Array.prototype.range = function(a, b, step) {
step = !step ? 1 : step;
b = b / step;
for(var i = a; i <= b; i++) {
this.push(i*step);
}
return this;
};
const spiral = function(dimen) {
"use strict";
const dim = dimen;
const dimw = dim;
const dimh = dim;
var steps = [1, dimh, -1, -dimh];
var stepIndex = 0;
var count = 1;
var countMax = dimw
var dec = 0
var index = 0;
var arr = [];
arr = arr.range(1, dimh * dimw)
const newArr = arr.reduce((coll, x, idx) => {
index += steps[stepIndex]
coll[index-1] = idx+1;
if (count === countMax) {count = 0; stepIndex++; dec++;}
if (dec === 1) {dec = -1; countMax--}
if (stepIndex == steps.length) {stepIndex = 0}
count++;
return coll;
}, []);
var ret = []
while (newArr.length) {
ret.push(newArr.splice(0,dimw))
}
return ret
}
console.log(spiral(3))
console.log(spiral(4))
console.log(spiral(5))
var n=14; // size of spiral
var s=[]; // empty instruction string
function emp() {} // no move
function xpp() {xp++;} // go right
function xpm() {xp--;} // go left
function ypp() {yp++;} // go down
function ypm() {yp--;} // go up
var r=[xpp,ypp,xpm,ypm]; // instruction set
s.push(emp); // push 'no move' (used for starting point)
var c=n-1;
while (c-->0) s.push(r[0]); // push first line - uses a different rule
for (var i=1;i<2*n-1;i++) { // push each leg
c=Math.floor((2*n-i)/2);
while (c-->0) s.push(r[i%4]);
}
var sp=new Array(n); // spiral array
for (var i=0;i<n;i++) sp[i]=new Array(n);
var xp=0; // starting position
var yp=0;
for (var i=0;i<n*n;i++) {
s[i](); // execute next instruction
sp[yp][xp]=i+1; // update array
}
for (var i=0;i<n;i++) console.log(sp[i].toString()); // log to console
This code makes a macro of functions to generate a run sequence, for example:
'right4, down4, left4, up3, right3, down2, left2, up1, right1
and then implements it.
Here is a solution to Spiral Matrix from leetcode, maybe this can help
https://leetcode.com/problems/spiral-matrix/
var spiralOrder = function(matrix) {
if (matrix.length == 0) {
return [];
}
let result = [];
let rowStart = 0;
let rowEnd = matrix.length - 1;
let colStart = 0;
let colEnd = matrix[0].length - 1;
while (true) {
// top
for (let i = colStart; i <= colEnd; i++) {
result.push(matrix[rowStart][i]);
}
rowStart++;
if (rowStart > rowEnd) {
return result;
}
// right
for (let i = rowStart; i <= rowEnd; i++) {
result.push(matrix[i][colEnd]);
}
colEnd--;
if (colEnd < colStart) {
return result;
}
// bottom
for (let i = colEnd; i >= colStart; i--) {
result.push(matrix[rowEnd][i]);
}
rowEnd--;
if (rowEnd < rowStart) {
return result;
}
// left
for (let i = rowEnd; i >= rowStart; i--) {
result.push(matrix[i][colStart]);
}
colStart++;
if (colStart > colEnd) {
return result;
}
}
return result;
};
console.log(
spiralOrder([[2, 3, 4], [5, 6, 7], [8, 9, 10], [11, 12, 13], [14, 15, 16]])
);
console.log(spiralOrder([[7], [9], [6]]));
console.log(spiralOrder([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]]));
console.log(spiralOrder([[1, 2, 3], [4, 5, 6], [7, 8, 9]]));
Here's my answer using only one for loop -
function matrix(n) {
const arr = [];
let row = 0;
let column = 0;
let counter = 1;
let edge = n - 1;
let leftToRightRow = false;
let topToBottomCol = false;
let rightToLeftRow = false;
let bottomToTopCol = false;
for (i = 0; i < n * n; i++) {
if (column <= edge && !leftToRightRow) {
if (!Array.isArray(arr[row])) {
arr[row] = []; // if array is not present at this index, then insert one
}
arr[row][column] = counter;
if (column == edge) {
row = row + 1;
leftToRightRow = true;
} else {
column = column + 1;
}
counter = counter + 1;
} else if (column === edge && !topToBottomCol) {
if (!Array.isArray(arr[row])) {
arr[row] = []; // if array is not present at this index, then insert one
}
arr[row][column] = counter;
if (row === edge) {
column = column - 1;
topToBottomCol = true;
} else {
row = row + 1;
}
counter = counter + 1;
} else if (column >= 0 && !rightToLeftRow) {
arr[row][column] = counter;
if (column === 0) {
row = row - 1;
rightToLeftRow = true;
} else {
column = column - 1;
}
counter = counter + 1;
} else if (row >= n - edge && !bottomToTopCol) {
arr[row][column] = counter;
if (row === n - edge) {
column = column + 1;
bottomToTopCol = true;
//setting these to false for next set of iteration
leftToRightRow = false;
topToBottomCol = false;
rightToLeftRow = false;
edge = edge - 1;
} else {
row = row - 1;
}
counter = counter + 1;
}
}
return arr;
}
Solution is implemented in C++, but only logic matter then you can do it in any language:
vector<vector<int> > Solution::generateMatrix(int A) {
vector<vector<int>> result(A,vector<int>(A));
int xBeg=0,xEnd=A-1;
int yBeg=0,yEnd=A-1;
int cur=1;
while(true){
for(int i=yBeg;i<=yEnd;i++)
result[xBeg][i]=cur++;
if(++xBeg>xEnd) break;
for(int i=xBeg;i<=xEnd;i++)
result[i][yEnd]=cur++;
if(--yEnd<yBeg) break;
for(int i=yEnd;i>=yBeg;i--)
result[xEnd][i]=cur++;
if(--xEnd<xBeg) break;
for(int i=xEnd;i>=xBeg;i--)
result[i][yBeg]=cur++;
if(++yBeg>yEnd) break;
}
return result;
}
Solition in c#:
For solving this problem we use loops for each moving directions
public IList<int> SpiralOrder(int[][] matrix) {
var result = new List<int>();
var n = matrix[0].Length;
var m = matrix.Length;
var i = 0;
var j = 0;
var x = 0;
var y = 0;
while (true)
{
//left to right moving:
while (x <= n - 1 - i)
{
result.Add(matrix[y][x]);
x++;
}
if (result.Count == n * m)
return result;
x--;y++;
//up to down moving:
while (y <= m - 1 - j)
{
result.Add(matrix[y][x]);
y++;
}
if (result.Count == n * m)
return result;
y--;x--;
//right to left moving:
while (x >= j)
{
result.Add(matrix[y][x]);
x--;
}
if (result.Count == n * m)
return result;
x++;y--;
//down to up moving:
while (y > j)
{
result.Add(matrix[y][x]);
y--;
}
if (result.Count == n * m)
return result;
y++;x++;
i++;
j++;
}
}
I am trying to find the indexes of element in an array equal to a specified sum.
I only want 2 indexes.
function sumArrayHashTable(arr, sum) {
var result = [];
var hashTable = {};
for (var i = 0; i < arr.length; i++) {
var S = sum - arr[i];
if (hashTable[S] !== undefined) {
result.push([arr[i], S]);
} else {
hashTable[arr[i]] = arr[i]
}
}
return result;
}
console.log(sumArrayHashTable([5, 2, 6, 1, 3, 9, 0], 9));
//Result should be [[2,4], [5,6]]
I am able to print the numbers but not the indexes. Please advice
Use your hash table to store the indices instead of the values. Also, push the indices in your result array:
function sumArrayHashTable(arr, sum) {
const result = [];
const hashTable = {};
for (let i = 0; i < arr.length; i++) {
const S = sum - arr[i];
if (hashTable[S] !== undefined) {
result.push([i, hashTable[S]]);
} else {
hashTable[arr[i]] = i;
}
}
return result;
}
console.log(sumArrayHashTable([5, 2, 6, 1, 3, 9, 0], 9));
//Result should be [[2,4], [5,6]]
function sumArrayHashTable(arr, sum) {
var result = [];
for (var i = 0; i < arr.length; i++) {
var test = 0;
for (var j = 0; j < arr.length; j++) {
if (i === j) {
continue ;
}
test = arr[i] + arr[j];
if (test === sum) {
result.push([i, j]);
}
}
}
return result;
}
console.log(sumArrayHashTable([5, 2, 6, 1, 3, 9, 0], 9));
Result will be [[2,4], [5,6]] plus [[4,2], [6,5]] you can take it from now
alghorithm is simple, O(n^2) - loop over array once, then again and find sum that equals 9
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 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;
}
}
}