I have the following array as an example;
let arr = [['red','blue','pink],['dog','cat','bird'],['loud', 'quiet']]
I need to write a generalized function that prints all combinations of one word from the first vector, one word from the second vector, etc. I looked up some codes on here but they are all recursion or working only with the specific array. How can I write this code without recursion?
let allComb = function(arr) {
if (arr.length == 1) {
return arr[0];
} else {
let result = [];
let arrComb = allComb(arr.slice(1));
for (let i = 0; i < arrComb.length; i++) {
for (let j = 0; j < arr[0].length; j++) {
result.push(arr[0][j] + ' ' + arrComb[i]);
}
}
return result;
}
}
allComb(arr)
This version uses a single increment per cycle technique with no recursion.
let arr = [
['red', 'blue', 'pink'],
['dog', 'cat', 'bird'],
['loud', 'quiet']
];
function allComb(arr) {
var total = 1;
var current = [];
var result = [];
for (var j = 0; j < arr.length; j++) {
total *= arr[j].length;
current[j] = 0;
}
for (var i = 0; i < total; i++) {
var inc = 1;
result[i] = "";
for (var j = 0; j < arr.length; j++) {
result[i] += arr[j][current[j]] + ' ';
if ((current[j] += inc) == arr[j].length)
current[j] = 0;
else
inc = 0;
}
}
return (result);
}
console.log(allComb(arr));
You may do as follows;
var arr = [['red','blue','pink'],['dog','cat','bird'],['loud', 'quiet']],
res = arr.reduce((p,c) => p.reduce((r,x) => r.concat(c.map(y => x + " " + y)),[]));
console.log(res);
Related
I am currently doing a codewars problem, and I think I almost got it however, I ran across a problem when sorting index values with the same letter. link to problem is here. https://www.codewars.com/kata/5782dd86202c0e43410001f6
function doMath(s) {
let strSplit = s.split(' ');
let clonedArr = strSplit.slice();
for (let i = 0; i < strSplit.length; i++) {
for (let j = 0; j < strSplit[i].length; j++) {
let current = strSplit[i][j];
if (isNaN(current)) {
let letter = current;
strSplit[i] = strSplit[i].replace(letter, '');
strSplit[i] = letter + strSplit[i];
}
}
}
let sortedArr = strSplit.sort();
console.log(sortedArr);
// ["b900", "y369", "z123", "z246", "z89"]
let noLetterArr = sortedArr.map(x => {
return x.slice(1);
});
let numberArr = noLetterArr.map(y => {
return +y;
})
let firstEl = numberArr[0];
for (let i = 1; i < numberArr.length; i++) {
if (numberArr.indexOf(numberArr[i]) % 4 == 1) {
firstEl += numberArr[i];
}
if (numberArr.indexOf(numberArr[i]) % 4 == 2) {
firstEl -= numberArr[i];
}
if (numberArr.indexOf(numberArr[i]) % 4 == 3) {
firstEl *= numberArr[i];
}
}
return firstEl;
}
console.log(doMath('24z6 1z23 y369 89z 900b'));
I would like to sort the sortedArr the ones with the same letter by how they first appeared in string. So since "z246" appeared first in the original string. I would like to have that before "1z23". I had a hard time creating a function for that.
var al = [];
function doMath(s) {
var ar = s.split(" ");
for (let i = 0; i < ar.length; i++) {
for (let char of ar[i]) {
let temp = char.match(/[a-z]/i);
if (temp) {
al[i] = char;
ar[i] = ar[i].replace(char, '');
ar[i] = char + ar[i];
}
}
}
al = al.sort();
//New Sort Logic to pass above test case and others too
var n = [];
for (let i = 0; i < al.length; i++) {
for (let j = 0; j < ar.length; j++) {
if (ar[j].startsWith(al[i]) && !n.includes(ar[j])) {
n.push(ar[j]);
}
}
}
var result = parseInt(n[0].substr(1)),
count = 1;
for (let i = 1; i < n.length; i++) {
if (count == 1) {
result = result + parseInt(n[i].substr(1));
count++;
} else if (count == 2) {
result = result - parseInt(n[i].substr(1));
count++;
} else if (count == 3) {
result = result * parseInt(n[i].substr(1));
count++;
} else if (count == 4) {
result = result / parseInt(n[i].substr(1));
count = 1;
}
}
return Math.round(result);
}
How do I display the input parameter like this:
displayText("you and me");
expected output:
["you and me", "you and", "and me", "you", "and", "me"]
I have tried like the following code, but the results are still wrong. My code is like this now.
let displayText = (txt) => {
let output= []
for(let i = 0; i < txt.length; i++) {
for(j = i + 1; j < txt.length + 1; j++) {
output.push(txt.slice(i, j))
}
}
return output
}
First of all you can split the sentence:
const sentence = 'you and me'
const parts = sentence.split(' ')
The difficult part is to extract the elements by pairs. Use slice for this:
const numberOfPairs = parts.length - 1
for (let i=0; i<numberOfPairs ; i++) {
console.log(parts.slice(i, i+2).join(' '))
}
Then, print the individual parts:
parts.forEach(part => console.log(part))
first you have to split sentence
Example:-
var strArr = string.split(" ");
var output= []
var temp = "";
for (var i = 0; i < strArr.length; i++) {
temp = strArr[i];
output.push(temp)
for (var j = i + 1; j < strArr.length; j++) {
temp += strArr[j];
output.push(temp)
}
}
console.log(output)
You may do like this.
You will need to split() the string on the word breaks to have your function work as is (and also declare j).
let displayText = (txt) => {
txt = txt.split(' ');
let output = []
for (let i = 0; i < txt.length; i++) {
for (let j = i + 1; j < txt.length + 1; j++) {
output.push(txt.slice(i, j).join(' '))
}
}
return output
}
console.log(displayText("you and me"))
To return an array sorted by combination length, you can group each slice() by word count and flat() the output array before returning.
let displayText = (txt) => {
txt = txt.split(' ');
let
len = txt.length, output = [];
for (let i = 0; i < len; i++) {
for (let j = i + 1; j < len + 1; j++) {
(output[len - (j - i)] ??= []).push(txt.slice(i, j).join(' '))
}
}
// before flat() call
// [ [ 'you and me' ], [ 'you and', 'and me' ], [ 'you', 'and', 'me' ] ]
return output.flat()
}
console.log(displayText("you and me"));
I have the following function to get all of the substrings from a string in JavaScript. I know it's not correct but I feel like I am going about it the right way. Any advice would be great.
var theString = 'somerandomword',
allSubstrings = [];
getAllSubstrings(theString);
function getAllSubstrings(str) {
var start = 1;
for ( var i = 0; i < str.length; i++ ) {
allSubstrings.push( str.substring(start,i) );
}
}
console.log(allSubstrings)
Edit: Apologies if my question is unclear. By substring I mean all combinations of letters from the string (do not have to be actual words) So if the string was 'abc' you could have [a, ab, abc, b, ba, bac etc...] Thank you for all the responses.
You need two nested loop for the sub strings.
function getAllSubstrings(str) {
var i, j, result = [];
for (i = 0; i < str.length; i++) {
for (j = i + 1; j < str.length + 1; j++) {
result.push(str.slice(i, j));
}
}
return result;
}
var theString = 'somerandomword';
console.log(getAllSubstrings(theString));
.as-console-wrapper { max-height: 100% !important; top: 0; }
A modified version of Accepted Answer. In order to give the minimum string length for permutation
function getAllSubstrings(str, size) {
var i, j, result = [];
size = (size || 0);
for (i = 0; i < str.length; i++) {
for (j = str.length; j - i >= size; j--) {
result.push(str.slice(i, j));
}
}
return result;
}
var theString = 'somerandomword';
console.log(getAllSubstrings(theString, 6));
Below is a recursive solution to the problem
let result = [];
function subsetsOfString(str, curr = '', index = 0) {
if (index == str.length) {
result.push(curr);
return result;
}
subsetsOfString(str, curr, index + 1);
subsetsOfString(str, curr + str[index], index + 1);
}
subsetsOfString("somerandomword");
console.log(result);
An answer with the use of substring function.
function getAllSubstrings(str) {
var res = [];
for (let i = 0; i < str.length; i++) {
for (let j = i + 1; j <= str.length; j++) {
res.push(str.substring(i, j));
}
}
return res;
}
var word = "randomword";
console.log(getAllSubstrings(word));
function generateALlSubstrings(N,str){
for(let i=0; i<N; i++){
for(let j=i+1; j<=N; j++){
console.log(str.substring(i, j));
}
}
}
Below is a simple approach to find all substrings
var arr = "abcde";
for(let i=0; i < arr.length; i++){
for(let j=i; j < arr.length; j++){
let bag ="";
for(let k=i; k<j; k++){
bag = bag + arr[k]
}
console.log(bag)
}
}
function getSubstrings(s){
//if string passed is null or undefined or empty string
if(!s) return [];
let substrings = [];
for(let length = 1 ; length <= s.length; length++){
for(let i = 0 ; (i + length) <= s.length ; i++){
substrings.push(s.substr(i, length));
}
}
return substrings;
}
I'm having a little trouble with my attempt at this problem. Code Below:
function pasc(n){
var result = [[1]];
for (var row = 1; row < n; row++){
for (var col = 1; col <= row; col++){
result[row][col] = result[row - 1][col] + result[row - 1][col - 1];
}
}
return result;
}
pasc(10)
for (var i = 0; i < result.length; i++){
document.write(result[i]+"<br>");
}
It seems the problem hinges on assigning values to an array using an expression like myArray[1][1] = "foo"
I'm confused about this because I can do this: var myArray = []; myArray[4] = "foo" which seems to suggest that an element can be created at an arbitrary position in a 1 dimensional array, but not with 2 dimensions.
Any help with clearing up my misconceptions appreciated.
The Pascal's Triangle can be printed using recursion
Below is the code snippet that works recursively.
We have a recursive function pascalRecursive(n, a) that works up till the number of rows are printed. Each row is a element of the 2-D array ('a' in this case)
var numRows = 10,
triangle,
start,
stop;
// N is the no. of rows/tiers
// a is the 2-D array consisting of the row content
function pascalRecursive(n, a) {
if (n < 2) return a;
var prevRow = a[a.length-1];
var curRow = [1];
for (var i = 1; i < prevRow.length; i++) {
curRow[i] = prevRow[i] + prevRow[i-1];
}
curRow.push(1);
a.push(curRow);
return pascalRecursive(n-1, a); // Call the function recursively
}
var triangle = pascalRecursive(numRows, [[1]]);
for(var i = 0; i < triangle.length; i++)
console.log(triangle[i]+"\n");
JavaScript doesn't have two-dimensional arrays. What it does have is arrays that happen to contain other arrays. So, yes, you can assign a value to any arbitrary position in an array, and the array will magically make itself big enough, filling in any gaps with 'undefined'... but you can't assign a value to any position in a sub-array that you haven't explicitly created yet. You have to assign sub-arrays to the positions of the first array before you can assign values to the positions of the sub-arrays.
Replacing
for (var row = 1; row < n; row++){
for (var col = 1; col <= row; col++){
with
for (var row = 1; row < n; row++){
result[row] = [];
for (var col = 1; col <= row; col++){
should do it. Assuming all of your indexing logic is correct, anyway. You've got some problems there, too, since your initial array only contains a single value, so result[row][col] = result[row - 1][col] + result[row - 1][col - 1]; is accessing at least one cell that has never been defined.
Thanks Logan R. Kearsley. I have now solved it:
function pasc(n){
var result = [];
result[0] = [1];
result[1] = [1,1];
for (var row = 2; row < n; row++){
result[row] = [1];
for (var col = 1; col <= row -1; col++){
result[row][col] = result[row-1][col] + result[row-1][col-1];
result[row].push(1);
}
}
return result;
}
for (var i = 0; i < pasc(10).length; i++){
document.write(pasc(10)[i]+"<br>");
console.log(pasc(10)[i]+"<br>");
}
you can create Pascal's triangle using below code:
function pascal(n) {
var arr = [];
if (n == 1) {
arr[0] = [];
arr[0][0] = 1;
} else if (n == 2) {
arr[0] = [];
arr[0][0] = 1;
arr[1] = [];
arr[1][0] = 1;
arr[1][1] = 1;
} else if (n > 2) {
arr[0] = [];
arr[1] = [];
arr[0][0] = 1;
arr[1][0] = 1;
arr[1][1] = 1;
for (i = 2; i < n; i++) {
arr[i] = [];
arr[i][0] = 1;
for (j = 1; j < i; j++) {
arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];
}
arr[i][j] = 1;
}
}
console.log(arr);
for (i = 0; i < arr.length; i++) {
console.log(arr[i].join(' '))
}
}
function pascal(n) {
var arr = [];
if (n == 1) {
arr[0] = [];
arr[0][0] = 1;
} else if (n == 2) {
arr[0] = [];
arr[0][0] = 1;
arr[1] = [];
arr[1][0] = 1;
arr[1][1] = 1;
} else if (n > 2) {
arr[0] = [];
arr[1] = [];
arr[0][0] = 1;
arr[1][0] = 1;
arr[1][1] = 1;
for (i = 2; i < n; i++) {
arr[i] = [];
arr[i][0] = 1;
for (j = 1; j < i; j++) {
arr[i][j] = arr[i - 1][j - 1] + arr[i - 1][j];
}
arr[i][j] = 1;
}
}
console.log(arr);
for (i = 0; i < arr.length; i++) {
console.log(arr[i].join(' '))
}
}
pascal(5)
This function will calculate Pascal's Triangle for "n" number of rows. It will create an object that holds "n" number of arrays, which are created as needed in the second/inner for loop.
function getPascalsTriangle(n) {
var arr = {};
for(var row = 0; row < n; row++) {
arr[row] = [];
for(var col = 0; col < row+1; col++) {
if(col === 0 || col === row) {
arr[row][col] = 1;
} else {
arr[row][col] = arr[row-1][col-1] + arr[row-1][col];
}
}
}
return arr;
}
console.log(getPascalsTriangle(5));
Floyd triangle
You can try the following code for a Floyd triangle
var prevNumber=1,i,depth=10;
for(i=0;i<depth;i++){
tempStr = "";j=0;
while(j<= i){
tempStr = tempStr + " " + prevNumber;
j++;
prevNumber++;
}
console.log(tempStr);
}
You can create arbitrary 2d arrays and store it in there and return the correct Pascal.
JavaScript does not have a special syntax for creating multidimensional arrays. A common workaround is to create an array of arrays in nested loops.
source
Here is my version of the solution
function pascal(input) {
var result = [[1], [1,1]];
if (input < 0) {
return [];
}
if (input === 0) {
return result[0];
}
for(var j = result.length-1; j < input; j++) {
var newArray = [];
var firstItem = result[j][0];
var lastItem = result[j][result[j].length -1];
newArray.push(firstItem);
for (var i =1; i <= j; i++) {
console.log(result[j][i-1], result[j][i]);
newArray.push(sum(result[j][i-1], result[j][i]));
}
newArray.push(lastItem);
result.push(newArray);
}
return result[input];
}
function sum(one, two) {
return one + two;
}
Here is the code i created for pascal triangle in javascript
'use strict'
let noOfCoinFlipped = 5
let probabiltyOfnoOfHead = 2
var dataStorer = [];
for(let i=0;i<=noOfCoinFlipped;i++){
dataStorer[i]=[];
for(let j=0;j<=i;j++){
if(i==0){
dataStorer[i][j] = 1;
}
else{
let param1 = (j==0)?0:dataStorer[i-1][j-1];
let param2 = dataStorer[i-1][j]?dataStorer[i-1][j]:0;
dataStorer[i][j] = param1+param2;
}
}
}
let totalPoints = dataStorer[noOfCoinFlipped].reduce((s,n)=>{return s+n;})
let successPoints = dataStorer[noOfCoinFlipped][probabiltyOfnoOfHead];
console.log(successPoints*100/totalPoints)
Here is the link as well
http://rextester.com/TZX59990
This is my solve:
function pascalTri(n){
let arr=[];
let c=0;
for(let i=1;i<=n;i++){
arr.push(1);
let len=arr.length;
if(i>1){
if(i>2){
for(let j=1;j<=(i-2);j++){
let idx=(len-(2*i)+j+2+c);
let val=arr[idx]+arr[idx+1];
arr.push(val);
}
c++;
}
arr.push(1);
}
}
return arr;
}
let pascalArr=pascalTri(7);
console.log(pascalArr);
here is the pattern for n = 3
#
##
###
here is js code to print this.
function staircase(n) {
for(var i=0 ; i<n ; i++) {
for(var j=n-1 ; j>i ; j--)
process.stdout.write(" ");
for(var k=0 ; k<=i; k++) {
process.stdout.write("#");
}
process.stdout.write("\n");
}
}
class PascalTriangle {
constructor(n) {
this.n = n;
}
factoriel(m) {
let result = 1;
if (m === 0) {
return 1;
}
while (m > 0) {
result *= m;
m--;
}
return result;
}
fill() {
let arr = [];
for (let i = 0; i < this.n; i++) {
arr.push([]);
}
for (let i = 0; i < arr.length; i++) {
for (let j = 0; j <= i; j++) {
arr[i].push(this.factoriel(i) / (this.factoriel(j) * this.factoriel(i - j)));
}
}
return arr;
}
}
var m = prompt("enter number:");
var arrMain = new Array();
for (var i = 0; i < m; i++) {
arrMain[i] = [];
}
for (var i = 0; i < m; i++) {
if (i == 0) {
arrMain[i] = [1];
} else if (i == 1) {
(arrMain[i]) = [1, 1];
} else {
for (var j = 0; j <= i; j++) {
if (j == 0 || j == arrMain[i - 1].length) {
arrMain[i][j] = 1;
} else {
arrMain[i][j] = arrMain[i - 1][j] + arrMain[i - 1][j - 1];
}
}
}
document.write(arrMain[i] + "<br>");
}
This is my take on this problem by gaining access to the previous row.
const generate = numRows => {
const triangle = [[1]]
for (let i = 1; i < numRows; i++) {
// Previous row
const previous = triangle[i - 1]
// Current row
const current = new Array(i + 1).fill(1)
// Populate the current row with the previous
// row's values
for (let j = 1; j < i; j++) {
current[j] = previous[j - 1] + previous[j]
}
// Add to triangle result
triangle.push(current)
}
return triangle
}
As an exercise, I'm trying to create a function that returns the palindromic numbers resulting from multiplying three-digit numbers. As far as I can tell, the function is running through numbers correctly, however, the resulting array is incorrect. I don't need the solution to the palindrome problem...just an idea of what I might be missing. Have I run into some limitation?
var palindromic = function() {
var a = [];
var res = [];
for (var i = 100; i < 1000; i++) {
a.push(i);
}
var ar = a.slice(0);
a.map(function(x) {
for (var j = 0; j < ar.length; j++) {
var result = x * ar[j];
if (result.toString() === result.toString().split("").reverse().join("")) {
res.push(result);
}
}
})
return res;
};
Pretty sure it's just trying to call console.log() 810,000 times. If you comment the console.log line, it works just fine.
var palindromic = function() {
var a = [];
var res = [];
for (var i = 100; i < 1000; i++) {
a.push(i);
}
var ar = a.slice(0);
a.map(function(x) {
for (var j = 0; j < ar.length; j++) {
var result = x * ar[j];
//console.log(x + " : " + ar[j] + ' = ' + result);
if (result.toString() === result.toString().split("").reverse().join("")) {
res.push(result);
}
}
});
return res;
};
console.log(palindromic());