I've finished this challenge from Coderbyte, but inelegantly:
Have the function PrimeChecker(num) take num and return 1 if any
arrangement of num comes out to be a prime number, otherwise return 0.
For example: if num is 910, the output should be 1 because 910 can be
arranged into 109 or 019, both of which are primes.
My solution works by producing an array of all possible permutations of the digits in the num argument, then scanning it for primes:
function PrimeChecker(num) {
// Accounting for 1 not being a prime number by definition
if (num == 1) {
return 0;
}
// Defining an empty array into which all permutations of num will be put
var resultsArray = [];
// Breaking num into an array of single-character strings
var unchangedArray = num.toString().split('');
// Function to push all permutations of num into resultsArray using recursion
function getAllCombos (array, count) {
if (count === num.toString().length) {
return;
}
else {
for (var i = 0; i < array.length; i++) {
var temp = array[count];
array[count] = array[i];
array[i] = temp;
resultsArray.push(array.join(''));
}
return getAllCombos(array, count+1) || getAllCombos(unchangedArray, count+1);
}
}
getAllCombos(unchangedArray, 0);
// Converting the results from strings to numbers and checking for primes
var numArr = [];
resultsArray.forEach(function(val, indx) {
return numArr[indx] = Number(val);
});
for (var i = 0; i < numArr.length; i++) {
var prime = 1;
for (var j = 2; j < numArr[i]; j++) {
if (numArr[i] % j == 0) {
prime = 0;
}
}
if (prime == 1) {
return prime;
}
}
return 0;
}
The problem is that the array of permutations of the num argument which I'm producing is full of duplicates.. I feel like this can be done more efficiently.
For example, running PrimeChecker(123) results in a permutations array of 20 entries, when there only need be 6.
Does anyone have an idea of how to do this more efficiently?
This is how I would do it:
var permute = (function () {
return permute;
function permute(list) {
return list.length ?
list.reduce(permutate, []) :
[[]];
}
function permutate(permutations, item, index, list) {
return permutations.concat(permute(
list.slice(0, index).concat(
list.slice(index + 1)))
.map(concat, [item]));
}
function concat(list) {
return this.concat(list);
}
}());
function isPrime(n) {
for (var i = 2, m = Math.sqrt(n); i <= m; i++)
if (n % i === 0) return false;
return true;
}
function PrimeChecker(num) {
return permute(num.toString(10).split("")).some(function (xs) {
return isPrime(parseInt(xs.join(""), 10));
}) ? 1 : 0;
}
alert(PrimeChecker(910));
Here's the explanation of the algorithm for finding the permutations of the elements of an array.
Hope that helps.
Related
The problem at hand is :
Given an array nums of distinct integers, return all the possible permutations. You can return the answer in any order.
I have this code with an ok run time but it seems that the push function is not working, I am not sure why bc it all makes sense in general
Example 1:
Input: nums = [1,2,3]
Output: [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
Constraints:
1 <= nums.length <= 6
-10 <= nums[i] <= 10
All the integers of nums are unique.
var permute = function(nums) {
if (nums == null) {
return []
}
return getPermutations(nums);
};
function getPermutations(arr) {
var set = {}
var size = factorial(arr.length)
var result = []
while (Object.keys(set).length < size) {
for (var i = 0; i < arr.length && result.length < size; i++) {
for (var j = arr.length - 1; j >= 0 && result.length < size; j--) {
if (!set[arr]) {
set[arr] = 1
console.log(arr) //clearly see the permutations printed
result.push(arr) //why is this not working...
}
arr = swap(arr,i,j)
}
}
}
return result
}
function swap(arr,i,j) {
var tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
return arr
}
function factorial(n) {
if (n == 0 || n == 1) {
return 1
}
return n*factorial(n-1)
}
You're pushing the same array multiple times into the result array.
You can fix this by creating a copy of the arr array before pushing it.
(So the code after it can't mutate the array again)
So instead of result.push(arr) you could use one of those examples:
// using splash operator
result.push([...arr]);
// Array#slice()
result.push(arr.slice());
// Array.from()
result.push(Array.from(arr));
// etc...
Working Example:
var permute = function(nums) {
if (nums == null) {
return []
}
return getPermutations(nums);
};
function getPermutations(arr) {
var set = {}
var size = factorial(arr.length)
var result = []
while (Object.keys(set).length < size) {
for (var i = 0; i < arr.length && result.length < size; i++) {
for (var j = arr.length - 1; j >= 0 && result.length < size; j--) {
if (!set[arr]) {
set[arr] = 1
console.log(arr) //clearly see the permutations printed
result.push([...arr]) //why is this not working...
}
arr = swap(arr,i,j)
}
}
}
return result
}
function swap(arr,i,j) {
var tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
return arr
}
function factorial(n) {
if (n == 0 || n == 1) {
return 1
}
return n*factorial(n-1)
}
console.log(permute([1,2,3]));
This question could also be a good read, it contains a lot of examples for how to efficiently calculate permutations in javascript.
I want to find all possible arrays -of non-negative numbers- that sum up to -at most- N in JavaScript:
function findArrays(maxSize, maxSum){}
Example input: findArrays(3, 10)
Some acceptable outputs: (not writing all as it would be too long)
[[0], [0,0,0], [10,0,0], [1,9], [1,2,3] /*, ... */]
What I tried so far:
I know it looks like homework but it's not :) I can think of a solution that simply generates all (size*maxSum) possible arrays of acceptable sizes and then iterate through them to check if sum is greater than maxSum. However, I think this solution is very bad in terms of performance as maxSum gets bigger. I'm looking for a more efficient implementation but I just don't know where to start.
My "bad" solution
function getNextArray(r,maxVal){
for(var i=r.length-1;i>=0;i--){
if(r[i]<maxVal){
r[i]++;
if(i<r.length-1){
r[i+1]=0;
}
break;
}
}
return r;
}
function getAllArraysOfSize(size, maxVal){
var arrays=[],r=[],i;
for(i=0;i<size;i++){
r[i]=0;
}
while(r.reduce((a, b) => a + b, 0) < (maxVal*size)){
r = getNextArray(r.slice(),maxVal);
arrays.push(r);
}
return arrays;
};
function findArrays(maxSize, maxSum){
var allArrays=[],arraysOfFixedSize=[],acceptableArrays=[],i,j;
for(i=1; i<=maxSize; i++){
arraysOfFixedSize=getAllArraysOfSize(i,maxSum);
for(j=0; j<arraysOfFixedSize.length; j++){
allArrays.push(arraysOfFixedSize[j]);
}
}
for(i=0; i<allArrays.length; i++){
if(allArrays[i].reduce((a, b) => a + b, 0) <= maxSum){
acceptableArrays.push(allArrays[i]);
}
}
return acceptableArrays;
};
You can use recursion and a generator. The number of outputs grows quickly for higher valued arguments, so I keep them low here:
function * findArrays(maxSize, maxSum) {
let arr = [];
function * recur(maxSum) {
let k = arr.length;
yield [...arr]; // or: if (k) yield [...arr]
if (k === maxSize) return;
for (let i = 0; i <= maxSum; i++) {
arr[k] = i;
yield * recur(maxSum - i);
}
arr.length = k;
}
yield * recur(maxSum);
}
// demo
for (let arr of findArrays(2, 4))
console.log(JSON.stringify(arr));
NB: this also produces the empty array, which makes sense. If you want to avoid this, then just check that you don't yield an empty array.
If you prefer working with plain functions instead of generators, then translate the innermost yield expression to a push unto a result array, as follows:
function findArrays(maxSize, maxSum) {
let arr = [];
let result = []; // <--- will collect all the subarrays
function recur(maxSum) {
let k = arr.length;
result.push([...arr]);
if (k === maxSize) return;
for (let i = 0; i <= maxSum; i++) {
arr[k] = i;
recur(maxSum - i);
}
arr.length = k;
}
recur(maxSum);
return result;
}
// demo
for (let arr of findArrays(2, 4))
console.log(JSON.stringify(arr));
i hope this is helpful
const data = [[0],[0,0,0],[10,0,0],[1,9],[1,2,3]];
function findArrays(maxSize, maxSum){
return data.reduce(
(acc, value) => {
if (value.length <= maxSize) {
const tempValue = value;
const sum = tempValue.reduce((acc, val) => val >= 0 ? acc + val : 0, 0);
if (sum <= maxSum && sum > 0) acc.push(value);
}
return acc
}, []
)
}
console.log(findArrays(3, 10));
So I've gotten the function to return the divisors beside 1 and itself, but my loop is still adding empty array elements since it increments the int argument. Can someone see where I'm going wrong?
function divisors(integer) {
var result = [];
for(let i = 0; i < integer; i++) {
if(i !== 1 && i !== integer && integer % i == 0) {
result[i] = i;
}
}
return result;
};
console.log(divisors(12)); // returns array length of 7, 3 emtpy slot and 2, 3, 4, 6 ??
You should add the elements using push()
function divisors(integer) {
var result = [];
for(let i = 0; i < integer; i++) {
if(i !== 1 && i !== integer && integer % i == 0) {
result.push(i)
}
}
return result;
};
console.log(divisors(12));
A better way is to start the loop from 2. And no need to check i !== integer because i will be always less than integer.
function divisors(integer) {
var result = [];
for(let i = 2; i < integer; i++) {
if(integer % i == 0) {
result.push(i)
}
}
return result;
};
console.log(divisors(12));
A more cleaner on-liner will be using filter()
const divisors = num => [...Array(num)]
.map((x,i) => i)
.filter(x => num % x === 0).slice(1)
console.log(divisors(12));
So the question reads:
Given an array a that contains only numbers in the range from 1 to a.length, find the first duplicate number for which the second occurrence has the minimal index. In other words, if there are more than 1 duplicated numbers, return the number for which the second occurrence has a smaller index than the second occurrence of the other number does. If there are no such elements, return -1.
Write a solution with O(n) time complexity and O(1) additional space complexity.
I have a solution, but apparently it's not fast enough and stalls when there are over a thousand items in the array.
This is what I have:
function firstDuplicate(arr) {
let dictionary = {};
for(let i = 0, ii = arr.length; i < ii; i++) {
for(let z = i+1, zz = arr.length; z < zz; z++) {
if(arr[i] === arr[z]) {
if(dictionary.hasOwnProperty(arr[i])) {
if(dictionary[arr[i]] !== 0 && dictionary[arr[i]] > z) {
dictionary[i] = z;
}
} else {
dictionary[arr[i]] = z;
}
}
}
}
let answer = [];
for(key in dictionary) {
// [array number, position];
answer.push([key, dictionary[key]]);
};
if(answer.length > 0) {
return Number(answer.sort((a, b) => {
return a[1]-b[1];
})[0][0]);
}
return -1;
}
I think converting the object into an array and then sorting the array after the answers are complete slows down the whole function. Using built in JS methods like forEach, map and sort (like I did above), slows the code/function down even more. There is obviously a better and more accurate way to do this, so I'm asking for some JS masterminds to help me out on this.
you can keep adding numbers to a dictionary as keys with values as their index, and as soon as you find a repeating key in the dictionary return its value. This will be O(n) time complexity and O(n) space complexity.
function firstDuplicate(arr) {
var dictionary = {};
for(var i = 0; i < arr.length; i++) {
if(dictionary[arr[i]] !== undefined)
return arr[i];
else
dictionary[arr[i]] = i;
}
return -1;
}
console.log(firstDuplicate([2, 3, 3, 1, 5, 2]));
Since the numbers are between 1 to arr.length you can iterate on the array. For each arr[i] use arr[i] as index and make the element present and arr[arr[i]] negative, then the first arr[arr[i]] negative return arr[i]. This give O(1) space complexity and O(n) time complexity you can do this:
function firstDuplicate(arr) {
for(var i = 0; i < arr.length; i++) {
if(arr[Math.abs(arr[i])] < 0)
return Math.abs(arr[i]);
else
arr[Math.abs(arr[i])] = 0 - arr[Math.abs(arr[i])];
}
return -1;
}
console.log(firstDuplicate([2, 3, 3, 1, 5, 2]));
function firstDuplicate(arr) {
for(var i = 0; i < arr.length; i++) {
var num = Math.abs(arr[i]);
if(arr[num] < 0)
return num;
arr[num] = - arr[num];
}
return -1;
}
console.log(firstDuplicate([2,2,3,1,2]));
function firstDuplicate(arr) {
var numMap = {};
for (var i = 0; i < arr.length; i++) {
if (numMap[arr[i]]) {
return arr[i];
}
numMap[arr[i]] = true;
}
return -1;
}
Answer mentioned by #dij is great, but will fail for [2,2] or [2,3,3],
a little change
for input [2,2], i=0 we get a[ Math.abs[a[0] ] = a[2] = undefined
so we remove 1 from a[ Math.abs[a[0] -1 ] this will work now
function firstDuplicate(arr) {
for(var i = 0; i < arr.length; i++) {
if(arr[Math.abs(arr[i])-1] < 0)
return Math.abs(arr[i]);
else
arr[Math.abs(arr[i])-1] = 0 - arr[Math.abs(arr[i])-1];
}
return -1;
}
Please try the code below:
function getCountOccurence(A) {
let result = [];
A.forEach(elem => {
if (result.length > 0) {
let isOccure = false;
result.some((element) => {
if (element.element == elem) {
element.count++;
isOccure = true;
}
});
if (!isOccure) {
result.push({element: elem, count: 1});
}
} else {
result.push({element: elem, count: 1});
}
});
return result;
}
function getFirstRepeatingElement(A) {
let array = getCountOccurence(A);
array.some((element)=> {
if (element.count > 1) {
result = element.element;
return true;
}
});
return result;
}
I'm trying to script a function that takes two numbers and returns the smallest common multiple that is also divisible by all the numbers between those numbers, what I've got only works for 1,1 through 1,12, but for some reason stops working at 1,13. Other set like 12,14 work but I can't figure out why or what the pattern is.
function smallestCommons(arr) {
arr.sort(function(a, b) {
return a-b;
});
var arr1 = [];
var arr2 = [];
for (var k = arr[0]; k<=arr[1]; k++) {
arr1.push(k);
}
function remainder(val1, val2) {
return val1%val2;
}
var b = arr1.reduce(function(a, b) {
return a*b;
});
var i = arr1[arr1.length-1]*arr1[arr1.length-2];
while (i<=b) {
for (var m = 0; m<arr1.length; m++) {
var a = remainder(i, arr1[m]);
arr2.push(a);
}
var answer = arr2.reduce(function(c, d) {
return c+d;
});
if (answer === 0) {
return i;
} else {
arr2 = [];
i++;
}
}
}
I guess you can do as follows in JavaScript; It can calculate the common LCM up to an 216 item array, such as [1,2,3,...,216] in less than 0.25 ms.
function gcd(a,b){
var t = 0;
a < b && (t = b, b = a, a = t); // swap them if a < b
t = a%b;
return t ? gcd(b,t) : b;
}
function lcm(a,b){
return a/gcd(a,b)*b;
}
var arr = [1,2,3,4,5,6,7,8,9,10,11,12,13],
brr = Array(216).fill().map((_,i) => i+1), // limit before Infinity
result = arr.reduce(lcm);
console.log(result);
console.time("limit");
result = brr.reduce(lcm);
console.timeEnd("limit");
console.log(result);
A way is to keep multiplying the largest number in your range with an increasing number and check if all the others are divisible by that. If yes, return that or continue the loop.
Here is my solution in typescript...
function findLowestCommonMultipleBetween(start: number, end: number): number {
let numbers: number[] = [];
for (let i = start; i <= end; i++) {
numbers.push(i);
}
for (let i = 1; true; i++) {
let divisor = end * i;
if (numbers.every((number) => divisor % number == 0)) {
return divisor;
}
}
}
...but for larger ranges, this is a more efficient answer :)
As far as I can tell your algorithm is giving you a correct answer.
I am far from being a professional programmer so anyone who wants please give options to improve my code or its style :)
If you want to be able to check for the answer yourself you can check this fiddle:
https://jsfiddle.net/cowCrazy/Ld8khrx7/
function multiplyDict(arr) {
arr.sort(function (a, b) {
return a - b;
});
if (arr[0] === 1) {
arr[0] = 2;
}
var currentArr = [];
for (var i = arr[0]; i <= arr[1]; i++) {
currentArr.push(i);
}
var primeDivs = allPrimes(arr[1]);
var divsDict = {};
for (var j = currentArr[0]; j <= currentArr[currentArr.length -1]; j++){
divsDict[j] = [];
if (primeDivs.indexOf(j) > -1) {
divsDict[j].push(j);
} else {
var x = j;
for (var n = 2; n <= Math.floor(j / 2); n++) {
if (x % n === 0) {
divsDict[j].push(n);
x = x / n;
n--;
continue;
}
}
}
}
return divsDict;
}
function allPrimes(num) {
var primeArr = [];
var smallestDiv = 2;
loopi:
for (var i = 2; i <= num; i++) {
loopj:
for (var j = smallestDiv; j <= largestDiv(i); j++) {
if (i % j === 0) {
continue loopi;
}
}
primeArr.push(i);
}
return primeArr;
}
function largestDiv (a) {
return Math.floor(Math.sqrt(a));
}
multiplyDict([1,13]);
it gives a dictionary of the requested array and the divisors of each element.
from there you can go on your own to check that your algorithm is doing the right job or you can check it here:
https://jsfiddle.net/cowCrazy/kr04mas7/
I hope it helps
It is true! The result of [1, 13] is 360360. and after this we have [1, 14].
14 = 2 * 7 and we now 360360 is dividable to 2 and 7 so the answer is 360360 again.
[1, 15]: 15 = 3 * 5 and result is same.
[1, 16]: result is 720720.
[1, 17]: result is: 12252240
[1, 18]: 18 = 2 * 9 and result is 12252240 same as 17
[1, 19]: for my computer this process is so heavy and can not do this. But in a strong machine it will work. I promise. But your code is not good in performance.
To find the LCM in N numbers.
It is Compatible with ES6, and consider that is there is no control for boundaries in case that we need to find for large numbers.
var a = [10, 40, 50, 7];
console.log(GetMinMultiple(a));
function GetMinMultiple(data) {
var maxOf = data.reduce((max, p) => p > max ? p : max, 0);
var incremental = maxOf;
var found = false;
do {
for (var j = 0; j < data.length; j++) {
if (maxOf % data[j] !== 0) {
maxOf += incremental;
break;
}
else {
if (j === data.length - 1) {
found = true;
break;
}
}
}
} while (!found);
return maxOf;
}
https://jsfiddle.net/djp30gfz/
Here is my solution in Typescript
function greatestCommonDivider(x: number, y: number): number {
if (y === 0) {
return x;
}
return greatestCommonDivider(y, x % y);
}
function singleLowestCommonMultiply(x: number, y: number): number {
return (x * y) / greatestCommonDivider(x, y);
}
function lowestCommonMultiply(...numbers: number[]): number {
/**
* For each number, get it's lowest common multiply with next number.
*
* Then using new number, compute new lowest common multiply
*/
return numbers.reduce((a, b) => {
return singleLowestCommonMultiply(a, b);
});
}
lowestCommonMultiply(2, 3); // Outputs 6
lowestCommonMultiply(2, 3, 5); // Outputs 30
Playground - click here