How to get numbers of variants numerical combinations in javascript? - javascript

I need to implement a function that accepts two parameters - the number of 0 and the number of 1 and determine how many ways to place these 0 and 1 so that there are no two zeros in a row.
For example, I need to find all methods of placing two 0 and two 1.
There are six possible ways to place them: 0011, 0101, 0110, 1001, 1010, 1100.
In three cases there are two zeros in a row: 0011, 1001 and 1100.
I subtract them from the total number and get three possible ways: 0101, 0110 and 1010. So the answer is 3.
First, I'm trying to write script to recognize which cases I need
let arr = ["1100","1010","1001","0011","0101","0110"]
let result = [];
for (let i = 0; i < arr.length; i++){
let expVal = arr[i];
for (let p = 0; p < expVal.length; p++){
if (expVal[p] === expVal[p++] || expVal[p] === "0"){
result.push(expVal)
}
}
}
console.log(result);
It's does not work. I don't know how to fix it.
And I don't understand what I need to do later

The problem you are solving is equivalent to the Fibonacci sequence.
filter for includes '00'
recursive function x generates binary number strings using n0 zeroes, n1 ones. Works by branching to (add a zero, find combos for one less zero)'0'+x(n0-1,n1) and (add a one, find combos for one less 1)'1'+x(n0,n1-1).
let arr = ["1100","1010","1001","0011","0101","0110"]
const x = (n0,n1) =>
!(n0 === 0 || n1 === 0) ?
x(n0-1,n1).map(x=>'0'+x).concat(
x(n0,n1-1).map(x=>'1'+x))
: ['0'.repeat(n0)||'1'.repeat(n1)]
arr = x(2,2)
console.log(
x(2,2)
)
console.log(arr.filter(x=>x.includes('00')))

There are 2 issues with the code, as far as I can see:
You're incrementing p two times: once in the for definition, and once in the execution itself (p++). The second one should be replaced by p+1
Also, you're checking if the next character is equal to the current, OR the current character is 0. This should be AND.
let arr = ["1100","1010","1001","0011","0101","0110"]
let result = [];
for (let i = 0; i < arr.length; i++){
let expVal = arr[i];
for (let p = 0; p < expVal.length; p++){
if (expVal[p] === expVal[p+1] && expVal[p] === "0"){
result.push(expVal)
}
}
}
console.log(result);
Changing those two things fixes it.

Related

my indexOf keeps finding the character at the same postition how can i solve it?

for(let j=0; j<=z;j++)
{
if( text[i+1].indexOf(y[j]) > -1 )
{
con++;
temp = (text[i+1].indexOf(y[j]));
text[i+1].substring(temp);
console.log(text[i+1].indexOf(y[j]))
this is code counts whenever a letter of the array y is in the string of text[i+1] but instead of going through the whole word to find the letter in other position it stays in the first position it finds.
Example:
text[i+1] = 'debefeene'
y = ['e','e','e','e']
// My expected output would be:
1
3
5
6
8
// The actual output is:
1
1
1
1
I tried using substring but is not really working.
z equals the length of the string text[i+1]
I didn't understand what did you try. That's why I can't tell what is the issue with your code. Look at my solution. It's quite self-explanatory. The 2nd "if" is not necessary. I just added it to stop the loop if there isn't any item left to search for.
let text = "debefeene"
let find = ['e','e','e','e', 'e'];
let positions = [];
for(let i = 0; i < text.length; i++) {
if (text[i] == find[0]) {
positions.push(i);
find.pop();
}
if (!find.length) break;
}
console.log(positions);

Changing a number to a string and then adding up the individual integers?

I have tried multiple functions so far but the piece that I am stuck on is how to setup the function so that the argument will be 1) split into an array, 2) added together then 3) returned. I am not sure if I am looking at the question right, but I am assuming to use string.Split. Any help is welcome! Thanks!
Write a function that takes a number an an argument and returns the sum of each individual digit. So an input of 998 would return 26 (because 9 + 9 + 8) is 26.
Write the same function above, but that takes an input from the built-in browser function, prompt().
Check this below code.
let num = 998;
function individualSum(inputNum) {
let numStr = String(inputNum);
let numStrArray = numStr.split("");
let result = 0;
let len = numStrArray.length;
for (let i = 0; i < len; i++) {
result = result + Number(numStrArray[i]);
}
console.log(result);
}
individualSum(num);
Another option, with comments
let n = 998;
// To string
n = String(n);
// Split
n = n.split("");
// Summ
n = n.reduce((a, b) => Number(a) + Number(b), 0);
// Log
console.log(n)

Minimum number of swaps to sort an array

I need to do something like this: Let's say I have an array:
[3, 4, 1, 2]
I need to swap 3 and 4, and 1 and 2, so my array looks like [4, 3, 2, 1]. Now, I can just do the sort(). Here I need to count how many iterations I need, to change the initial array to the final output. Example:
// I can sort one pair per iteration
let array = [3, 4, 1, 2, 5]
let counter = 0;
//swap 3 and 4
counter++;
// swap 1 and 2
counter++;
// 5 goes to first place
counter++
// now counter = 3 <-- what I need
EDIT: Here is what I tried. doesn't work always tho... it is from this question: Bubble sort algorithm JavaScript
let counter = 0;
let swapped;
do {
swapped = false;
for (var i = 0; i < array.length - 1; i++) {
if (array[i] < array[i + 1]) {
const temp = array[i];
array[i] = array[i + 1];
array[i + 1] = temp;
swapped = true;
counter++;
}
}
} while (swapped);
EDIT: It is not correct all the time because I can swap places from last to first, for example. Look at the example code above, it is edited now.
This is most optimal code I have tried so far, also the code is accepted as optimal
answer by hackerrank :
function minimumSwaps(arr) {
var arrLength = arr.length;
// create two new Arrays
// one record value and key separately
// second to keep visited node count (default set false to all)
var newArr = [];
var newArrVisited = [];
for (let i = 0; i < arrLength; i++) {
newArr[i]= [];
newArr[i].value = arr[i];
newArr[i].key = i;
newArrVisited[i] = false;
}
// sort new array by value
newArr.sort(function (a, b) {
return a.value - b.value;
})
var swp = 0;
for (let i = 0; i < arrLength; i++) {
// check if already visited or swapped
if (newArr[i].key == i || newArrVisited[i]) {
continue;
}
var cycle = 0;
var j = i;
while (!newArrVisited[j]) {
// mark as visited
newArrVisited[j] = true;
j = newArr[j].key; //assign next key
cycle++;
}
if (cycle > 0) {
swp += (cycle > 1) ? cycle - 1 : cycle;
}
}
return swp;
}
reference
//You are given an unordered array consisting of consecutive integers [1, 2, 3, ..., n] without any duplicates.
//still not the best
function minimumSwaps(arr) {
let count = 0;
for(let i =0; i< arr.length; i++){
if(arr[i]!=i+1){
let temp = arr[i];
arr[arr.indexOf(i+1)] =temp;
arr[i] = i+1;
count =count+1;
}
}
return count;
}
I assume there are two reasons you're wanting to measure how many iterations a sort takes. So I will supply you with some theory (if the mathematics is too dense, don't worry about it), then some practical application.
There are many sort algorithms, some of them have a predicable number of iterations based on the number of items you are sorting, some of them are luck of the draw simply based on the order of the items to be sorted and which item how you select what is called a pivot. So if optimisation is very important to you, then you'll want to select the right algorithm for the purpose of the sort algorithm. Otherwise go for a general purpose algorithm.
Here are most popular sorting algorithms for the purpose of learning, and each of them have least, worst and average running-cases. Heapsort, Radix and binary-sort are worth looking at if this is more than just an theoretical/learning exercise.
Quicksort
Worst Case: Θ(n 2)
Best case: Θ(n lg n)
Average case: Θ(n lg n)
Here is a Quicksort implementation by Charles Stover
Merge sort
Worst case: Θ(n lg n)
Best case: Θ(n lg n)
Average Case: Θ(n lg n)
(note they're all the same)
Here is a merge sort implementation by Alex Kondov
Insertion sort
Worst case: Θ(n2)
Best case: Θ(n)
Average case:Θ(n2)
(Note that its worst and average case are the same, but its best case is the best of any algorithm)
Here is an insertion sort implementation by Kyle Jensen
Selection sort
Worst case: Θ(n2)
Best case: Θ(n2)
Average case: Θ(n2)
(note they're all the same, like a merge sort).
Here is a selection sort algorithm written by #dbdavid updated by myself for ES6
You can quite easily add an iterator variable to any of these examples to count the number of swaps they make, and play around with them to see which algorithms work best in which circumstance.
If there's a very good chance the items will already be well sorted, insertion sort is your best choice. If you have absolutely no idea, of the four basic sorting algorithms quicksort is your best choice.
function minimumSwaps(arr) {
var counter = 0;
for (var i = arr.length; i > 0; i--) {
var minval = Math.min(...arr); console.log("before", arr);
var minIndex = arr.indexOf(minval);
if (minval != = arr[0]) {
var temp = arr[0];
arr[0] = arr[minIndex];
arr[minIndex] = temp; console.log("after", arr);
arr.splice(0, 1);
counter++;
}
else {
arr.splice(0, 1); console.log("in else case")
}
} return counter;
}
This is how I call my swap function:
minimumSwaps([3, 7, 6, 9, 1, 8, 4, 10, 2, 5]);
It works with Selection Sort. Logic is as follows:
Loop through the array length
Find the minimum element in the array and then swap with the First element in the array, if the 0th Index doesn't have the minimum value founded out.
Now remove the first element.
If step 2 is not present, remove the first element(which is the minimum value present already)
increase counter when we swap the values.
Return the counter value after the for Loop.
It works for all values.
However, it fails due to a timeout for values around 50,000.
The solution to this problem is not very intuitive unless you are already somewhat familiar with computer science or real math wiz, but it all comes down to the number of inversions and the resulting cycles
If you are new to computer science I recommend the following resources to supplement this solution:
GeeksforGeeks Article
Informal Proof Explanation
Graph Theory Explanation
If we define an inversion as:
arr[i]>arr[j]
where "i" is the current index and "j" is the following index --
if there are no inversions the array is already in order and requires no sorting.
For Example:
[1,2,3,4,5]
So the number of swaps is related to the number of inversions, but not directly because each inversion can lead to a series of swaps (as opposed to a singular swap EX: [3,1,2]).
So if one consider's the following array:
[4,5,2,1,3,6,10,9,7,8]
This array is composed of three cycles.
Cycle One- 4,1,3 (Two Swaps)
Cycle Two- 5,2 (One Swap)
Cycle Three- 6 (0 Swaps)
Cycle Four- 10,9,7,8 (3 Swaps)
Now here's where the CS and Math magic really kicks in: each cycle will only require one pass through to properly sort it, and this is always going to be true.
So another way to say this would be-- the minimum number of swaps to sort any cycle is the number of element in that cycle minus one, or more explicitly:
minimum swaps = (cycle length - 1)
So if we sum the minimum swaps from each cycle, that sum will equal the minimum number of swaps for the original array.
Here is my attempt to explain WHY this algorithm works:
If we consider that any sequential set of numbers is just a section of a number line, then any set starting at zero should be equal to its own index should the set be expressed as a Javascript array. This idea becomes the criteria to programmatically determined if in element is already in the correct position based on its own value.
If the current value is not equal to its own index then the program should detect a cycle start and recording its length. Once the while loop reaches the the original value in the cycle it will add the minimum number of swaps in the cycle to a counter variable.
Anyway here is my code-- it is very verbose but should work:
export const minimumSwaps = (arr) => {
//This function returns the lowest value
//from the provided array.
//If one subtracts this value the from
//any value in the array it should equal
//that value's index.
const shift = (function findLowest(arr){
let lowest=arr[0];
arr.forEach((val,i)=>{
if(val<lowest){
lowest=val;
}
})
return lowest;
})(arr);
//Declare a counter variable
//to keep track of the swaps.
let swaps = 0;
//This function returns an array equal
//in size to the original array provided.
//However, this array is composed of
//boolean values with a value of false.
const visited = (function boolArray(n){
const arr=[];
for(let i = 0; i<n;i++){
arr.push(false);
}
return arr;
})(arr.length);
//Iterate through each element of the
//of the provided array.
arr.forEach((val, i) => {
//If the current value being assessed minus
//the lowest value in the original array
//is not equal to the current loop index,
//or, if the corresponding index in
//the visited array is equal to true,
//then the value is already sorted
if (val - shift === i || visited[i]) return;
//Declare a counter variable to record
//cycle length.
let cycleLength = 0;
//Declare a variable for to use for the
//while loop below, one should start with
//the current loop index
let x = i;
//While the corresponding value in the
//corresponding index in the visited array
//is equal to false, then we
while (!visited[x]) {
//Set the value of the current
//corresponding index to true
visited[x] = true;
//Reset the x iteration variable to
//the next potential value in the cycle
x = arr[x] - shift;
//Add one to the cycle length variable
cycleLength++;
};
//Add the minimum number of swaps to
//the swaps counter variable, which
//is equal to the cycle length minus one
swaps += cycleLength - 1;
});
return swaps
}
This solution is simple and fast.
function minimumSwaps(arr) {
let minSwaps = 0;
for (let i = 0; i < arr.length; i++) {
// at this position what is the right number to be here
// for example at position 0 should be 1
// add 1 to i if array starts with 1 (1->n)
const right = i+1;
// is current position does not have the right number
if (arr[i] !== right) {
// find the index of the right number in the array
// only look from the current position up passing i to indexOf
const rightIdx = arr.indexOf(right, i);
// replace the other position with this position value
arr[rightIdx] = arr[i];
// replace this position with the right number
arr[i] = right;
// increment the swap count since a swap was done
++minSwaps;
}
}
return minSwaps;
}
Here is my solution, but it timeouts 3 test cases with very large inputs. With smaller inputs, it works and does not terminate due to timeout.
function minimumSwaps(arr) {
let swaps = 0;
for (let i = 0; i < arr.length; i++) {
if (arr[i] === i + 1) continue;
arr.splice(i, 1, arr.splice(arr.indexOf(i + 1), 1, arr[i])[0]); //swap
swaps++;
}
return swaps;
}
I'm learning how to make it more performant, any help is welcome.
This is my solution to the Main Swaps 2 problem in JavaScript. It passed all the test cases. I hope someone finds it useful.
//this function calls the mainSwaps function..
function minimumSwaps(arr){
let swaps = 0;
for (var i = 0; i < arr.length; i++){
var current = arr[i];
var targetIndex = i + 1;
if (current != targetIndex){
swaps += mainSwaps(arr, i);
}
}
return swaps;
}
//this function is called by the minimumSwaps function
function mainSwaps(arr, index){
let swapCount = 0;
let currentElement = arr[index];
let targetIndex = currentElement - 1;
let targetElement = arr[currentElement - 1];
while (currentElement != targetElement){
//swap the elements
arr[index] = targetElement;
arr[currentElement - 1] = currentElement;
//increase the swapcount
swapCount++;
//store the currentElement, targetElement with their new values..
currentElement = arr[index];
targetElement = arr[currentElement - 1];
}
return swapCount;
}
var myarray = [2,3,4,1,5];
var result = console.log(minimumSwaps(myarray));
you can also do it with a map. But its O(nlogn)
const minSwaps = (arr) =>{
let arrSorted = [...arr].sort((a,b)=>a-b);
let indexMap = new Map();
// fill the indexes
for(let i=0; i<arr.length; i++){
indexMap.set(arr[i],i);
}
let count = 0;
for(let i=0; i<arrSorted.length;i++){
if(arr[i] != arrSorted[i]){
count++;
// swap the index
let newIdx = indexMap.get(arrSorted[i]);
indexMap.set(arr[i],newIdx);
indexMap.set(arrSorted[i],i);
// sawp the values
[arr[i],arr[newIdx]] =[arr[newIdx],arr[i]];
}
}
return count;
}

I want to take the first number from an array and depending on its index put equal amount of 0 next to it

edit: without giving me too much of the answer to how i can do it in a for loop. Could you give me the logic/pseudocode on how to achieve this? the one part i am stuck at is, ok i know that i have to take the first index number and add array.length-1 zeros to it (2 zeros), but i am confused as to when it arrives at the last index, do i put in a if statement in the for loop?
In the below example 459 would be put in an array [4,5,9]
Now I want to take 4 add two zeros to the end because it has two numbers after it in the array
Then I want to take 5 and add one zero to it because there is one number after it in the array.
then 9 would have no zeros added to it because there are no numbers after it.
So final output would be 400,50,9
how can i best achieve this?
var num=459;
var nexint=num.toString().split("");
var finalint=nexint.map(Number);
var nextarr=[];
You need to use string's repeat method.
var num=459;
var a = (""+num).split('').map((c,i,a)=>c+"0".repeat(a.length-i-1))
console.log(a);
Here's another possible solution using a loop.
var num = 459;
var a = ("" + num).split('');
var ar = [];
for (var i = 0; i < a.length; i++) {
var str = a[i];
str += "0".repeat(a.length-i-1);
ar.push(str);
}
console.log(ar);
You could use Array#reduce and Array#map for the values multiplied by 10 and return a new array.
var num = 459,
result = [...num.toString()].reduce((r, a) => r.map(v => 10 * v).concat(+a), []);
console.log(result);
OP asked for a solution using loops in a comment above. Here's one approach with loops:
var num = 459
var numArray = num.toString().split('');
var position = numArray.length - 1;
var finalArray = [];
var i;
var j;
for(i = 0; i < numArray.length; i++) {
finalArray.push(numArray[i]);
for(j = 0; j < position; j++) {
finalArray.push(0);
}
position--;
}
console.log(finalArray);
The general flow
Loop over the original array, and on each pass:
Push the element to the final array
Then push X number of zeros to the final array. X is determined by
the element's position in the original array, so if the original
array has 3 elements, the first element should get 2 zeros after it.
This means X is the original array's length - 1 on the first pass.
Adjust the variable that's tracking the number of zeros to add before
making the next pass in the loop.
This is similar to #OccamsRazor's implementation, but with a slightly different API, and laid out for easier readability:
const toParts = n => String(n)
.split('')
.map((d, i, a) => d.padEnd(a.length - i, '0'))
.map(Number)
console.log(toParts(459))

Why does this loop terminate partway through?

I'm trying to write a program to find the smallest common multiple of the provided parameters that can be evenly divided by both, as well as by all sequential numbers in the range between these parameters.
The range will be an array of two numbers that will not necessarily be in numerical order.
For example, for 1 and 3 - find the smallest common multiple of both 1 and 3 that is evenly divisible by all numbers between 1 and 3.
Why does the loop stop at i = 510,000 (or something close to that) instead of 7,000,000, as I set it?
I also have a screenshot with the output:
function smallestCommons(arr) {
var start;
var finish;
var something;
if(arr[0] < arr[1]){start = arr[0]; finish = arr[1];}else{
start = arr[1]; finish = arr[0];
}
for(var i = finish;i <= 7000000;i++){
var boolea = true;
for(var j = start;j <= finish;j++){
if(i % j !== 0){boolea = false;break;} // 2 % 1
}
if(boolea)return i;
something = i;
}
console.log("final i = " + i);
return 0;
}
Try to add this at the beginning of your loop
// noprotect
it must be that jsbin is forcing your code to exit from the loop. See source

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