I'm trying to create a function which takes 3 parameters – start, end and bannedNumber. It should print all the numbers from start to end but skip all multiples of the banned number and any number that has a banned number in it.
Here is my code:
<script>
var arr = [];
var str = "";
var newarr = [];
var str1
function Zumbaniaa(a, b, c) {
for (var i = a; i < b; i++) {
arr.push(i);
}
for (var j = 0; j < arr.length; j++) {
if (arr[j] % c == 0) {
arr.splice(j, 1);
}
}
for (var m = 0; m < arr.length; m++) {
str = arr[m].toString();
str1=str.split("");
if (str1.indexOf(c) >=0) {
arr.splice(m, 1);
}
}
return arr
}
document.write(Zumbaniaa(1, 90, 8))
console.log(str1)
</script>
For some reason the third loop is not working. It's not filtering numbers with 8 at all.
It's simplest to just not push the banned values into the array to begin with. So check if the numbers to be pushed are a multiple of the banned digit, or contain the banned digit before you push:
function filtered_range(start, end, banned) {
let nums = [];
for (let i = start; i <= end; i++) {
if (i % banned != 0 && i.toString().indexOf(banned) == -1) nums.push(i);
}
return nums;
}
console.log(filtered_range(1, 90, 8).join(' '));
Remove this line: str1 = str.split("");. It is causing indexOf() to fail. Call str.indexOf(c) instead. Then reverse the final loop.
What is happening: indexOf() works on strings and arrays. When used on a string, the character being searched for is always parsed first to a string, because strings only contain strings. When used on an array, the character being searched for isn't parsed at all, because arrays can contain strings, numbers, etc.
var arr = [];
var str = "";
var newarr = [];
var str1
function Zumbaniaa(a, b, c) {
for (var i = a; i < b; i++) {
arr.push(i);
}
for (var j = 0; j < arr.length; j++) {
if (arr[j] % c == 0) {
arr.splice(j, 1);
}
}
for (var m = arr.length - 1; m > 0; m--) {
str = arr[m].toString();
//str1 = str.split("");
if (str.indexOf(c) >= 0) {
arr.splice(m, 1);
}
}
return arr
}
Zumbaniaa(1, 90, 8);
console.log(arr);
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 am trying to push all first indexes that point to a start of a sequence of consecutive 0s from the array A into a new array arr.
var C determines the amount of 0s in the sequence. For example, if C is 2, the algorithm will look for 00s, if C is 3 it will look for 000s and so on. N is the length of an array A. The algorithm seems to work, but for some reason the values in the new array arr are duplicated
var A = [1, 0, 0, 1];
var N = 4;
var C = 1;
function S(A, N, C) {
var arr = [];
for (var i = 0; i < N; i++) {
for (var j = 0; j <= C; j++) {
if ((A[i] == 0) && (A[i + j] == 0)) {
arr.push(i);
}
}
}
console.log(arr);
return -1;
}
/// console result:
Array(5)
0: 1
1: 1
2: 2
3: 2
//Expected:
0: 1
1: 2
First I would like to recommend that you use more descriptive variable names. The fact that you need to describe what each of them means, means that they are not descriptive enough.
Also your variable N seems redundant, because arrays already have a .length property that you can use to see how many elements are in there.
The source of your error seems to be that you use a nested loop. There is no need to use nested loops. You only need to go through all elements once and keep track of the repeated zeroes. Every time you encounter a non-zero value, you reset the sequence count to 0. If do encounter a zero you increment the sequence count and afterwards you check if the sequence count is equal to the number of zeroes you passed as an argument. In that case you want to push the first index to the resulting array and reset the sequence count to 0 again.
function getFirstIndexesOfSequenceOfConsecutiveZeroes(input, numberOfRepeatedZeroes) {
if (numberOfRepeatedZeroes <= 0) {
throw new Error("numberOfRepeatedZeroes need to be 1 or higher");
}
var firstIndexes = [];
let sequenceStartIndex;
let sequenceCount = 0;
for (var i = 0; i < input.length; i++) {
if (input[i] !== 0) {
sequenceCount = 0;
} else {
if (sequenceCount == 0) {
sequenceStartIndex = i;
}
sequenceCount++;
}
if (sequenceCount === numberOfRepeatedZeroes) {
firstIndexes.push(sequenceStartIndex);
sequenceCount = 0;
}
}
return firstIndexes;
}
let input = [1, 0, 0, 1];
let numberOfRepeatedZeroes = 1;
console.log(getFirstIndexesOfSequenceOfConsecutiveZeroes(input, numberOfRepeatedZeroes));
Try:
function S(A, B, C) {
var arr = [];
for (var i = 0; i < B; i++) {
for (var j = 0; j <= C; j++) {
if ((A[i] == 0) && (A[i + j] == 0) && !arr.includes(i)) {
arr.push(i);
}
}
}
console.log(arr);
return -1;
}
With this simple add in the if, you check if the value is already in your array.
Please help me to solve this leetcode problem using javascript as I am a beginner and dont know why this code is not working
Ques: Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.
Find all the elements of [1, n] inclusive that do not appear in this array.
var findDisappearedNumbers = function (nums) {
var numLength = nums.length;
nums.sort(function (a, b) { return a - b });
for (var i = 0; i < nums.length - 1; i++) {
if (nums[i + 1] === nums[i]) {
nums.splice(i, 1);
}
}
for (var k = 0; k < nums.length; k++) {
for (var j = 1; j <= numLength; j++) {
if (nums[k] !== j) {
return j;
}
}
}
};
if there is any error in my code please let me know;
i have done the following thing
first i have sorted the array in ascending order
then i have cut all the duplicate elements
then i have created loop that will check if nums[k] !== j ;
and it will return j which is the missing number;
for example this is the testcase [4,3,2,7,8,2,3,1]
first my code will sort this in ascending order [1,2,2,3,3,4,7,8]
then it will remove all duplicate elements and it will return [1,2,3,4,,7,8]
and then it will check nums[k] is not equal to j and it will print j
I think it'd be easier to create a Set of numbers from 1 to n, then just iterate through the array and delete every found item from the set:
var findDisappearedNumbers = function(nums) {
const set = new Set();
for (let i = 0; i < nums.length; i++) {
set.add(i + 1);
}
for (const num of nums) {
set.delete(num);
}
return [...set];
};
console.log(findDisappearedNumbers([4,3,2,7,8,2,3,1]));
To fix your existing code, I'm not sure what the logic you're trying to implement in the lower section, but you can iterate from 1 to numLength (in the outer loop, not the inner loop) and check to see if the given number is anywhere in the array. Also, since you're mutating the array with splice while iterating over it in the upper loop, make sure to subtract one from i at the same time so you don't skip an element.
var findDisappearedNumbers = function(nums) {
var numLength = nums.length;
nums.sort(function(a, b) {
return a - b
});
for (var i = 0; i < nums.length - 1; i++) {
if (nums[i + 1] === nums[i]) {
nums.splice(i, 1);
i--;
}
}
const notFound = [];
outer:
for (var j = 1; j < numLength; j++) {
for (var k = 0; k < nums.length; k++) {
if (nums[k] === j) {
continue outer;
}
}
notFound.push(j);
}
return notFound;
};
console.log(findDisappearedNumbers([4, 3, 2, 7, 8, 2, 3, 1]));
#CertainPerformance certainly cracked it again using the modern Set class. Here is a slighly more conservative approach using an old fashioned object:
console.log(missingIn([4,3,2,7,8,2,3,1]));
function missingIn(arr){
const o={};
arr.forEach((n,i)=>o[i+1]=1 );
arr.forEach((n) =>delete o[n] );
return Object.keys(o).map(v=>+v);
}
My solution for the problem to find the missing element
var findDisappearedNumbers = function(nums) {
const map={};
const result=[];
for(let a=0;a<nums.length;a++){
map[nums[a]]=a;
}
for(let b=0;b<nums.length;b++){
if(map[b+1]===undefined){
result.push(b+1)
}
}
return result;
};
Example 1:
Input: nums = [4,3,2,7,8,2,3,1]
Output: [5,6]
Example 2:
Input: nums = [1,1]
Output: [2]
EDITED:
Can someone help me with the problem below further. I have a class and an array inside the class. I want now use a for loop to sum the length of the previous array length, but for each iteration. If i == 1 I want sum the length of arr[0].x.length, If i == 2 I want sum the length of arr[0].x.length+arr[1].x.length, ect. It will be a lot of code to check each iteration.
Is there a simple way to do this? Instead allways use a new line like
if (i == 1) n = n + arr[i-1].x.length;
if (i == 2) n = n + arr[i-1].x.length+arr[i-2].x.length;
if (i == 3) n = n + arr[i-1].x.length+arr[i-2].x.length+arr[i-3].x.length;
function Class() {
var x = [];
}
for (var i = 0; i < 9; i++) {
arr[i] = new Class();
}
I add 4 items to each object.
arr[0].x.push(...)
arr[0].x.push(...)
...
arr[1].x.push(...)
arr[1].x.push(...)
...
var n = 0;
for (var i = 0; i < arr.length; i++) {
if (i == 1) {
n = n + arr[i-1].x.length;
} else if (i == 2) {
n = n + arr[i-1].x.length+arr[i-2].x.length;
} else if (i == 3) {
n = n + arr[i-1].x.length+arr[i-2].x.length+arr[i-3].x.length;
}
// ect.
}
You could use reduce to get a total of all the lengths of your sub-arrays. For example:
const arrs = [[1, 2, 3], [4, 5, 6]];
const sum = arrs.reduce((acc, arr) => acc += arr.length, 0);
console.log(sum);
// 6
Just nest the loop two times: go over the indexes once then go up to that index from 0 in an inner loop:
for (var i = 0; i < arr1.length; i++) {
for(var j = 0; j <= i; j++) {
n = n + arr1[j].length;
}
}
Edit: benvc's answer is what you are looking for if you want to use reduce.
var arr = [[1,2,3], [4,5,6], [7]];
var n = 0;
for (var i = 0; i < arr.length; i++){
n += arr[i].length;
}
console.log(n);
I have a combinatoric script that's working fine, actually got most of it from the IBM dev website. But I want to be able to not just show the possible combinations, but also extract the numbers on each combination and get the product of the entire numbers. The project am working on mixes numbers (quantity) with strings (codename). So after combining them, i extract the number from each string and get the product of all the numbers in each combination. As shown;
[A2,B4,C5] = 2*4*5 = 40
Here is my javascript code that gets the combination, not to worry, I ran it with a test array of numbers 1-6, without the characters as shown above.
var Util = function() {
};
Util.getCombinations = function(array, size, start, initialStuff, output) {
if (initialStuff.length >= size) {
output.push(initialStuff);
} else {
var i;
for (i = start; i < array.length; ++i) {
Util.getCombinations(array, size, i + 1, initialStuff.concat(array[i]), output);
}
}
}
Util.getAllPossibleCombinations = function(array, size, output) {
Util.getCombinations(array, size, 0, [], output);
}
// Create an array that holds numbers from 1 ... 6.
var array = [];
for (var i = 1; i <= 6; ++i) {
array[i - 1] = i;
}
var output = [];
var resultArray = [];
Util.getAllPossibleCombinations(array, 4, output);
for(var j=0; j<output.length; j++) {
resultArray += output[j] + "=" + "<br />";
}
document.getElementById("test").innerHTML = resultArray;
});
I tried running this code inside the last for loop to get my multiplication, but it's just not executing, i must be doing something wrong. Here is the code;
var inputval = output[j].replace(/[^,.0-9]/g, '');
inputval = inputval.slice(0, -1);
var hoArray = inputval.split(',');
var cunt= hoArray.length;
var ans=1;
for(var m=0; m<cunt; m++)
{
ans *= hoArray[m];
}
Thanks for your assistance in advance.
walk the array then walk the string, then cast and see if it is an integer then tally and sum the product.
let array = ['A20', 'B11', 'C5'];
function getProduct(ar) {
let product = 1;
for (let x of ar) {
let semiProduct = [];
for (let i of x) {
if (Number.isInteger(+i)) {
semiProduct.push(i);
}
}
product *= semiProduct.join('');
}
return product;
}
console.log(getProduct(array))
You could also use a regular expression.
let array = ['A20', 'B11', 'C5'];
function getProduct(ar) {
let product = 1;
for (let x of ar) {
product *= x.match(/\d+/)[0];
}
return product;
}
console.log(getProduct(array))
If you want a way to generate permutations, you can utilize a generator to make things more concise.
let array = ['A20', 'B11', 'C5'];
function* permu(arr, l = arr.length) {
if (l <= 0) yield arr.slice();
else
for (var i = 0; i < l; i++) {
yield* permu(arr, l - 1);
const j = l % 2 ? 0 : i;
[arr[l - 1], arr[j]] = [arr[j], arr[l - 1]];
}
}
console.log(
Array.from(permu(array))
);
When I run that code in the console it throws an error because output[j] is an array [1,2,3,4] and it looks like you're expecting it to be a string. Arrays do not have a replace method in JS.
You should run this:
var count= hoArray.length;
var ans=1;
for(var m=0; m<count; m++)
{
ans *= hoArray[m];
}
And put output[j] instead of hoArray. And don't do any of this:
var inputval = output[j].replace(/[^,.0-9]/g, '');
inputval = inputval.slice(0, -1);
var hoArray = inputval.split(',');