I'm trying to loop through a range of numbers in NodeJS, but not by using integers, by using a string.
For example, I want to loop through from 000000 to 500000. Ie. 000001, 000002, all the way to 500000. When percieved as an integer, NodeJS will just go 1, 2, all the way up to 500000. I want to keep if so the number always has 6 digits and loops through every possible number.
Edit: I need it to loop through like 000001, 000002, ..., 000010, ... , 000100, ... , 001000, ... , 010000, ... , 100000, ... , 500000. Filling in every number in between though
Thanks in advance
If you're using a relatively new version of node you can use string.padStart():
for(var i = 0; i <= 500000; i++) {
str = i.toString().padStart(6, 0)
// 000001, etc
}
More docs here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/padStart
Iterating through the number as usual but padding 0's before it.
function pad (str, max) {
str = str.toString();
return str.length < max ? pad("0" + str, max) : str;
}
for(var i = 0; i <= 500000; i++) {
console.log(pad("6", i)) ;
}
Related
I found a script that converts binary to string but how can I input a string and get the binary representation? so say I put in "P" I want it to output 01010000 as a string.
I have this but it is not what I am trying to do - it converts a string containing a binary number into a real value of that number:
///string_to_binary(string)
var str = argument0;
var output = "";
for(var i = 0; i < string_length(str); i++){
if(string_char_at(str, i + 1) == "0"){
output += "0";
}
else{
output += "1";
}
}
return real(output);
Tip: search for GML or other language term, these questions answered many times. Also please check your tag as it is the IDE tag, not language tag.
Im not familiar with GML myself, but a quick search showed this:
At least semi-official method for exactly this: http://www.gmlscripts.com/script/bytes_to_bin
/// bytes_to_bin(str)
//
// Returns a string of binary digits, 1 bit each.
//
// str raw bytes, 8 bits each, string
//
/// GMLscripts.com/license
{
var str, bin, p, byte;
str = argument0;
bin = "";
p = string_length(str);
repeat (p) {
byte = ord(string_char_at(str,p));
repeat (8) {
if (byte & 1) bin = "1" + bin else bin = "0" + bin;
byte = byte >> 1;
}
p -= 1;
}
return bin;
}
GML forum (has several examples) https://www.reddit.com/r/gamemaker/comments/4opzhu/how_could_i_convert_a_string_to_binary/
///string_to_binary(string)
var str = argument0;
var output = "";
for(var i = 0; i < string_length(str); i++){
if(string_char_at(str, i + 1) == "0"){
output += "0";
}
else{
output += "1";
}
}
return real(output);
And other language examples:
C++ Fastest way to Convert String to Binary?
#include <string>
#include <bitset>
#include <iostream>
using namespace std;
int main(){
string myString = "Hello World";
for (std::size_t i = 0; i < myString.size(); ++i)
{
cout << bitset<8>(myString.c_str()[i]) << endl;
}
}
Java: Convert A String (like testing123) To Binary In Java
String s = "foo";
byte[] bytes = s.getBytes();
StringBuilder binary = new StringBuilder();
for (byte b : bytes)
{
int val = b;
for (int i = 0; i < 8; i++)
{
binary.append((val & 128) == 0 ? 0 : 1);
val <<= 1;
}
binary.append(' ');
}
System.out.println("'" + s + "' to binary: " + binary);
JS: How to convert text to binary code in JavaScript?
function convert() {
var output = document.getElementById("ti2");
var input = document.getElementById("ti1").value;
output.value = "";
for (var i = 0; i < input.length; i++) {
output.value += input[i].charCodeAt(0).toString(2) + " ";
}
}
I was looking around for a simple GML script to convert a decimal to binary and return the bits in an array. I didn't find anything for my need and to my liking so I rolled my own. Short and sweet.
The first param is the decimal number (string or decimal) and the second param is the bit length.
// dec_to_bin(num, len);
// argument0, decimal string
// argument1, integer
var num = real(argument0);
var len = argument1;
var bin = array_create(len, 0);
for (var i = len - 1; i >= 0; --i) {
bin[i] = floor(num % 2);
num -= num / 2;
}
return bin;
Usage:
dec_to_bin("48", 10);
Output:
{ { 0,0,0,0,1,1,0,0,0,0 }, }
i think the binary you mean is the one that computers use, if thats the case, just use the common binary and add a kind of identification.
binary is actually simple, instead of what most people think.
every digit represents the previous number *2 (2¹, 2², 2³...) so we get:
1, 2, 4, 8, 16, 32, 64, 128, 256, 512...
flip it and get:
...512, 256, 128, 64, 32, 16, 8, 4, 2, 1
every digit is "activated" with 1's, plus all the activated number ant thats the value.
ok, so binary is basically another number system, its not like codes or something. Then how are letters and other characters calculated?
they arent ;-;
we just represent then as their order on their alphabets, so:
a=1
b=2
c=3
...
this means that "b" in binary would be "10", but "2" is also "10". So thats where computer's binary enter.
they just add a identification before the actual number, so:
letter_10 = b
number_10 = 2
signal_10 = "
wait, but if thats binary there cant be letter on it, instead another 0's and 1's are used, so:
011_10 = b
0011_10 = 2
001_10 = "
computers also cant know where the number starts and ends, so you have to always use the same amount of numbers, which is 8. now we get:
011_00010 = b
0011_0010 = 2
001_00010 = "
then remove the "_" cuz again, computers will only use 0's and 1's. and done!
so what i mean is, just use the code you had and add 00110000 to the value, or if you want to translate these numbers to letters as i wanted just add 01100000
in that case where you have the letter and wants the binary, first convert the letter to its number, for it just knows that the letters dont start at 1, capitalized letters starts at 64 and the the non-capitalized at 96.
ord("p")=112
112-96=16
16 in binary is 10000
10000 + 01100000 = 01110000
"p" in binary is 01110000
ord("P")=80
80-64=16
16 in binary is 10000
10000 + 01000000 = 01010000
"P" in binary is 01010000
thats just a explanation of what the code should do, actually im looking for a simple way to turn binary cuz i cant understand much of the code you showed.
(011)
1000 1111 10000 101 1001 1000 101 1100 10000 101 100
We need to find the maximum element in an array which is also equal to product of two elements in the same array. For example [2,3,6,8] , here 6=2*3 so answer is 6.
My approach was to sort the array and followed by a two pointer method which checked whether the product exist for each element. This is o(nlog(n)) + O(n^2) = O(n^2) approach. Is there a faster way to this ?
There is a slight better solution with O(n * sqrt(n)) if you are allowed to use O(M) memory M = max number in A[i]
Use an array of size M to mark every number while you traverse them from smaller to bigger number.
For each number try all its factors and see if those were already present in the array map.
Here is a pseudo code for that:
#define M 1000000
int array_map[M+2];
int ans = -1;
sort(A,A+n);
for(i=0;i<n;i++) {
for(j=1;j<=sqrt(A[i]);j++) {
int num1 = j;
if(A[i]%num1==0) {
int num2 = A[i]/num1;
if(array_map[num1] && array_map[num2]) {
if(num1==num2) {
if(array_map[num1]>=2) ans = A[i];
} else {
ans = A[i];
}
}
}
}
array_map[A[i]]++;
}
There is an ever better approach if you know how to find all possible factors in log(M) this just becomes O(n*logM). You have to use sieve and backtracking for that
#JerryGoyal 's solution is correct. However, I think it can be optimized even further if instead of using B pointer, we use binary search to find the other factor of product if arr[c] is divisible by arr[a]. Here's the modification for his code:
for(c=n-1;(c>1)&& (max==-1);c--){ // loop through C
for(a=0;(a<c-1)&&(max==-1);a++){ // loop through A
if(arr[c]%arr[a]==0) // If arr[c] is divisible by arr[a]
{
if(binary_search(a+1, c-1, (arr[c]/arr[a]))) //#include<algorithm>
{
max = arr[c]; // if the other factor x of arr[c] is also in the array such that arr[c] = arr[a] * x
break;
}
}
}
}
I would have commented this on his solution, unfortunately I lack the reputation to do so.
Try this.
Written in c++
#include <vector>
#include <algorithm>
using namespace std;
int MaxElement(vector< int > Input)
{
sort(Input.begin(), Input.end());
int LargestElementOfInput = 0;
int i = 0;
while (i < Input.size() - 1)
{
if (LargestElementOfInput == Input[Input.size() - (i + 1)])
{
i++;
continue;
}
else
{
if (Input[i] != 0)
{
LargestElementOfInput = Input[Input.size() - (i + 1)];
int AllowedValue = LargestElementOfInput / Input[i];
int j = 0;
while (j < Input.size())
{
if (Input[j] > AllowedValue)
break;
else if (j == i)
{
j++;
continue;
}
else
{
int Product = Input[i] * Input[j++];
if (Product == LargestElementOfInput)
return Product;
}
}
}
i++;
}
}
return -1;
}
Once you have sorted the array, then you can use it to your advantage as below.
One improvement I can see - since you want to find the max element that meets the criteria,
Start from the right most element of the array. (8)
Divide that with the first element of the array. (8/2 = 4).
Now continue with the double pointer approach, till the element at second pointer is less than the value from the step 2 above or the match is found. (i.e., till second pointer value is < 4 or match is found).
If the match is found, then you got the max element.
Else, continue the loop with next highest element from the array. (6).
Efficient solution:
2 3 8 6
Sort the array
keep 3 pointers C, B and A.
Keeping C at the last and A at 0 index and B at 1st index.
traverse the array using pointers A and B till C and check if A*B=C exists or not.
If it exists then C is your answer.
Else, Move C a position back and traverse again keeping A at 0 and B at 1st index.
Keep repeating this till you get the sum or C reaches at 1st index.
Here's the complete solution:
int arr[] = new int[]{2, 3, 8, 6};
Arrays.sort(arr);
int n=arr.length;
int a,b,c,prod,max=-1;
for(c=n-1;(c>1)&& (max==-1);c--){ // loop through C
for(a=0;(a<c-1)&&(max==-1);a++){ // loop through A
for(b=a+1;b<c;b++){ // loop through B
prod=arr[a]*arr[b];
if(prod==arr[c]){
System.out.println("A: "+arr[a]+" B: "+arr[b]);
max=arr[c];
break;
}
if(prod>arr[c]){ // no need to go further
break;
}
}
}
}
System.out.println(max);
I came up with below solution where i am using one array list, and following one formula:
divisor(a or b) X quotient(b or a) = dividend(c)
Sort the array.
Put array into Collection Col.(ex. which has faster lookup, and maintains insertion order)
Have 2 pointer a,c.
keep c at last, and a at 0.
try to follow (divisor(a or b) X quotient(b or a) = dividend(c)).
Check if a is divisor of c, if yes then check for b in col.(a
If a is divisor and list has b, then c is the answer.
else increase a by 1, follow step 5, 6 till c-1.
if max not found then decrease c index, and follow the steps 4 and 5.
Check this C# solution:
-Loop through each element,
-loop and multiply each element with other elements,
-verify if the product exists in the array and is the max
private static int GetGreatest(int[] input)
{
int max = 0;
int p = 0; //product of pairs
//loop through the input array
for (int i = 0; i < input.Length; i++)
{
for (int j = i + 1; j < input.Length; j++)
{
p = input[i] * input[j];
if (p > max && Array.IndexOf(input, p) != -1)
{
max = p;
}
}
}
return max;
}
Time complexity O(n^2)
I'm trying to write an algorithm that will generate all strings of length nm, with exactly n of each number 1, 2, ... m,
For instance all strings of length 6, with exactly two 1's, two 2's and two 3's e.g. 112233, 121233,
I managed to do this with just 1's and 2's using a recursive method, but can't seem to get something that works when I introduce 3's.
When m = 2, the algorithm I have is:
generateAllStrings(int len, int K, String str)
{
if(len == 0)
{
output(str);
}
if(K > 0)
{
generateAllStrings(len - 1, K - 1, str + '2');
}
if(len > K)
{
generateAllStrings(len - 1, K, str + '1');
}
}
I've tried inserting similar conditions for the third number but the algorithm doesn't give a correct output. After that I wouldn't even know how to generalise for 4 numbers and above.
Is recursion the right thing to do? Any help would be appreciated.
One option would be to list off all distinct permutations of the string 111...1222...2...nnn....n. There are nice algorithms for enumerating all distinct permutations of a string in time proportional to the length of the string, and they'd probably be a good way to go about solving this problem.
To use a simple recursive algorithm, give each recursion the permutation so far (variable perm), and the number of occurances of each digit that is still available (array count).
Run the code snippet to generate all unique permutations for n=2 and m=4 (set: 11223344).
function permutations(n, m) {
var perm = "", count = []; // start with empty permutation
for (var i = 0; i < m; i++) count[i] = n; // set available number for each digit = n
permute(perm, count); // start recursion with "" and [n,n,n...]
function permute(perm, count) {
var done = true;
for (var i = 0; i < count.length; i++) { // iterate over all digits
if (count[i] > 0) { // more instances of digit i available
var c = count.slice(); // create hard copy of count array
--c[i]; // decrement count of digit i
permute(perm + (i + 1), c); // add digit to permutation and recurse
done = false; // digits left over: not the last step
}
}
if (done) document.write(perm + "<BR>"); // no digits left: complete permutation
}
}
permutations(2, 4);
You can easily do this using DFS (or BFS alternatively). We can define an graph such that each node contains one string and a node is connected to any node that holds a string with a pair of int swaped in comparison to the original string. This graph is connected, thus we can easily generate a set of all nodes; which will contain all strings that are searched:
set generated_strings
list nodes
nodes.add(generateInitialString(N , M))
generated_strings.add(generateInitialString(N , M))
while(!nodes.empty())
string tmp = nodes.remove(0)
for (int i in [0 , N * M))
for (int j in distinct([0 , N * M) , i))
string new = swap(tmp , i , j)
if (!generated_strings.contains(new))
nodes.add(new)
generated_strings.add(new)
//generated_strings now contains all strings that can possibly be generated.
I have recently come across with this problem,
you have to find an integer from a sorted two dimensional array. But the two dim array is sorted in rows not in columns. I have solved the problem but still thinking that there may be some better approach. So I have come here to discuss with all of you. Your suggestions and improvement will help me to grow in coding. here is the code
int searchInteger = Int32.Parse(Console.ReadLine());
int cnt = 0;
for (int i = 0; i < x; i++)
{
if (intarry[i, 0] <= searchInteger && intarry[i,y-1] >= searchInteger)
{
if (intarry[i, 0] == searchInteger || intarry[i, y - 1] == searchInteger)
Console.WriteLine("string present {0} times" , ++cnt);
else
{
int[] array = new int[y];
int y1 = 0;
for (int k = 0; k < y; k++)
array[k] = intarry[i, y1++];
bool result;
if (result = binarySearch(array, searchInteger) == true)
{
Console.WriteLine("string present inside {0} times", ++ cnt);
Console.ReadLine();
}
}
}
}
Where searchInteger is the integer we have to find in the array. and binary search is the methiod which is returning boolean if the value is present in the single dimension array (in that single row).
please help, is it optimum or there are better solution than this.
Thanks
Provided you have declared the array intarry, x and y as follows:
int[,] intarry =
{
{0,7,2},
{3,4,5},
{6,7,8}
};
var y = intarry.GetUpperBound(0)+1;
var x = intarry.GetUpperBound(1)+1;
// intarry.Dump();
You can keep it as simple as:
int searchInteger = Int32.Parse(Console.ReadLine());
var cnt=0;
for(var r=0; r<y; r++)
{
for(var c=0; c<x; c++)
{
if (intarry[r, c].Equals(searchInteger))
{
cnt++;
Console.WriteLine(
"string present at position [{0},{1}]" , r, c);
} // if
} // for
} // for
Console.WriteLine("string present {0} times" , cnt);
This example assumes that you don't have any information whether the array is sorted or not (which means: if you don't know if it is sorted you have to go through every element and can't use binary search). Based on this example you can refine the performance, if you know more how the data in the array is structured:
if the rows are sorted ascending, you can replace the inner for loop by a binary search
if the entire array is sorted ascending and the data does not repeat, e.g.
int[,] intarry = {{0,1,2}, {3,4,5}, {6,7,8}};
then you can exit the loop as soon as the item is found. The easiest way to do this to create
a function and add a return statement to the inner for loop.
I am looking for a function which counts how many numbers in a range of cells are in the set of numbers
For example I have the set of numbers(1,2,3) and my cells contains 1 | 2 | 3 | 4 | 5 | 3 , the count should return 4
I have tried using countif but no success, I would like to have an excel function Ex.: =countif(A1:D5,...)
How about this? Assume data is in range A1:D5 and you want to count cells with a value of 1, 2 or 3:
=SUM(COUNTIF(A1:D5, {"1","2","3"}))
I hope my pseudo-code would be understandable
int count(int *set, int set_size, int *cells, int cells_size)
{
int v = 0;
// For every number in set
for(int i = 0; i < set_size; ++i)
{
// Loop through every number in cells
for(int j = 0; j < cells_size; ++j)
{
// If number in cells equals number in set, increment v
if(cells[j] == set[i])
{
v++;
}
}
}
// Result is in v, return it
return v;
}
Of course you can optimize a bit with using better containers than just arrays and sizes of them, but I hope you get the basics from this.
Note I used C-like language for pseudo-code, if anything is unclear I can explain further.