I looked for a function to determine if a number is prime and found this
for (var i = 2; i <= Math.sqrt(num); i++)
if (num % i === 0) {
return false;
}
return true;
and I don't understand why that works, yet this doesn't
for (var i = 2; i <= Math.sqrt(num); i++) {
if (num % i === 0) {
return false;
}
return true;
}
What is it about the (lack of the) block statement that is functioning differently
Your first code looks like this:
for (var i = 2; i <= Math.sqrt(num); i++){
if (num % i === 0) {
return false;
}
}
return true;
Notice how return true is on the outside. Your second code doesn't work because it prematurely returns true when there are more numbers to check. Your entire for loop is equivalent to
return num % 2 !== 0;
which is clearly incorrect.
Let me tell you something about blocks that you might not have known (It took me a while to discover this at least).
When you use a loop or if-else statement, you can ignore the using braces { and }.
Example 1
if (a === b){
c = 0;
}
is actually the same as
if (a === b)
c = 0;
Example 2
for (i = 0; i < 10; i++){
a += 1;
}
is actually the same as
for (i = 0; i < 10; i++)
a += 1;
However 1
if (a === b){
c = 0;
d = 1;
}
is not the same with
if (a === b)
c = 0;
d = 1;
However 2
for (i = 0; i < 10; i++){
a += 1;
b += 1;
}
is not the same with
for (i = 0; i < 10; i++)
a += 1;
b += 1;
Explanation
In loops and if-else statement, the block statement (codes surrounded by { and } groups the codes within it and execute it.
However, in the absence of { and }, the loops or if-else statement will only execute the single line of code after it.
Meaning,
var a = 0,
b = 0;
for (i = 0; i < 10; i++)
a += 1;
b += 1;
In this case, a === 10 but b === 1.
Also, in the following case,
var a = 0,
b = 0;
if (false)
a = 10;
b = 10;
a === 0 but b === 10.
Related
I have to make a pattern like this:
=========1=========
=======22122=======
====33322122333====
4444333221223334444
I have not found the logic yet. I tried to code it, but the output is different.
Here is the output of my working code snippet:
----1-----
---123----
--12345---
-1234567--
123456789-
function nomor3(input){
let temp = '';
for (let x = 1; x <= input; x++){
for (let y = input ; y > x; y--){
temp += "-";
}
for (let z = 1; z <= (x * 2) - 1; z++){
temp += z;
}
for (let k = input; k >= x; k--){
temp += "-";
}
temp += '\n';
}
return temp
}
console.log(nomor3(5));
The logic for each level - say 4th level - it begins with the digit of the level to the count of the digit, then one less and so on. So line 4 looks like 4444-333-22-1 and backwards (dashes added for demonstration).
So here we build each line like that, starting from the biggest so we know its length so we can center other lines with dashes. We use arrays here and reversing them because it's easier than strings. But lastly we join so we have a string.
function pyramide(level) {
var len = null;
var result = [];
while (level > 0) {
var arr = [];
for (var i = level; i > 1; i--) {
for (var repeat = 0; repeat < i; repeat++) {
arr.push(i)
}
}
var str_level = arr.join("") + "1" + arr.reverse().join("");
if (len === null) {
len = str_level.length;
}
while (str_level.length < len) {
str_level = "-" + str_level + "-";
}
result.push(str_level);
level--;
}
return result.reverse().join("\n");
}
console.log(pyramide(5))
I have an issue with a recursive algorithm, that solves the problem of finding the happy numbers.
Here is the code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
//return false;
}
}
}
}
CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Algorithm is working ALMOST fine. I've tested it out by calling function with different numbers, and console was displaying correct results. The problem is that I almost can't get any value from the function. There are only few cases in which I can get any value: If the number is build out of ,,0", and ,,1", for example 1000.
Because of that, I figured out, that I have problem with returning any value when the function is calling itself again.
Now I ended up with 2 results:
Returning the
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
which is giving an infinity looped number. For example when the number is happy, the function is printing in the console number one again and again...
Returning the
//return true
or
//return false
which gives me an undefined value
I'm a little bit in check by this problem, and I'm begging you guys for help.
I would take a step back and reexamine your problem with recursion in mind. The first thing you should think about with recursion is your edge cases — when can you just return a value without recursing. For happy numbers, that's the easy case where the sum of squares === 1 and the harder case where there's a cycle. So test for those and return appropriately. Only after that do you need to recurse. It can then be pretty simple:
function sumSq(num) {
/* simple helper for sums of squares */
return num.toString().split('').reduce((a, c) => c * c + a, 0)
}
function isHappy(n, seen = []) {
/* seen array keeps track of previous values so we can detect cycle */
let ss = sumSq(n)
// two edge cases -- just return
if (ss === 1) return true
if (seen.includes(ss)) return false
// not an edge case, save the value to seen, and recurse.
seen.push(ss)
return isHappy(ss, seen)
}
console.log(isHappy(23))
console.log(isHappy(22))
console.log(isHappy(7839))
Here's a simplified approach to the problem
const digits = x =>
x < 10
? [ x ]
: [ ...digits (x / 10 >> 0), x % 10 ]
const sumSquares = xs =>
xs.reduce ((acc, x) => acc + x * x, 0)
const isHappy = (x, seen = new Set) =>
x === 1
? true
: seen.has (x)
? false
: isHappy ( sumSquares (digits (x))
, seen.add (x)
)
for (let n = 1; n < 100; n = n + 1)
if (isHappy (n))
console.log ("happy", n)
// happy 1
// happy 7
// happy 10
// ...
// happy 97
The program above could be improved by using a technique called memoization
Your code is almost correct. You just forgot to return the result of the recursive call:
function TestingFunction(number){
sumNumberContainer = new Array(0);
if (CheckIfNumberIsHappy(number))
console.log(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
for (let i=0; i<100; ++i)
TestingFunction(i.toString()); // 1 7 10 13 ... 91 94 97
I've got the solution, which was given to me in the comments, by the user: Mark_M.
I just had to use my previous
return true / return false
also I had to return the recursive statement in the function, and return the value of the CheckIfTheNumberIsHappy function, which was called in TestingFunction.
The working code:
function TestingFunction(number){
sumNumberContainer = new Array(0);
return CheckIfNumberIsHappy(number);
}
function CheckIfNumberIsHappy(number){
var sumOfTheNumbers = 0;
for (var i = 0; i < number.length; i++) {
sumOfTheNumbers += Math.pow(parseInt(number[i]), 2);
}
console.log(sumOfTheNumbers);
if(sumOfTheNumbers == 1){
return true;
} else {
sumNumberContainer.push(sumOfTheNumbers);
if(sumNumberContainer.length > 1){
for (var i = 0; i < sumNumberContainer.length - 1; i++) {
for (var j = i + 1; j < sumNumberContainer.length; j++) {
if(sumNumberContainer[i] == sumNumberContainer[j]){
return false;
}
}
}
}
return CheckIfNumberIsHappy(sumOfTheNumbers.toString());
}
}
Thanks for the great support :)
I'm trying to check if string b contains any 5-character substring a.
This works, but is a little messy:
var a = "1eabcde";
var b = "12abcde12fg";
for(var i=0; i<a.length; i++){
for(var j=i;j<a.length-i;j++){
if(a.charAt(j) == b.charAt(i) && a.charAt(j+1) == b.charAt(i+1) && a.charAt(j+2) == b.charAt(i+2) && a.charAt(j+3) == b.charAt(i+3) && a.charAt(j+4) == b.charAt(i+4)){
alert("ya");
}
}
}
Are there any other cleaner options?
You can use substring and indexOf:
var a = "1eabcde";
var b = "12abcde12fg";
for (var i = 0; i <= a.length - 5; i++) {
if (b.indexOf(a.substring(i, i + 5)) >= 0) {
alert("ya");
}
}
(You could use a.substr(i, 5) instead of a.substring(i, i + 5). Those two calls behave identically.)
Note that if you loop from 0 to a.length (as in your original code), then all suffixes of a of length 5 or less will be searched for in b.
In one respect, this code does not behave the same as your original: it will alert only once for each matching substring of a, regardless of how many times that particular substring may occur in b. Thus, if a = 'abcde' and b = '01abcde23abcde45, your original code would pop up two alerts (one for each occurrence of 'abcde'), whereas the above will only alert once. If you want the original behavior, change the if to a while like this:
for (var i = 0; i <= a.length - 5; i++) {
var j = -1;
while ((j = b.substring(j+1).indexOf(a.substr(i, 5))) >= 0) {
alert("ya");
}
}
This is the cleanest approach :
var a = "1eabcde";
var b = "12abcde12fg";
for (var i = 0; i <= a.length - 5; i++) {
if(b.indexOf(a.substr(i, 5)) > -1) {
alert("ya");
}
}
I'm trying to write a program in JavaScript that generates 100 random numbers and checks the primality of each. The program does just that, except for some reason it doesn't stop at 100 and just loops infinitely. I'm sure I made some simple novice mistake, but for some reason I can't see it. Any advice?
My code:
function isPrime(n) {
if (n < 2 || n % 1)
return false;
var r = Math.sqrt(n);
for (i = 2; i <= r; i++)
if (n % i === 0)
return false;
return true;
}
for (i = 0; i < 100; i++) {
var temp = Math.floor((Math.random() * 100) + 1);
if (isPrime(temp))
console.log(temp + " is a prime number!");
else
console.log(temp + " is not a prime number.");
}
Thanks!
You need to declare i variable in for-loops:
(var i = 0; i < 100; i++) ...
otherwise it is defined in global scope and it is shared between for-loop and isPrime function.
madox2 is correct that you should declare i in the for loop, however I think the reason the loop itself is infinite is because by only doing i=0 in the loop, and then for (i = 2; i <= r; i++) in the function the loop calls, you are resetting i every iteration
You should change your code to declare i within the scope of both loops separately, like so:
function isPrime(n) {
if (n < 2 || n % 1)
return false;
var r = Math.sqrt(n);
for (var i = 2; i <= r; i++)
if (n % i === 0)
return false;
return true;
}
for (var i = 0; i < 100; i++) {
var temp = Math.floor((Math.random() * 100) + 1);
if (isPrime(temp))
console.log(temp + " is a prime number!");
else
console.log(temp + " is not a prime number.");
}
I made a function that counts the occurances of x's and o's in a given string and returns true if they are equal.
function ExOh(str) {
var x_count = 0;
var o_count = 0;
for (var i = 0;i < str.length-1;i++){
if (str[i] === 'x'){
x_count = x_count + 1;
}
else if (str[i] === 'o'){
o_count = o_count + 1;
}
}
console.log(o_count);
console.log(x_count);
if (x_count === o_count){
return true;}
else{
return false;
}
}
// keep this function call here
// to see how to enter arguments in JavaScript scroll down
ExOh(readline());
I added the lines of code
console.log(o_count);
console.log(x_count);
To see if it was counting correctly and I discovered that was the issue. After testing it I realized that this function is not testing the last element in the string. I tried changing the length of the for loop, but I can't think of what else could be wrong.
Any advice?
Thanks mates
JavaScript arrays are 0 index based objects. So, your loop should be like this
for (var i = 0; i < str.length; i++) {
otherwise the last element will be skipped.
Consider that the length of the string is 5. So, i starts from 0 and if you had your original condition
for (var i = 0; i < str.length - 1; i++) {
following are the comparisons happening in the loop
0 < 4
1 < 4
2 < 4
3 < 4
4 < 4 -- Fails
So it breaks out of the loop. But the last element will be at index 4. But when you have the condition like this
for (var i = 0; i < str.length; i++) {
the comparisons go like this
0 < 5
1 < 5
2 < 5
3 < 5
4 < 5
5 < 5 -- Fails
It breaks out of the loop only after comparing all the elements.
So, your actual program can be written like this
function ExOh(str) {
var x_count = 0, o_count = 0;
for (var i = 0; i < str.length; i++) {
if (str[i] === 'x') {
x_count = x_count + 1;
} else if (str[i] === 'o') {
o_count = o_count + 1;
}
}
return x_count === o_count;
}
alternate method to count characters:
var s = 'example';
s.split('').filter(function (i) { return i === 'e'; }).length; // 2
Your for loop is running one too short. Try this instead.
for (var i = 0;i < str.length;i++){
if (str[i] === 'x'){
x_count = x_count + 1;
}
else if (str[i] === 'o'){
o_count = o_count + 1;
}
}
Your problem is in for loop. Try changing to this.
for (var i = 0; i < str.length; i++) {
If you want to avoid using for loops, you can use this much shorter version of ExOh function.
function ExOh(str) {
return str.match(/o/g).length == str.match(/x/g).length
}
Rather than looping over the whole String with for, I'd see if using indexOf achieves a faster result
function countOccurance(haystack, needle) {
var total = 0, pos = -1;
while (-1 !== (pos = haystack.indexOf(needle, pos + 1)))
total += 1;
return total;
}
Then
var x_count = countOccurance(str, 'x'),
o_count = countOccurance(str, 'o');
return x_count === o_count;
EDIT looks like I might have been wrong about it being faster! jsperf
function indexOfMethod(haystack, needle) {
var total = 0, pos = -1;
while (-1 !== (pos = haystack.indexOf(needle, pos + 1)))
total += 1;
return total;
}
function splitMethod(haystack, needle) {
return haystack.split(needle).length - 1;
}
function forMethod(haystack, needle) {
var total = 0, i;
for (i = 0; i < haystack.length; ++i)
if (haystack.charAt(i) === needle)
total += 1;
return total;
}
The forMethod will only work with char needle, whereas the other two should work with any String as needle, if that matters.