I have a problem about recursive in JavaScript.
This is my code when I use loop, but I want to use recursive.
function triangleNumber(num) {
for(var i = num; i >= 1 ; i--){
var str = ''
for(var j = i; j >= 1; j--){
str += j
}
console.log(str)
}
//DRIVER CODE
console.log(triangleNumber(5));
// 54321
// 4321
// 321
// 21
// 1
Try this solution, where you build the string using the for loop appending numbers till you reach 1. Then you call the function recursively for the next smaller numbern - 1:
function triangleNumber(num) {
if(num <= 0){ //Base case return if number reaches '0'
return;
}
var str = ''
for(var i = num; i >= 1; i--){
str += i; //Create number string by appending successive decreasing value
}
console.log(str);
triangleNumber(num - 1) //Recursively call the function for the next lower number
}
triangleNumber(5);
Pretty straightforward :)
function triangleNumber(num) {
if(num<1) return;
var str = '';
for(var i = num; i >= 1 ; i--) str += i;
console.log(str);
triangleNumber(num-1)
}
triangleNumber(5);
Related
The code works but I don't want the inner for loop to take me to the new line.
for (i = 0; i < 5; i++) {
for (j = 1; j <= i; j++) {
console.log('*');
}
console.log();
}
console.log('-----------------');
console.log will automatically break the line. Concatenate to a string instead of a log. Log at the end.
let str = '';
for(i = 0; i <= 5 ; i++) {
for(j = 1; j <= i; j++) {
str += '*';
}
str += '\n';
}
console.log(str);
You can do this way, with the help of a string variable:
for (i = 0; i < 5; i++) {
var str = '';
for (j = 1; j <= i; j++) {
str+='*';
}
console.log(str);
}
console.log('-----------------');
If you want to print at the page, use like below
for (i = 0; i < 5; i++) {
let j=0;
do{document.write("*");j++;}while(j < i)
document.write("<br/>")
}
You need to break the line with the console.log you can also controle the space between * with output+='*' + " ";
function pyramid() {
var total = 5;
var output="";
for (var i = 1; i <= total; i++) {
for (var j = 1; j <= i; j++) {
output+='*' + " ";
}
console.log(output);
output="";
}
}
pyramid()
You can get rid of second for loop as follows:
var str = '';
for (i = 1; i <= 5; i++) {
str +=Array(i).join("*");
str +='\n';
}
console.log(str);
let string = "";
for (let i = 0; i < 5; i++){
string += '*';
console.log(string);
}
Output:
*
**
***
****
*****
A simple way to solve this "exercise" in JavaScript:
let width = ""
while(width.length < 6) console.log(width += `#` );
Basically, we create a string (width) and increment its value using the while loop till we hit a restriction.
I found the more typical method "bulky"(?)...plus there's the issue of not getting the exact picture of a half pyramid.
let i,j
for (i= 0; i < 6; i++){
for (j = 0; j<=i; j++){
console.log("#")
}
console.log("\n")
}
function pyramid(n){
let result = "";
for(let i=0; i<=n; i++){
result += "*".repeat(i);
result += "\n"
}
return result;
}
console.log(pyramid(5));
//OutPut
*
**
***
****
*****
As we need n number of pyramid structure with '' / '#' / any symbol. by using above code we can achieve. Here you can see we just created a function called pyramid with one parameter 'n'. and inside function we declare a variable 'result'. So inside for loop the length of 'i' is "<=n" and also you can use "repeat() method to print '' 'i' times. So if you call that function like console.log(pyramid(5)); You can able to see your Answer as expected..
shortest code:
console.log('*\n**\n***\n****\n*****');
/Write a function called weave that accepts an input string and number. The function should return the string with every xth character replaced with an 'x'./
function weave(word,numSkip) {
let myString = word.split("");
numSkip -= 1;
for(let i = 0; i < myString.length; i++)
{
numSkip += numSkip;
myString[numSkip] = "x";
}
let newString = myString.join();
console.log(newString);
}
weave("weave",2);
I keep getting an infinite loop. I believe the answer I am looking for is "wxaxe".
Here's another solution, incrementing the for loop by the numToSkip parameter.
function weave(word, numToSkip) {
let letters = word.split("");
for (let i=numToSkip - 1; i < letters.length; i = i + numToSkip) {
letters[i] = "x"
}
return letters.join("");
}
Well you need to test each loop to check if it's a skip or not. Something as simple as the following will do:
function weave(word,numSkip) {
var arr = word.split("");
for(var i = 0; i < arr.length; i++)
{
if((i+1) % numSkip == 0) {
arr[i] = "x";
}
}
return arr.join("");
}
Here is a working example
Alternatively, you could use the map function:
function weave(word, numSkip) {
var arr = word.split("");
arr = arr.map(function(letter, index) {
return (index + 1) % numSkip ? letter : 'x';
});
return arr.join("");
}
Here is a working example
Here is a more re-usable function that allows specifying the character used for substitution:
function weave(input, skip, substitute) {
return input.split("").map(function(letter, index) {
return (index + 1) % skip ? letter : substitute;
}).join("");
}
Called like:
var result = weave('weave', 2, 'x');
Here is a working example
You dont need an array, string concatenation will do it, as well as the modulo operator:
function weave(str,x){
var result = "";
for(var i = 0; i < str.length; i++){
result += (i && (i+1)%x === 0)?"x":str[i];
}
return result;
}
With arrays:
const weave = (str,x) => str.split("").map((c,i)=>(i&&!((i+1)%x))?"x":c).join("");
You're getting your word greater in your loop every time, so your loop is infinite.
Try something like this :
for(let k = 1; k <= myString.length; k++)
{
if(k % numSkip == 0){
myString[k-1]='x';
}
}
Looking at what you have, I believe the reason you are getting an error is because the way you update numSkip, it eventually becomes larger than
myString.length. In my code snippet, I make i increment by numSkip which prevents the loop from ever executing when i is greater than myString.length. Please feel free to ask questions, and I will do my best to clarify!
JSFiddle of my solution (view the developer console to see the output.
function weave(word,numSkip) {
let myString = word.split("");
for(let i = numSkip - 1; i < myString.length; i += numSkip)
{
myString[i] = "x";
}
let newString = myString.join();
console.log(newString);
}
weave("weave",2);
Strings are immutable, you need a new string for the result and concat the actual character or the replacement.
function weave(word, numSkip) {
var i, result = '';
for (i = 0; i < word.length; i++) {
result += (i + 1) % numSkip ? word[i] : 'x';
}
return result;
}
console.log(weave("weave", 2));
console.log(weave("abcd efgh ijkl m", 5));
You can do this with fewer lines of code:
function weave(word, numSkip) {
word = word.split("");
for (i = 0; i < word.length; i++) {
word[i] = ((i + 1) % numSkip == 0) ? "x" : word[i];
}
return word.join("");
}
var result = weave("weave", 2);
console.log(result);
how to transform every 3rd character to uppercase. for example this string pifedvcbtauzkwgnxyjrhmq converted into piFedVcbTauKkwGnxYjrHmq.
here is what i've done so far
function up3rdletter(str){
var i, result = '';
for(i = 0; i < str.length; i++){
if(i > 0 && i % 3 == 0){
result += str[i].toUpperCase();
}else{
result += str[i];
}
}
return result;
}
the script above return pifEdvCbtAuzKwgNxyJrhMq, it start converting from 4th letter. if I change the modulo number to 2 it become piFeDvCbTaUzKwGnXyJrHmQ
It's rather trivial: indexing in JS starts from 0, but you want to change letters based on 1-based index, natural for human beings. ) Solution? Either shift the remainder check:
result += i % 3 === 2 ? str[i].toUpperCase() : str[i];
... or go on checking against 0, but advance i instead:
result += (i + 1) % 3 ? str[i] : str[i].toUpperCase();
But actually, I'd probably write it as follows:
result = str.replace(/(..)(.)/g, function(_, m1, m2) {
return m1 + m2.toUpperCase();
});
function up3rdletter(str){
var i, result = '';
var counter = 1;
for(i = 0; i < str.length; i++){
if(i > 0 || counter % 3 == 0){
result += str[i].toUpperCase();
}else{
result += str[i];
}
counter++;
}
return result;
}
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.
Any ideas on the following? I want to input a number into a function and insert dashes "-" between the odd digits. So 4567897 would become "456789-7". What I have so far is to convert the number into a string and then an array, then look for two odd numbers in a row and use the .splice() method to add the dashes where appropriate. It does not work and I figure I may not be on the right track anyway, and that there has to be a simpler solution.
function DashInsert(num) {
var numArr = num.toString().split('');
for (var i = 0; i < numArr.length; i++){
if (numArr[i]%2 != 0){
if (numArr[i+1]%2 != 0) {
numArr.splice(i, 0, "-");
}
}
}
return numArr;
}
The problem is you're changing the thing you're iterating over. If instead you maintain a separate output and input...
function insertDashes(num) {
var inStr = String(num);
var outStr = inStr[0], ii;
for (ii = 1; ii < inStr.length; ii++) {
if (inStr[ii-1] % 2 !== 0 && inStr[ii] % 2 !== 0) {
outStr += '-';
}
outStr += inStr[ii];
}
return outStr;
}
You can try using regular expressions
'4567897'.replace(/([13579])(?=[13579])/g, '$1-')
Regex Explained
So, we find an odd number (([13579]) is a capturing group meaning we can use it as a reference in the replacement $1) ensure that it is followed by another odd number in the non-capturing positive lookahead ((?=[13579])) and replace the matched odd number adding the - prefix
Here is the function to do it:
function dashes(number){
var numString = '';
var numArr = number.toString().split('');
console.log(numArr);
for(i = 0; i < numArr.length; i++){
if(numArr[i] % 2 === 1 && numArr[i+1] % 2 === 1){
numString += numArr[i] + '-';
}else{
numString += numArr[i];
}
}
console.log(numString);
}
dashes(456379);
Tested and everything.
Edit: OrangeDog's answer was posted earlier (by nearly a full half hour), I just wanted to make an answer which uses your code since you're almost there.
Using another array instead of splicing into one you were looping through (this happens to return a string using join):
var num = 4567897;
function DashInsert(num) {
var numArr = num.toString().split('');
var len = numArr.length;
var final = [];
for (var i = 0; i < len; i++){
final.push(numArr[i]);
if (numArr[i]%2 != 0){
if (i+1 < len && numArr[i+1]%2 != 0) {
final.push("-")
}
}
}
return final.join("");
}
alert(DashInsert(num));
function dashInsert(str) {
var arrayNumbers = str.split("");
var newString = "";
for (var i = 0; i < arrayNumbers.length; i++){
if(arrayNumbers[i] % 2 === 1 && arrayNumbers[i + 1] % 2 === 1){
newString = newString + arrayNumbers[i] + "-";
} else {
newString = newString + arrayNumbers[i];
}
}
return newString;
}
var result = dashInsert("3453246");
console.log(result);