Develop Reference

Counting the occurrence of chars in a javascript string

I have written down code to calculate the count of each of the character in a string.
It seems to be working correctly for some of the words where as for some it fails.
It fails for the last character, as I see the length of the string becomes smaller than the iteration count (but for some of the words)
var str1 = "america"
function noofchars(str1) {
for (var m = 0; m < str1.length + 1; m++) {
var countno = 1;
if (m != 0) {
str1 = str1.slice(1)
}
str2 = str1.substr(0, 1)
for (var i = 0; i < str1.length; i++) {
if (str2 === str1.charAt(i + 1)) {
countno += 1
str1 = str1.slice(0, i + 1) + str1.slice(i + 2) + " "
}
}
console.log(str1.charAt(0) + "=" + countno)
}
}
var findnoofchar = noofchars(str1)
It passes for london, philadelphia, sears, happy
But fails for america, chicago etc
london = l=1, o=2, n=2, d=1

It'd be easier to use an object. First reduce into character counts, then iterate through the key/value pairs and console.log:
function noofchars(str1) {
let r = [...str1].reduce((a, c) => (a[c] = (a[c] || 0) + 1, a), {});
Object.entries(r).forEach(([k, v]) => console.log(`${k}=${v}`));
}
noofchars("america");
ES5 syntax:
function noofchars(str1) {
var r = str1.split("").reduce(function(a, c) {
a[c] = (a[c] || 0) + 1;
return a;
}, {});
Object.keys(r).forEach(function(k) {
console.log(k + "=" + r[k]);
});
}
noofchars("america");
It's easier to understand what reduce is doing in the above snippet.
First, we take a function with two parameters a and c. These can be called anything, I just use a and c for the accumulator and the current item.
Now, the second line:
a[c] = (a[c] || 0) + 1;
This is kind hard, so let's break it down. First let's look at what's in the parentheses:
a[c] || 0
This checks if a has a key/value pair with the key c (as in, the value of c, not the key literally being c). If that doesn't exist, it returns 0. So if a[c] exists, save it as the value of the expression, otherwise use 0.
Now, we add 1, to increment the value.
Finally, we assign the result to a[c]. So if a contained c, the value of a[c] would be incremented. If a didn't contain c, the value of a[c] would be 1.
Then, we return a to be used in the next iteration of reduce.
In the next line:
}, {});
We assign a default value for a. If we didn't do this, the first time reduce ran, a would be "a", and c would be "m" (the first two characters of america). This way, a is {} (an empty object), and c is "a". If we didn't have this second argument, our function wouldn't work.
In this line:
Object.keys(r).forEach(function(k) {...});
We're getting an array of all the keys in r, and looping through them with forEach, with k being the key.
Then, we're logging k (the key), then an equals sign =, then the value of r[k].

You can split the string by each character and then count the number of occurrence of each character by reduce function as below
function noofchars(str1) {
const chArray = str1.split('');
return chArray.reduce(function(acc, ch) {
if (acc[ch]) {
acc[ch]++;
} else {
acc[ch] = 1;
}
return acc;
}, {});
}
var str1 = "america";
var findnoofchar = noofchars(str1);
console.log(findnoofchar);
In your solution you are mutating str1 in this line
str1 = str1.slice(0, i + 1) + str1.slice(i + 2) + " "
which actually changes the length of the string and also checking str1.length in the first loop. In you case you can take the length in the first place. Working version of your code snippet is here
var str1 = "america"
function noofchars(str1) {
const len = str1.length;
for (var m = 0; m < len; m++) {
var countno = 1;
if (m !== 0) {
str1 = str1.slice(1)
}
if (str1.charAt(0) === ' ') {
break;
}
str2 = str1.substr(0, 1)
for (var i = 0; i < str1.length; i++) {
if (str2 === str1.charAt(i + 1)) {
countno += 1
str1 = str1.slice(0, i + 1) + str1.slice(i + 2) + " "
}
}
console.log(str1.charAt(0) + "=" + countno)
}
}
var findnoofchar = noofchars(str1)

Not really knowing what you are trying to accomplish with your code, one solution is to utilize the charAt and Set functions. CharAt is a more direct way to iterate over the string and the Set function eliminates duplicate characters from the set automatically.
var str1 = "america";
function noofchars(str1) {
var charList = new Set();
for (var m = 0; m < str1.length; m++) {
var charX = str1.charAt(m).substr(0, 1);
var countno = 1;
for (var i = 0; i < str1.length; i++) {
if (str1.slice(0, i + 1) == charX) {
countno++;
}
charList.add(charX + "=" + countno);
}
}
// you may need to expand set on Chrome console
console.log(charList);
// a hack to display more easily display on your screen
var listOnScreen = Array.from(charList);
document.getElementById('displaySet').innerHTML=listOnScreen;
}
noofchars(str1);
<div id="displaySet"></div>

How do you iterate over an array every x spots and replace with letter?

/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);

Split string at nth occurance of delimiter into array in javascript

Is there a javascript/jquery method or function that will let me split a string at nth occurrence of a selected delimiter? I'd like it work just like the regular str.split(delimiter) except instead of splitting at every occurrence of the delimiter it could be instructed to skip n number of them each time.
var str = "A,BB,C,DDD,EEEE,F";
var strAry = str.split(",");
Would result in strAry looking like {"A","BB","C","DDD","EEEE","F"}
What I want would be {"A,BB","C,DDD","EEEE,F"} assuming I set nth occurance to 2.
I wrote a small function that appears to work but hoping there was a simpler way to do this:
function splitn(fullString, delimiter, n){
var fullArray = fullString.split(delimiter);
var newArray = [];
var elementStr = "";
for(var i = 0; i < fullArray.length; i++) {
if (i == fullArray.length-1) {
if (elementStr.length == 0) {
elementStr = fullArray[i];
} else {
elementStr += (delimiter + fullArray[i]);
}
newArray.push(elementStr);
} else {
if (((i + 1) % n) == 0) {
if (elementStr.length == 0) {
elementStr = fullArray[i];
} else {
elementStr += (delimiter + fullArray[i]);
}
newArray.push(elementStr);
elementStr = "";
} else {
if (elementStr.length == 0) {
elementStr = fullArray[i];
} else {
elementStr += (delimiter + fullArray[i]);
}
}
}
};
return newArray;
};
Thanks.

You could simply use Array.prototype.reduce() to modify the array returned by split to your liking. The idea is similar to your code, just shorter.
function modify(str, n, delim) {
return str.split(delim).reduce(function(output, item, i) {
if (!(i % n)) {
output.push(item);
} else {
output[i / n | 0] += delim + item;
};
return output;
}, []);
};
modify("A,BB,C,DDD,EEEE,F", 3, ','); //["A,BB,C", "DDD,EEEE,F"]
modify("A,BB,C,DDD,EEEE,F", 2, ','); //["A,BB", "C,DDD", "EEEE,F"]

EDIT: Just noticed you wanted to use an arbitrary "nth" value. I updated it so you can simply change the nth to whatever positive integer you like.
Here's a way that takes advantage of the second argument to .indexOf() so that you can anchor your searches from after the last ending point in the string:
function splitn(fullString, delimiter, n) {
var lastIdx = 0
, idx = -1
, nth = 0
, result = [];
while ((idx = fullString.indexOf(delimiter, idx + delimiter.length)) !== -1) {
if ((nth = ++nth % n) === 0) {
result.push(fullString.slice(lastIdx, idx));
lastIdx = idx + 1;
}
}
result.push(fullString.slice(lastIdx));
return result;
}
var result = splitn("A,BB,C,DDD,EEEE,F", ",", 2);
document.body.innerHTML = "<pre>" + JSON.stringify(result, null, 4) + "</pre>";

all valid combinations of n-pair of parenthesis

I am learning js now..
I am trying to write a simple js programme..
what I am trying to do is to print all valid combinations of n-pair
of parenthesis(properly opened and closed)
eg (), (()()),(())
i have written the logic can you tell me whether its correct or not
https://jsfiddle.net/e7mcp6xb/
module.exports = Parentheses = (function() {
var _isParenthesesMatch = function(str) {
var parentheses = str.length;
var rightParentheses = '(';
var leftParentheses = ')';
var rightCount = 0;
var leftCount = 0;
for(i=0;i<=str.length;i++){
if(rightParentheses == str.charAt(i))
{
rightCount++;
}
else if(leftParentheses == str.charAt(i))
{
leftCount++;
}
}
if(rightCount == leftCount){
return true;
}
else(rightCount != leftCount){
return false;
}
}
}());

The check is wrong, but You can fix it easily: In each step of the for loop the number of opening parenthesis cannot be smaller than the number of closing ones:
if (rightCount < leftCount)
return false;
The whole function should look like this:
function(str) {
var rightParentheses = '(';
var leftParentheses = ')';
var rightCount = 0;
var leftCount = 0;
for (var i = 0; i <= str.length; i++) {
if (rightParentheses == str.charAt(i))
rightCount++;
else if (leftParentheses == str.charAt(i))
leftCount++;
if (rightCount < leftCount)
return false;
}
return rightCount == leftCount;
}
If You'd like to generate all valid strings, you can use this function:
function nPair(n) {
if (n == 0)
return [""];
var result = [];
for (var i = 0; i < n; ++i) {
var lefts = nPair(i);
var rights = nPair(n - i - 1);
for (var l = 0; l < lefts.length; ++l)
for (var r = 0; r < rights.length; ++r)
result.push("(" + lefts[l] + ")" + rights[r]);
}
return result;
}
// result of nPair(3):
// ["()()()", "()(())", "(())()", "(()())", "((()))"]

Try this, i have modified your code a little bit. Modification and its explanation is marked in comments.
module.exports = Parentheses = (function() {
var _isParenthesesMatch = function(str) {
var parentheses = str.length;
var rightParentheses = '(';
var leftParentheses = ')';
var count=0;
for(i=0;i<str.length;i++){
//this is to check valid combination start always from ( and end with )
if(str.charAt(0)==rightParentheses && str.length-1==leftParentheses)
{
if(rightParentheses == str.charAt(i))
{
count++; //this will calculate how many times rightParentheses is present & increment count by 1
}
else if(leftParentheses == str.charAt(i))
{
count--; //this will simply decrement count to match valid sequence
}
}
if(count==0){
return true;
}
}
}());

Your function is wrong, try checking if left and right parenthesis and balanced:
function isValid(str){
var stripedStr = str.replace(/[^\(\)]+/g, '');
return stripedStr.split('').reduce(function(a, b){
return a > -1 ? b === '(' ? a + 1 : a - 1 : -1;
}, 0) === 0;
}
stripedStr - use replace() to remove any characters that are not ( or ).
split('') - returns an array so we can use reduce.
reduce() - applies a function against an accumulator and each value of the array (from left-to-right) has to reduce it to a single value.
The reduce starts with 0 as initial value and in the reduce function we count parenthesis
(+1 for (, -1 for ) )
Our string is valid if our counter never goes below 0 and we end up with 0.
You can write the reduce function like this too:
function(previousValue, currentValue){
if (previousValue > -1){
if (currentValue === '('){
return previousValue + 1;
} else {
return previousValue - 1;
}
}
return -1;
}
This is equivalent to:
function(a, b){
return a > -1 ? b === '(' ? a + 1 : a - 1 : -1;
}

It is wrong, because your function will return true for this example ))(( or this ())(()

Is there a way I can change the contents of a variable in Camel Case to space separated words?

I have a variable which contains this:
var a = "hotelRoomNumber";
Is there a way I can create a new variable from this that contains: "Hotel Room Number" ? I need to do a split on the uppercase character but I've not seen this done anywhere before.

Well, you could use a regex, but it's simpler just to build a new string:
var a = "hotelRoomNumber";
var b = '';
if (a.length > 0) {
b += a[0].toUpperCase();
for (var i = 1; i != a.length; ++i) {
b += a[i] === a[i].toUpperCase() ? ' ' + a[i] : a[i];
}
}
// Now b === "Hotel Room Number"

var str = "mySampleString";
str = str.replace(/([A-Z])/g, ' $1').replace(/^./, function(str){ return str.toUpperCase(); });
http://jsfiddle.net/PrashantJ/zX8RL/1/

I have made a function here:
http://jsfiddle.net/wZf6Z/2/
function camelToSpaceSeperated(string)
{
var char, i, spaceSeperated = '';
// iterate through each char
for (i = 0; i < string.length; i++) {
char = string.charAt(i); // current char
if (i > 0 && char === char.toUpperCase()) { // if is uppercase
spaceSeperated += ' ' + char;
} else {
spaceSeperated += char;
}
}
// Make the first char uppercase
spaceSeperated = spaceSeperated.charAt(0).toUpperCase() + spaceSeperated.substr(1);
return spaceSeperated;
}
The general idea is to iterate through each char in the string, check if the current char is already uppercased, if so then prepend a space to it.