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>";
Related
Trying to take an integer and have it return as a
string with the integers from 1 to the number passed.
Trying to use a loop to return the string but not sure how!
Example of how I want it to look:
count(5) => 1, 2, 3, 4, 5
count(3) => 1, 2, 3
Not really sure where to even start
I would do it with a recursive function. Keep concatenating the numbers until it reaches 1.
var sequence = function(num){
if(num === 1) return '1';
return sequence(num - 1) + ', ' + num;
}
Or just:
var sequence = (num) => num === 1 ? '1' : sequence(num - 1) + ', ' + num;
You can use a for loop to iterate the number of times that you pass in. Then, you need an if-statement to handle the comma (since you don't want a comma at the end of the string).
function count(num) {
var s = "";
for(var i = 1; i <= num; i++) {
s += i;
if (i < (num)) {
s += ', ';
}
}
return s;
}
JSBin
Try this:
function count(n) {
var arr = [];
for (var i = 1; i<=n; i++) {
arr.push(i.toString());
}
return arr.toString();
}
Here's a non-recursive solution:
var sequence = num => new Array(num).fill(0).map((e, i) => i + 1).toString();
here is a goofy way to do it
function count(i)
{
while (i--) {
out = (i + 1) + "," + this.out;
}
return (out + ((delete out) && "")).replace(",undefined", "");
}
Quite possibly the most ridiculous way, defining an iterator:
"use strict";
function count ( i ) {
let n = 0;
let I = {};
I[Symbol.iterator] = function() {
return { next: function() { return (n > i) ? {done:true}
: {done:false, value:n++} } } };
let s = "";
let c = "";
for ( let i of I ) {
s += c + i;
c = ", "
}
return s;
}
let s = count(3);
console.log(s);
On a recent interview, I was asked to return all possible combinations of order of operations on an input string, and the result. you should return all the ways/combinations in which you can "force" operations with parenthesis. I got the result (right hand side of the equation) but got stuck on the left side. how could I have done the left side and the right hand side together? Seems like two problems in one...
//input:
console.log(diffWaysToCompute("2 * 3 - 4 * 5"));
//output:
(2*(3-(4*5))) = -34
((2*3)-(4*5)) = -14
((2*(3-4))*5) = -10
(2*((3-4)*5)) = -10
(((2*3)-4)*5) = 10
'use strict'
function getNumbersAndOperators(str) {
var arr = str.split(" ");
var operators = [];
for (var i = 0; i < arr.length; i++) {
if (arr[i] === "-" || arr[i] === "*" || arr[i] === "+") {
operators.push(arr[i]);
arr.splice(i, 1);
// console.log(operators);
}
}
return [arr, operators];
}
// console.log(getNumbersAndOperators("2 - 1 - 1"))
var diffWaysToCompute = function (input) {
// var numbers = input.split(" ");
// console.log(numbers);
// // console.log(number);
var results = compute(input);
results.sort(function (a, b) {
return a - b;
});
//put the numbers length into valid parenthesis:
var NumbersAndOperators = getNumbersAndOperators(input);
var numbers = NumbersAndOperators[0];
console.log(numbers);
var operators = NumbersAndOperators[1];
console.log(operators);
var parens = validParentheses(numbers.length);
// console.log(numbers);
console.log(operators);
// for (var i = 0; i < parens.length; i++) {
// for (var j = 0; j < parens[i].length; j++) {
// var val = parens[i][j];
// console.log(val);
// if (val === " ") {
// var num = numbers.shift();
// parens.splice(val, 0, num);
// //starting running into infinite loops and out of time.
// j--;
// }
// }
// i--;
// }
console.log(parens);
return results;
};
function validParentheses(n) {
if (n === 1) {
return ['( )'];
}
var prevParentheses = validParentheses(n - 1);
var list = {};
prevParentheses.forEach(function (item) {
list['( ' + item + ' )'] = null;
list['( )' + item] = null;
list[item + '( )'] = null;
});
console.log(Object.keys(list))
return Object.keys(list);
}
function compute(str) {
var res = [];
var i;
var j;
var k;
var left;
var right;
var string = [];
var placed = true;
if (!/[+*-]/.test(str)) { // + - *
return [parseInt(str)];
}
for (i = 0; i < str.length; i++) {
if (/\+|\-|\*/.test(str[i])) { // + - *
left = compute(str.substring(0, i));
right = compute(str.substring(i + 1, str.length));
for (j = 0; j < left.length; j++) {
for (k = 0; k < right.length; k++) {
if (str[i] === '+') {
res.push(parseInt(left[j] + right[k]));
} else if (str[i] === '-') {
// string.push("(" + str[i-2], str[i+2] + ")");
res.push(parseInt(left[j] - right[k]));
} else if (str[i] === '*') {
res.push(parseInt(left[j] * right[k]));
}
}
}
}
}
// console.log(string);
return res;
}
console.log(diffWaysToCompute("2 - 1 - 1"));
console.log(diffWaysToCompute("2 * 3 - 4 * 5"));
I never had to do such silly things, so let me try my teeth at it now.
(Caveat as always: it's highly simplified and without any checks&balances!)
The parser is the simplest thing here:
/*
Use of strings instead of ASCII codes for legibility.
I changed x - y to x + (-y) not only for convenience
but for algebraic correctness, too.
#param a array number nodes
#param o array operator nodes
*/
function parse(s,a,o){
var fnum = 0;
var uminus = false
for(var i=0;i<s.length;i++){
switch(s[i]){
case '-': uminus = true;
a.push(fnum);
o.push('+');
fnum = 0;
break;
case '+':
case '*':
case '/': if(uminus){
uminus = false;
fnum *= -1;
}
a.push(fnum);
o.push(s[i]);
fnum = 0;
break;
case '0':
case '1':
case '2':
case '3':
case '4':
case '5':
case '6':
case '7':
case '8':
case '9': fnum = fnum * 10 + parseInt(s[i]);
break;
default: break;
}
}
//assuming symmetry
a.push(fnum);
}
The (-generation took me some time, too much time--I cheated here ;-)
/*
Found in an old notebook (ported from C)
Algo. is O(n^2) and can be done faster but I
couldn't be a...ehm, had no time, sorry.
#idx int index into individual result
#n int number of groups
#open int number of opening parentheses
#close int number of closing parentheses
#a array individual result
#all array space for all results
*/
function makeParens(idx,n,open,close,a,all){
if(close == n){
all.push(a.slice(0));
return;
} else {
if(open > close){
a[idx] = ')';
makeParens(idx+1,n,open,close+1,a,all);
}
if(open < n){
a[idx] = '(';
makeParens(idx+1,n,open+1,close,a,all);
}
}
}
And now? Yepp, that took me a while:
/*
The interesting part
Not very optimized but working
#s string the equation
#return array nicely formatted result
*/
function parenthesing(s){
var nums = [];
var ops = [];
var all = [];
var parens = [];
// parse input into numbers and operators
parse(input,nums,ops);
/*
Rules:
1) out-most parentheses must be open in direction to center
e.g.: (1+2+3), 1+(2+3), 1+(2+3)+4
but not: 1)+(2+3)+(4
so: first parenthesis on the left side must be open and
the last parenthesis on the right side must be close
2) parentheses in direct neighborhood to a number must be
open in direction to the number (multiplication is
not mutual)
e.g.: 1+(2+3)+4, but not: 1+2(+3+4)
3) parentheses in direct neighborhood to an operator must be
closed in direction to the operator (multiplication is
not mutual)
e.g.: 1+(2+3)+4, but not: 1+2(+3+)4
*/
// build combinations separately not in-line
// it's already a mess, no need to add more
makeParens(0,nums.length,0,0,[],parens);
// You may take a look at the raw material here
// console.log(parens.join("\n"));
for(var i= 0;i<parens.length;i++){
var term = [];
// work on copies to reduce pointer juggling
var _ops = ops.slice(0);
var _nums = nums.slice(0);
for(var j=0;j<parens[i].length;j++){
if(parens[i][j] === '('){
term.push("(");
// rule 3
if(parens[i][j+1] === ')'){
term.push(_nums.shift());
}
// rules 1,2
else {
term.push(_nums.shift());
term.push(_ops.shift());
}
}
if(parens[i][j] === ')'){
term.push(")");
// rules 2,3
if(parens[i][j+1] !== ')')
term.push(_ops.shift());
}
}
// some pretty printing
term = term.join("");
// eval() because I didn't want to write a parser
// but if you need one...
all.push(term + " = " + eval(term));
}
return all;
}
I'm not sure if I would get hired with that abomination. Ah, to be honest: I doubt it.
But I hope it is at least a little bit helpful.
Yikes. That was tricky. Good challenge. I'm sure this could be cut way down, but it works. I used lodash and broke the various functions down to make it more flexible. Here's a jsfiddle:
https://jsfiddle.net/mckinleymedia/3e8g22Lk/8/
Oops - had to add parseInt to the addition so it doesn't add as strings.
/*
//input:
diffWaysToCompute("2 * 3 - 4 * 5");
//output:
(2*(3-(4*5))) = -34 - 2,1,0
((2*3)-(4*5)) = -14 - 0,2,1 & 2,0,1
((2*(3-4))*5) = -10 - 1,0,2
(2*((3-4)*5)) = -10 - 1,2,0
(((2*3)-4)*5) = 10 - 0,1,2
*/
'use strict'
var diffWaysToCompute = function(str) {
var opsAvailable = ['+','-','/','*'],
numbers = [],
operators = [],
getNumbersAndOperators = function(str) {
var arr = str.split(" ");
for (var i in arr) {
if ( opsAvailable.indexOf( arr[i] ) > -1 ) {
operators.push( arr[i] );
} else {
numbers.push( arr[i] );
}
};
return;
},
permutator = function(range) {
var results = [];
function permute(arr, memo) {
var cur,
memo = memo || [];
for (var i in arr) {
cur = arr.splice(i, 1);
if (arr.length === 0) results.push(memo.concat(cur));
permute(arr.slice(), memo.concat(cur));
arr.splice(i, 0, cur[0]);
}
return results;
}
return permute(_.range(range));
},
equations = function( perms ) {
var results = [];
_.each(perms, function( perm, k ) {
results[k] = nest ( perm );
});
return results;
},
nest = function( perm ) {
var eqs = eqs || [],
ref = ref || _.range(perm.length).map(function () { return undefined }),
eq,
target = undefined;
for (var i in perm) {
var cur = perm[i],
next = perm[i] + 1,
n1 = numbers[ cur ],
n2 = numbers[ next ],
r1 = ref[ cur ],
r2 = ref[ next ];
if ( r1 !== undefined) n1 = eqs [ r1 ];
if ( r2 !== undefined) n2 = eqs [ r2 ];
var rNew;
rNew = eqs.length;
for (var x in ref ) {
if ( ( ref[ x ] !== undefined ) && ( ref[ x ] == r1 || ref[ x ] == r2 ) ) ref[ x ] = eqs.length;
};
ref[ cur ] = ref[ next ] = eqs.length;
eqs.push({
ops: operators[ cur ],
nums: [ n1, n2 ]
});
};
return eqs[ eqs.length - 1 ];
},
calculations = function ( eqs ) {
var results = []
_.each(eqs, function(equation) {
results.push(calculate( equation ));
});
return results;
},
calculate = function( eq ) {
var result = {
text: ""
};
// result.eq = eq;
result.text += "( ";
result.total = eq.nums[ 0 ];
if ( _.isObject(result.total) ) {
var result1 = calculate( result.total );
result.total = result1.total;
result.text += result1.text;
} else {
result.text += eq.nums[ 0 ];
}
_.each(eq.ops, function (op, k) {
var num = eq.nums[ k + 1 ];
result.text += " " + op + " ";
if ( _.isObject(num) ) {
var result2 = calculate( num );
num = result2.total;
result.text += result2.text;
} else {
result.text += num;
}
if ( op === '+') result.total = parseInt(result.total) + parseInt(num);
if ( op === '-') result.total = result.total - num;
if ( op === '/') result.total = result.total / num;
if ( op === '*') result.total = result.total * num;
});
result.text += " )";
return result;
},
display = function( as ) {
var target = document.getElementById('result');
target.innerHTML += '<h3 class="problem">String given: ' + str + '</h3>';
target.innerHTML += '<h4>Permutations</h4>';
_.each( as, function(a) {
target.innerHTML += '<div class="permutation">';
target.innerHTML += ' <span class="formula">' + a.text + '</span> = ';
target.innerHTML += ' <span class="total">' + a.total + '</span>';
target.innerHTML += '</div>';
});
},
perms,
eqs,
answers;
getNumbersAndOperators(str);
perms = permutator( operators.length );
eqs = equations( perms );
answers = calculations( eqs );
answers = _.uniq(answers, 'text');
display(answers);
return answers;
};
console.log(diffWaysToCompute("2 * 3 - 4 * 5"));
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 ())(()
I tried to rewrite this indexOf MDN example to practice recursion
var str = 'To be, or not to be, that is the question.';
var count = 0;
var pos = str.indexOf('e');
while (pos !== -1) {
count++;
pos = str.indexOf('e', pos + 1);
}
console.log(count); // displays 4
This was my solution:
var count = 0;
function countLetters(str, p) {
var pos = str.indexOf(p);
if (pos == -1) {
return count;
}
else {
count ++;
return countLetters(str.substr(pos + 1), p)
}
}
console.log(countLetters('To be, or not to be, that is the question.', 'e'));
It works, but is there anyway to get the count variable inside the function itself? Is it not really a true recursion if I have a count variable outside the function?
In a recursive function, if you want to keep a variable around from one "iteration" to the next, then you need to pass it as an argument:
function countLetters(str, p, count) {
count = count || 0;
var pos = str.indexOf(p);
if (pos == -1) {
return count;
}
else {
return countLetters(str.substr(pos + 1), p, count + 1);
}
}
console.log(countLetters('To be, or not to be, that is the question.', 'e'));
// => 4
However, this is not always necessary, as Arun P Johny's answer illustrates.
What you can do is to return the count value form the method, so if the item is not found you return 0, else you return 1 + value-of-recursive-call
function countLetters(str, p) {
var pos = str.indexOf(p);
if (pos == -1) {
return 0;
} else {
return 1 + countLetters(str.substr(pos + 1), p)
}
}
console.log(countLetters('To be, or not to be, that is the question.', 'e'));
Demo: Fiddle
How do I increment a string "A" to get "B" in Javascript?
function incrementChar(c)
{
}
You could try
var yourChar = 'A'
var newChar = String.fromCharCode(yourChar.charCodeAt(0) + 1) // 'B'
So, in a function:
function incrementChar(c) {
return String.fromCharCode(c.charCodeAt(0) + 1)
}
Note that this goes in ASCII order, for example 'Z' -> '['. If you want Z to go back to A, try something slightly more complicated:
var alphabet = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'.split('')
function incrementChar(c) {
var index = alphabet.indexOf(c)
if (index == -1) return -1 // or whatever error value you want
return alphabet[index + 1 % alphabet.length]
}
var incrementString = function(string, count){
var newString = [];
for(var i = 0; i < string.length; i++){
newString[i] = String.fromCharCode(string[i].charCodeAt() + count);
}
newString = newString.join('');
console.log(newString);
return newString;
}
this function also can help you if you have a loop to go through