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);
Related
function expandedForm(num) {
let len = num.toString().length;
let n = num.toString().split("");
let result = "";
for (let i = 0; i < len; i++) {
result += n[i] + "0".repeat(len -1 -i).join(" + ");
}
return result;
}
What I am trying to do is to separate numbers like this:
1220 = "1000 + 200 + 20"
221 = "200 + 20 + 1"
I have written the code (not the perfect one) where it gets me all the necessary values but I struggle with joining them together with "+". I tried using .join() but it did not work.
.join works on arrays only
function expandedForm(num) {
let len = num.toString().length;
let n = num.toString().split("");
let result = "";
let arr=[];
for (let i = 0; i < len; i++) {
arr[i] = n[i] + '0'.repeat(len-1-i);
console.log(arr[i]);
}
let ans=arr.join('+');
return ans;
}
console.log(expandedForm(1220))
Although there are a variety of approaches, here are some general tips for you:
Probably don't want to output a 0 term unless the input number is exactly 0 (only a leading 0 term is relevant, because it will be the only such term)
str.split('') can also be [...str]
No need to split a string into an array to access a character str.split('')[0] can also be just str[0]
Might want to assert that num is a whole number.
Make sure you provide enough test cases in your question to fully define the behaviour of your function. (How to handle trailing zeros, interstitial zeros, leading zeros, etc. Whether the input can be a string.)
function expandedForm(num) {
const s = num.toString();
const n = s.length - 1;
const result = [...s]
.map((char, index) => char + '0'.repeat(n - index))
.filter((str, index) => !index || +str)
.join(' + ');
return result;
}
console.log(expandedForm(1220));
console.log(expandedForm(221));
console.log(expandedForm(10203));
console.log(expandedForm(0));
console.log(expandedForm(2n**64n));
Join works with an array, not string. It stringifies two subsequent indexes for all indexes and you can decide what to add between them.
function expandedForm(num) { // num = 321
let len = num.toString().length; // len = 3
let n = num.toString().split(""); // [3,2,1]
let result = [];
for (let i = 0; i < len; i++) {
result.push(n[i] + "0".repeat(len -1 -i)); // pushing till result = ['300','20','10']
}
return num + ' = ' + result.join(' + ');
// connection result[0] + ' + ' result[1] + ' + ' result[2]
}
expandedForm(321); // output: "321 = 300 + 20 + 1"
Here's one way of doing it
let num = 221;
function expandedForm(num) {
let len = num.toString().length;
let n = num.toString().split("");
let result = "";
for (let i = 0; i < len; i++) {
let t = "0"
t = t.repeat(len-1-i)
if(result.length > 0){
n[i] !== '0'? result += '+'+ n[i] + t : result
} else {
n[i] !== '0'? result += n[i] + t : result
}
}
return result;
}
console.log(expandedForm(2200))
console.log(expandedForm(num))
below would be my approach in a more mathimatical but clean code that you can adjust to your needs.
let result = parseInt(num / 1000);
return result ;
}
function x100( num ) {
num = num % 1000;
let result = parseInt( num / 100);
return result;
}
function x10(num ) {
num = num % 1000;
num = num % 100;
let result = parseInt(num /10);
return result;
}
function x1( num ) {
num = num % 1000;
num = num % 100;
num = num % 10;
return num
}
num = 12150
console.log(num = `1000 x ${x1000(num)}, 100 x ${x100(num)}, 10 x ${x10(num)}`)```
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"));
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>";
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
It's been a while since I wrote any Javascript. Is there a more elegant way to do this. Specifically want to get rid of the second loop:
<script>
var number = 0;
for (var i=1; i<11; i++) {
for (var x=1; x<11; x++) {
if (i==1) {
number = x;
} else {
number = Math.pow(i, x);
}
document.write(number + " ");
if (x == 10) {
document.write("<br>");
}
}
}
</script>
I would stick with 2 loops but i would change one if statement and move it after the 2nd loop and avoid document.write and insert it all at once to reduce the number of time you change the DOM
let result = ''
for (let i = 1; i < 11; i++) {
for (let x = 1; x < 11; x++)
result += (i==1 ? x : Math.pow(i, x)) + ' '
result += '<br>'
}
document.body.insertAdjacentHTML('beforeend', result)
Edit If you really don't want the 2nd loop:
let result = ''
// you must swap the condition to check for x instead of i
for (let i = 1, x = 1; x < 11; i++) {
result += (x==1 ? i : Math.pow(x, i)) + ' '
// and reset i and increase x yourself
if (i == 10) {
i = 0
x++
result += '<br>'
}
}
document.body.insertAdjacentHTML('beforeend', result)
Edit2 just for the fun: No for loops.
Just a recursive function :P
function build(i = 1, x = 1, res = '') {
res += (x == 1 ? i : Math.pow(x, i)) + ' '
i == 10 ? (x++, i=1, res += '<br>') : i++
return x == 11 ? res : build(i, x, res)
}
document.body.insertAdjacentHTML('beforeend', build())
In terms of 'elegancy', I'd go for for... in loops or map function. That doesn't solve your nested loop though.
On a side note, nested loops are not necessarily bad. If that's the correct way to implement the specific algorithm, then that's how it is.
Using Math.pow() is un-necessary overhead. Nested loops are not necessarily bad.
var number = 0;
for (var i=1; i<11; i++) {
document.write(i + " ");
number = i;
for (var x=2; x<11; x++) {
number = (i == 1) ? x : number * i;
document.write(number + " ");
}
document.write("<br>");
}
Another way of doing it with 1 loop only, tho not as clean:
var number = 0;
var x = 1;
var calc = 0;
var calcx = 1;
var increment = false;
for (var i=1; i<101; i++) {
increment = false;
calc = i % 10;
if(calc == 0){
calc = 10;
increment = true;
}
if (calcx==1) {
number = calc;
} else {
number = Math.pow(calcx, calc);
console.log(calcx+" "+calc);
}
document.write(number + " ");
if (i % 10 == 0) {
document.write("<br>");
}
if(increment){
calcx++;
}
}
Here's another way with only one loop:
[...Array(100)].map((_,i) => {
document.write(((i>9)?Math.pow(Math.floor((i+10)/10),(i%10)+1):i+1) + ' ' + ((i%10==9)?'<br>':''));
});