The Coderbyte problem is:
Using the JavaScript language, have the function DashInsert(str) insert dashes ('-') between each two odd numbers in str. For example: if str is 454793 the output should be 4547-9-3. Don't count zero as an odd number.
So when the input is 99946, the output should be 9-9-946.
I had this solution, which wouldn't quite work:
function DashInsert(num) {
var arr = num.toString().split('');
var i = 0;
while(i < arr.length-1){
if( arr[i]%2 !==0 && arr[i+1]%2 !==0){
arr.splice(i+1,0,'-');
}
i++
}
return arr.join('');
}
Then I found this similar answer:
function DashInsert(num) {
var arr = num.toString().split('');
var i = 0
while(i < arr.length-1){
if( arr[i]%2===1 && arr[i+1]%2===1){
arr.splice(i+1,0,'-');
}
i++
}
return arr.join(''); }
str = 99946;
alert(DashInsert(str));
Can anyone explain why it should be arr[i]%2===1?
both are correct .
for example , take 9 : 9%2 != 0 and also 9%2 ==1 . think about it , all odd numbers can be split to
2n+1 . a modulo of 2 will always return 1 , which is not 0.
For anyone else who stumbles upon this in a frustrated googling haze...
After the first hyphen is added, it changes the length of the array, so it gets evaluated in the loop to see if the hyphen element !== 0.
Since a '-' !== 0, another hyphen is added.
This is also why you keep blowing your stack since the hyphen keeps changing the length of your array (side note, always cache your length in a variable outside of your for loop and use that in your loop instead).
To fix it, you could add a bunch more &&'s to your if statement i.e.
if(theArray[x] % 2 !== 0 && theArray[x+1]% 2 !== 0 && theArray[x] !== '-' && theArray[x+1] !== '-')
or you could just be more specific and only look for modulo results that evaluate to 1.
I tried this and it works..
HTH..
private static String createDashedString(String str) {
StringBuilder builder = new StringBuilder();
char[] chararray = str.toCharArray();
for (int i=0;i<chararray.length-1;i++) {
int firstInt = Character.getNumericValue(chararray[i]);
int nextInt = Character.getNumericValue(chararray[i+1]);
if ((firstInt%2 !=0) && (nextInt%2 !=0)) {
builder.append(firstInt);
builder.append("-");
}else {
builder.append(firstInt);
}
}
builder.append(chararray[chararray.length-1]);
return builder.toString();
}
public static void main(String args[]) {
String str = "999999";
//01234
System.out.println(createDashedString(str));
}
function DashInsert(str) {
let bil = 0;
while (bil < str.length-1) {
if (Number(str[bil]) % 2 === 1 && Number(str[bil+1]) % 2 === 1) {
str = str.slice(0,bil+1) + "-" + str.slice(bil+1);
bil = bil + 2;
}
else {
bil++;
}
}
return str;
}
console.log(DashInsert("454793"));
let x = '99946'
let z=[]
for(var i = 0;i<x.length;i++){
if(x[i]%2 == 0 && x[i+1]%2 == 1 ){
z+=x[ i ]
}else if (x[i]%2 == 1 && x[i+1]%2 == 0){
z+=x[i]
}else if (x[i]%2 == 1 && x[i+1]%2 == 1){
z+=x[i]+"-"
}else{
z+=x[i]
}
}
console.log(z)
Related
I am trying to insert dashes ('-') between each two odd numbers and insert asterisks ('*') between each two even numbers, but I am only getting the last result.
I want to print out all the elements in the array.
For example: if num is 4546793 the output should be 454*67-9-3. I Did not count zero as an odd or even number.
function StringChallenge(num) {
let result = "";
for (let i = 0; i < num.length; i++) {
if (num[i] === 0) {
continue;
}
if (num[i - 1] % 2 == 0 && num[i] % 2 == 0) {
result = num[i - 1] + "*" + num[i];
continue;
}
if (num[i - 1] % 2 == !0 && num[i] % 2 == !0) {
result = num[i - 1] + "-" + num[i];
continue;
}
}
return result;
}
console.log(StringChallenge([4,5,4,6,7,9,3]));
You do not need to check as if&continue. Inserting given numbers to the result string and only adding "-" when index and previous are odd, and "*" when index and previous are even.
function StringChallenge(num) {
let result = "";
for (let i = 0; i < num.length; i++) {
if (num[i]%2 ===0) {// even
if(i !== 0 && num[i-1]%2===0){// previous is even either
result+="*"+num[i];
}else{
result+=num[i];
}
}else{// odd
if(i !== 0 && num[i-1]%2===1){// previous is odd either
result+="-"+num[i];
}else{
result+=num[i];
}
}
}
return result;
}
console.log(StringChallenge([4,5,4,6,7,9,3]));
Try this :)
function test(a){
let result=""
for(let i=0; i < a.length; i++){
if(a[i] != 0 && a[i-1] % 2 == 0 && a[i] % 2 == 0){
result = result + '*' + a[i]
}
else if (a[i] != 0 && a[i-1] % 2 != 0 && a[i] % 2 != 0){
result = result + '-' + a[i]
}
else{
result = result + a[i]
}
}
return result
}
console.log(test([4,5,4,6,7,9,3]));
As everyone has identified, the problem is you are not adding to result.
But here is a suggestion to make your code easier to read
// These one line functions make your code easier to read
function IsEven(num){
return num % 2 === 0;
}
function IsOdd(num){
return num % 2 !== 0;
}
function StringChallenge(numArray) {
// return empty string if not an array or empty array
if(!Array.isArray(numArray) || numArray.length === 0) return "";
let result = "" + numArray[0]; // use "" to coerce first element of numArray from number to string
for (let i = 1; i < numArray.length; i++) {
// focus on the conditions to determine the separator you want between each element
separator = "";
if (numArray[i] !== 0) {
if (IsEven(numArray[i]) && IsEven(numArray[i - 1])) {
separator = "*";
} else if (IsOdd(numArray[i]) && IsOdd(numArray[i - 1])){
separator = "-";
}
}
// build the result
result += separator + numArray[i];
}
return result;
}
I will do that this way :
== some advices for 2 cents ==
1 - try to make your code as readable as possible.
2 - use boolean tests rather than calculations to simply do a parity test
3 - ES7 has greatly improved the writing of JS code, so take advantage of it
console.log(StringChallenge([4,5,4,6,7,9,3])); // 454*67-9-3
function StringChallenge( Nums = [] )
{
const
isOdd = x => !!(x & 1) // Boolean test on binary value
, isEven = x => !(x & 1) && x!==0 // zero is not accepted as Even value
;
let result = `${Nums[0]??''}`; // get first number as
// result if Nums.length > 0
for (let i=1; i<Nums.length; i++)
{
if ( isOdd(Nums[i-1]) && isOdd(Nums[i]) ) result += '-';
if ( isEven(Nums[i-1]) && isEven(Nums[i]) ) result += '*';
result += `${Nums[i]}`; // same as Nums[i].toString(10);
}
return result
}
I hope this helps. I tried to keep it as simple as possible.
function StringChallenge(num) {
//start with a string to concatenate, or else interpreter tries to do math
operations
let result = num[0].toString();
function checkOdd(num){ //helper function to check if odd
return num % 2
}
for (let i = 0; i < num.length - 1; i++) {
if (checkOdd(num[i]) && checkOdd(num[i+1])) { //checks if both odd
result += `-${num[i+1]}`; //adds - and next number
} else if (!checkOdd(num[i]) && !checkOdd(num[i+1])) { //checks if both even
result += `*${num[i+1]}`; //adds * and next number
} else { //otherwise
result += num[i+1]; //just add next number
}
}
return result;
}
console.log(StringChallenge([4,5,4,6,7,9,3]));
Use +=. And, change your logic, your code prints out "4*67-99-3".
The zero check was pretty hard for me I hope the variables in my code explain itself. If not, let me know.
function even(num) {
return num % 2 === 0;
}
function odd(num) {
return num % 2 !== 0;
}
function StringChallenge(num) {
let result = "";
for (let i = 0; i < num.length; i++) {
var currentZero = num[i] === 0
var previousZero = num[i-1] === 0
var bothEven = even(num[i]) && even(num[i-1])
var bothOdd = odd(num[i]) && odd(num[i-1])
var firstNumber = (i === 0)
if (!currentZero) {
if (firstNumber) {
result += num[i]
} else {
if (bothEven && !previousZero) {
result += "*" + num[i]
} else if (bothOdd && !currentZero) {
result += "-" + num[i]
} else {
result += num[i]
}
}
}
}
return result;
}
console.log(StringChallenge([0,4,5,0,4,6,7,9,3]));
I'm currently solving a problem at CoderByte. Here's the scenario.
Have the function SerialNumber(str) take the str parameter being passed and determine if it is a valid serial number with the following constraints:
It needs to contain three sets each with three digits (1 through 9) separated by a period.
The first set of digits must add up to an even number.
The second set of digits must add up to an odd number.
The last digit in each set must be larger than the two previous digits in the same set.
If all the above constraints are met within the string, the your program should return the string true, otherwise your program should return the string false. For example: if str is "224.315.218" then your program should return "true".
Examples
Input: "11.124.667"
Output: false
Input: "114.568.112"
Output: true
Here's my JS code for this problem.
function SerialNumber(str) {
// code goes here
if(str.length < 11) {
return "false";
}
for (let x = 0; x < str.length; x++) {
if(str[x] === '0') {
return "false";
}
}
const first = str[0] + str[1] + str[2];
const second = str[4] + str[5]+ str[6];
if (first % 2 !== 0 || second % 2 === 0) {
return "false";
}
if (str[2] < str[1] || str[2] < str[0] || str[6] < str[5] || str[6] < str[4] || str[10] < str[8] || str[10] < str[9]) {
return "false";
} else {
return "true";
}
}
console.log("11.124.667 : " + SerialNumber("11.124.667"));
console.log("114.568.112 : " + SerialNumber("114.568.112"));
When I input "11.124.667", it will return to false (which is correct). And when I input "114.568.112", it will also return to false (which is not correct). Any tips on how can I obtain this? Thanks for the feedback.
I tried your code and the problem comes from the third return false for the 2nd example, when you set second, you concatenate strings instead of adding numbers, thus %2 is always equal to 0 since
I added console.log before each return false to see which condition failed.
Here's the fixed code in which I added parseInt for first and second
function SerialNumber(str) {
if (str.length < 11) {
console.log("first")
return "false";
}
for (let x = 0; x < str.length; x++) {
if (str[x] === '0') {
console.log("second")
return "false";
}
}
const first = parseInt(str[0]) + parseInt(str[1]) + parseInt(str[2]);
const second = parseInt(str[4]) + parseInt(str[5]) + parseInt(str[6]);
if (first % 2 !== 0 || second % 2 === 0) {
console.log("third")
return "false";
}
if (str[2] < str[1] || str[2] < str[0] || str[6] < str[5] || str[6] < str[4] || str[10] < str[8] || str[10] < str[9]) {
console.log("fourth")
return "false";
} else {
return "true";
}
}
console.log("11.124.667 : " + SerialNumber("11.124.667"));
console.log("114.568.112 : " + SerialNumber("114.568.112"))
if
(previousNum == "." && currentNum == ".") {
screen.value = screen.value.substring(0, numchar - 1);
You are using javascript. you can split it on "." then count the length of the array which should be 2 or less.
var x = '12.34.56';
var arr = x.split(".")
var number_of_dots = arr.length - 1
console.log(number_of_dots)
I think you can check if the number is Integer or float with this sample function
function isFloat(n){
return Number(n) === n && n % 1 !== 0;
}
function isInt(n) {
return n % 1 === 0;
}
and write catch, so if had error the entered number is not valid for you calculator
as answer to an exercise in which I had to create a function that given an array of numbers return the number with most occurrences, and if more than one number had the max number of occurrences return the minor one. This is the implementation I made, but I'm pulling my hair figuring out why it return 10 instead of 9 in the example.
It appears to be evaluating 10 < 9 as true. What's wrong?
function maxOccurencies(arr) {
var aux = [], max = 0, final = null;
for (var i=0,t=arr.length; i<t; i++) {
aux[arr[i]] = (aux[arr[i]] || 0) + 1;
if (aux[arr[i]] > max) max = aux[arr[i]];
}
for (x in aux) {
if ( aux[x] == max && (x < final || final == null)) {
final = x;
}
}
return final;
}
document.write(maxOccurencies([10,10,10,9,9,9,8,7,4,5,1]));
Putting typeof(x) in your second loop reveals that some of your variables are being cast as type string! Still looking into exactly where this is occurring. You can replace
if ( aux[x] == max && (x < final || final == null)) {
with
if ( aux[x] == max && (parseInt(x) < parseInt(final) || final == null)) {
to return the correct value of 9.
Edit:
Very interesting, I was unaware of Javascript's exact handling of arrays in for...in loops. See the following other questions for more information:
JavaScript For-each/For-in loop changing element types
Why is using “for…in” with array iteration such a bad idea?
Also note that you can use arr.forEach(function(element){...}); and the elements are returned with their types intact.
I think the problem is just that the x in aux is not a number so the if statement isn't evaluating correctly. when converted to a number then it returns 9 (below).
(3 == 3 && ("10" < "9" || "9" == null)) evaluates to true
function maxOccurencies(arr) {
var aux = [], max = 0, final = null;
for (var i=0,t=arr.length; i<t; i++) {
aux[arr[i]] = (aux[arr[i]] || 0) + 1;
if (aux[arr[i]] > max) max = aux[arr[i]];
}
for (x in aux) {
if ( aux[x] == max && (parseInt(x) < final || final == null)) {
final = parseInt(x);
}
}
return final;
}
document.write(maxOccurencies([10,10,10,9,9,9,8,7,4,5,1]));
"I'm pulling my hair figuring out why it return 10 instead of 9 in the example."
That's because in this sort of comparison, 10 is smaller than 9,8,7,6,5,4,3, 2 but a bit grater than 1.
:)
This small type correction will fix it:
function maxOccurences(arr) {
aux = [], max = 0, final = null;
for (var i=0,t=arr.length; i<t; i++) {
aux[arr[i]] = (aux[arr[i]] || 0) + 1;
if (aux[arr[i]] > max) max = aux[arr[i]];
}
for (x in aux) {
if ( aux[x] == max && (+x < final || final == null)) {
final = x;
}
}
return final;
}
In javascript whats the best way to check if a character (length 1), is a number (i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9) or a letter (i.e. A to Z, a to z)?
Thanks
Why not:
function isNumber(i) {
return (i >= '0' && i <= '9');
}
function isLetter(i) {
return ((i >= 'a' && i <= 'z') || (i >= 'A' && i <= 'Z'));
}
I wrote a little test case for you, at least for the numeric checking function.
Considering the fact that all functions returns true with either a numberic 1 or a string '1' literal, using an Array seems to be the fastest way (at least in Chrome).
var isNumericChar = (function () {
var arr = Array.apply(null, Array(10)).map(function () { return true; });
return function (char) { return !!arr[char]; };
})();
However, if you accept that it might return false for 1, the switch statement is then significantly faster.
Try this.
function validate_string() {
var str = "a"; //change to desired value;
var regX = new RegExp("([0-9A-Za-z])");
var ans = false;
if(str.length == 1) {
ans = regX.test(str);
}
return ans;
}
Edit: Refactored my answer.
function validateString(char) {
let regx = new RegExp(/^[0-9A-Za-z]{1}$/g);
return regx.test(char);
}
validateString('4'); // true
validateString('as'); // false
validateString(''); // false
validateString(); // false
Maybe try something like this
var sum = 0; //some value
let num = parseInt(val); //or just Number.parseInt
if(!isNaN(num)) {
sum += num;
}
This blogpost sheds some more light on this check if a string is numeric in Javascript | Typescript & ES6
You can check for the type of the variable
function checkType(input){
console.log(typeof input)
}
checkType(1234); //number
checkType('Hello') //string
Here is an updated version
function checkType(i){
var input = i.toString(); //convert everything to strings to run .lenght() on it
for(var i=0; i<input.length; ++i){
if(input[i] >= '0' && input[i] <= '9'){
console.log(input[i]+' is a number');
}else if((input[i] >= 'a' && input[i] <= 'z') || (input[i] >= 'A' && input[i] <= 'Z')){
console.log(input[i]+' is a letter');
}
}
}
checkType('aa9fgg5')