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;
}
Related
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"))
Given an array of integers of any length, return an array that has 1 added to the value represented by the array.
the array can't be empty
only non-negative, single digit integers are allowed
Return nil (or your language's equivalent) for invalid inputs.
Examples
For example the array [2, 3, 9] equals 239, adding one would return the array [2, 4, 0].
My code so far:
function upArray(arr){
let i = parseInt(arr.join('')) + 1;
return arr.some(e => typeof e !== 'number' || e < 0) ?
null : i
.toString()
.split('')
.map(e => parseInt(e));
};
It seems to pass most basic test however fails with larger inputs. Where have I gone wrong?
Just like you have converted the array into a number, you have to convert the number back into an array.
function upArray(arr){
let i = parseInt(arr.join('')) + 1;
return i.toString().split('').map(x => parseInt(x));
};
console.log(upArray([2,3,9]));
Your code won't work if the array length is greater than 100k...
Number type of javascript or any language is not enough big to handle it.
It's better if we calculate the last element with 1. If result is larger than nice ( < 10 ),
we continue to calculate next element with 1 and assign current value to 0. If result is smaller or equal 9, just assign the result to current and exit loop.
Then we print the final array as result:
pseudo code:
for i from: n-1:0
result = arr[i] + 1;
if(result < 10) :
arr[i] = result;
exit loop;// no need to continue calculate
else:
arr[i] = 0;
endif;
endfor;
You can join final array as string.
Here's probably the fastest solution (performance wise) - also there's no need to deal with BigInt, NaN, or Infinity:
function upArray(arr) {
if (!isInputIsNonEmptyArray(arr)) {
return null;
}
const isNumber = num => typeof num === 'number';
const isIntSingleDigit = num => Number.isInteger(num) && num >= 0 && num <10;
let resultArr = [];
let i = arr.length;
let num;
while (i-- > 0) {
num = arr[i];
if (!isNumber(num) || !isIntSingleDigit(num)) {
return null;
}
if (num === 9) {
resultArr[i] = 0;
if (i === 0) { //means we're in the msb/left most digit, so we need to insert 1 to the left
resultArr.unshift(1);
break; //you can leave it out really, as the next check in the while will fail anyway
}
}
else {
resultArr[i] = num + 1; //No more + 1 should be made, just check for validity
//of the rest of the input and copy to the result arr
while (--i > -1) {
num = arr[i];
if (!isNumber(num) || !isIntSingleDigit(num)) {
return null;
}
resultArr[i] = arr[i];
}
break;
}
}
return resultArr;
function isInputIsNonEmptyArray(arr) {
return Array.isArray(arr) && arr.length > 0;
}
}
If the input arg is not an array or an empty array, or if you encounter invalid element during the main while loop you return null.
In the main while loop you go from the right most element (lsd), and add 1 to it (or insert 0 if the number is 9) up the the left most digit.
If a number which is less than 9 is incremented, no need to increment any more (this is the while loop in the else clause).
Background - I took an online coding test and was presented with questions similar to this, I did rather poorly on it compared to the hidden grading criteria and I was hoping to get another pair of eyes to look at it and maybe help point out some of my mistakes.
Practice Test questions -
Task: Given an integer inject the number 5 into it to make the largest possible integer
Conditions: (-80000...80000) range needed to handle
Expected input: int
Expected output: int
Testcase: -999 -> -5999
80 -> 850
var lrgInt = function(num) {
var stringInt = num.toString();
for (let i = 0; i < stringInt.length; i++) {
if (stringInt.charAt(i) === "-") {
return parseInt([stringInt.slice(0, 1), '5', stringInt.slice(1)].join(''));
}else if (stringInt.charAt(i) < 5) {
return parseInt([stringInt.slice(0, i), '5', stringInt.slice(i)].join(''));
}
}
return parseInt([stringInt.slice(0, stringInt.length), '5', stringInt.slice(stringInt.length)].join(''));
};
Task: Determine the number of operations done on a number following the conditions to reduce it to 0.
Conditions:
- If the number is odd, subtract 1
- If the number is even, divide by 2
Expected input: int
Expected output: int
var operations = 0;
var numberOfSteps = function(num) {
if (num === 0){
return operations;
}else if (num % 2 == 0) {
operations++;
return numberOfSteps(num/2);
} else {
operations++;
return numberOfSteps(num-1);
}
};
For the second question, you could add one plus the result of recursion with the adjusted number without having a global counter.
function numberOfSteps(number) {
if (!number) return 0;
if (number % 2) return 1 + numberOfSteps(number - 1);
return 1 + numberOfSteps(number / 2);
}
console.log(numberOfSteps(5)); // 5 4 2 1 0
For the first question, we make the observation that if the number is positive, we want to inject the 5 before the first digit less than 5, but if it's negative then we want to inject it before the first digit greater than 5. For the second problem, we can just use a simple while loop.
function largestNum(num) {
if (num == 0) {
// this edge case is weird but I'm assuming this is what they want
return 50;
}
var negative = num < 0;
var numAsStr = Math.abs(num).toString();
var inj = -1;
for (var i = 0; i < numAsStr.length; i++) {
var cur = parseInt(numAsStr[i], 10);
if ((!negative && cur < 5) || (negative && cur > 5)) {
// we found a place to inject, break
inj = i;
break;
}
}
if (inj == -1) {
// didn't inject anywhere so inject at the end
inj = numAsStr.length;
}
return (
(negative ? -1 : 1) *
parseInt(numAsStr.substr(0, inj) + "5" + numAsStr.substr(inj))
);
}
function numSteps(num) {
var steps = 0;
while (num != 0) {
if (num % 2) {
// it's odd
num--;
} else {
num /= 2;
}
steps++;
}
return steps;
}
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)
I'm checking for integer values in node.js using IsNaN function.
Unexpectedly, this function validates the strings like 1E267146, 1E656716 , 914E6583 to be numbers, as these strings are exponential values. Any way to work around this? In actual scenario i wont get any exponential values.
ECMA6 defines Number.isInteger as follows:
Javascript
function isInteger(nVal) {
return typeof nVal === "number" && isFinite(nVal) && nVal > -9007199254740992 && nVal < 9007199254740992 && Math.floor(nVal) === nVal;
}
but this will also accept scientific notation
console.log(isInteger(1e6));
console.log(isInteger(+"1e6"));
jsfiddle
You need to be clear as to what your definitions/expectations are.
My guess is that you may want something like this, if you are testing strings and have no limits on the max or min integer.
Javascript
function isStringNumericalInteger(testValue) {
return typeof testValue === "string" && /^[\-+]?[1-9]{1}\d+$|^[\-+]?0$/.test(testValue);
}
console.log(isStringNumericalInteger("9007199254740991"));
console.log(isStringNumericalInteger("-123216848516878975616587987846516879844651654847"));
console.log(isStringNumericalInteger("1.1"));
console.log(isStringNumericalInteger("-1.1"));
console.log(isStringNumericalInteger("1e10"));
console.log(isStringNumericalInteger("010"));
console.log(isStringNumericalInteger("0x9"));
console.log(isStringNumericalInteger(""));
console.log(isStringNumericalInteger(" "));
console.log(isStringNumericalInteger());
console.log(isStringNumericalInteger(null));
console.log(isStringNumericalInteger([]));
console.log(isStringNumericalInteger({}));
Output
true
true
false
false
false
false
false
false
false
false
false
false
false
jsfiddle
If you want to bound the range to what javascript can represent numerically as an integer then you will need to add a test for && +testValue > -9007199254740992 && +testValue < 9007199254740992
If you don't like using RegExs, you can also accomplish this with a parser. Something like this:
Javascript
function isCharacterDigit(testCharacter) {
var charCode = testCharacter.charCodeAt(0);
return charCode >= 48 && testCharacter <= 57;
}
function isStringNumericalInteger(testValue) {
var start = 0,
character,
index,
length;
if (typeof testValue !== "string") {
return false;
}
character = testValue.charAt(start);
if (character === "+" || character === "-") {
start += 1;
character = testValue.charAt(start);
}
start += 1;
length = testValue.length;
if ((length > start && character === "0") || !isCharacterDigit(character)) {
return false;
}
for (index = start; index < length; index += 1) {
if (!isCharacterDigit(testValue.charAt(index))) {
return false;
}
}
return true;
}
jsfiddle
I would use something like below code to validate number input. First I parse the given value to float and then check isNaN().
var isNumber = function (obj) {
return !isNaN(parseFloat(obj)) && isFinite(obj);
};
I think this is what you need in your case (i hate regex because this is not very good for the performance but..)
http://jsbin.com/EQiBada/1/
var NMAX = Math.pow(2, 53);
function isNumeric(n) {
n = n < 0 ? n * -1 : n;
var r = /^\d+$/.test(n);
if (r === true)
{
return parseInt(n, 10) >= (NMAX * -1) + 1 && parseInt(n, 10) <= NMAX;
}
return false;
}
Minified
var NMAX = Math.pow(2, 53);
function isNumericMin(n) {
n = n < 0 ? n * -1 : n;
return /^\d+$/.test(n) === true ? parseInt(n, 10) >= (NMAX * -1) + 1 && parseInt(n, 10) <= NMAX : false;
}
var i = '1E267146'
if(isNaN(i) || !isFinite(i) !! i=="")
{
// do stuff
}
else
{
// do stuff
}