I was doing this kata on codewars. The question wants the function to return true if the first argument (string) passed in ends with the 2nd argument (also a string). So I wrote my function and everything worked just fine until it compares ':-)' with ':-(' and returns true.
What is wrong? I'm so confident that my code should work that I don't even know what to search for.
function solution(str, ending){
if (!ending) return true; // if ending is a empty string return true (the question wants that)
let ok;
const strArr = str.split(''), endingArr = ending.split('');
for (let i = 0; i < endingArr.length; i++) strArr.reverse()[i] === endingArr.reverse()[i] ? ok = true : ok = false;
return ok;
}
console.log(solution(":-)",":-("));
Your problem is a misunderstanding of what reverse() does. It does not return a reversed copy of the old array, it reverses the existing array and returns that same array. As a result, you keep reversing the arrays back and forth every iteration of the loop, causing some elements to be skipped and some to be checked twice.
Array.prototype.reverse() on MDN
Edit:
As pointed out by others in the comments, both to the question and this answer, there are in fact multiple problems.
reverse() aside, the loop always sets ok to the result of the last comparison, making the function ignore all previous results.
The easier way to implement this is to remove ok altogether. Instead, return false as soon as a mismatch is detected. If the function runs long enough to exit the loop, it means no mismatch was detected and true can be returned.
Edit 2:
Just as a friendly suggestion:
While both reverse() and ok are real issues with the code, I only noticed the first one the first time around due to the formatting of the code. The ok problem was off-screen due to the line being too long. As such, once I spotted the reverse() issue, I assumed that was it and didn't bother scrolling sideways to see the rest of the code.
I am not going to demand that you write your own code in a certain way, but if you format it properly, it allows others to read it more easily. In essence, you help us to more easily help you.
For instance, this line:
for (let i = 0; i < endingArr.length; i++) strArr.reverse()[i] === endingArr.reverse()[i] ? ok = true : ok = false;
...would have been significantly easier to read as...
for (let i = 0; i < endingArr.length; i++) {
if(strArr.reverse()[i] === endingArr.reverse()[i])
ok = true;
else
ok = false;
}
...or some variation thereof. Here, the problem is significantly more visible and obvious.
The other answer explains many of the mistakes you've made. I wanted to point out just how much you've over-thought your solution.
function solution(str, ending){
if (ending === "") return true; // if ending is a empty string return true (the question wants that)
return str.endsWith(ending);
}
console.log(solution(":-)",":-("));
console.log(solution("foo",""));
console.log(solution("foo","bar"));
console.log(solution("foobar","bar"));
Even my solution above is overengineered, str.endsWith("") always returns true. So this can be simplified further.
function solution(str, ending){
return str.endsWith(ending);
}
console.log(solution(":-)",":-("));
console.log(solution("foo",""));
console.log(solution("foo","bar"));
console.log(solution("foobar","bar"));
Related
I'm creating a few specific functions for a compiler I'm working on, But certain restrictions within the compiler's nature will prevent me from using native JavaScript methods like Array.prototype.pop() to perform array pops...
So I decided to try and write some rudimentary pseudo-code to try and mimic the process, and then base my final function off the pseudo-code... But my tests seem to fail... based on the compiler's current behavior, it will only allow me to use array.length, array element assignments and that's about it... My code is below...
pop2 = function(arr) {
if(arr.length>0){
for(var w=undefined,x=[],y=0,z=arr.length;y<=z;y++){
y+1<z?(x[y]=arr[y]):(w=arr[y],arr=x);
}
}
return w;
}
Arr = [-1,0,1,2];
// Testing...
console.log(pop2(Arr)); // undefined... should be 2
console.log(Arr); // [-1,0,1,2]... should be [-1,0,1]
I'm trying to mimic the nature of the pop function but can't seem to put my finger on what's causing the function to still provide undefined and the original array... undefined should only return if an initial empty array is sent, just like you would expect with a [].pop() call...
Anyone have any clues as to how I can tailor this code to mimic the pop correctly?
And while I have heard that arr.splice(array.length-1,1)[0]; may work... the compiler is currently not capable of determining splice or similar methods... Is it possible to do it using a variation of my code?
Thanks in advance...
You're really over-thinking [].pop(). As defined in the specs, the process for [].pop() is:
Get the length of the array
If the length is 0
return undefined
If length is more than 0
Get the item at length - 1
Reduce array.length by 1
Return item.
(... plus a few things that the JavaScript engine needs to do behind the scenes like call ToObject on the array or ensure the length is an unsigned 32-bit integer.)
This can be done with a function as simple as the one below, there's not even a need for a loop.
function pop(array) {
var length = array.length,
item;
if (length > 0) {
item = array[length - 1];
array.length -= 1;
}
return item;
}
Edit
I'm assuming that the issue with the compiler is that Array.prototype.pop isn't understood at all. Re-reading your post, it looks like arrays have a pop method, but the compiler can't work out whether the variable is an array or not. In that case, an even simpler version of this function would be this:
function pop(array) {
return Array.prototype.pop.call(array);
}
Try that first as it'll be slightly faster and more robust, if it works. It's also the pattern for any other array method that you may need to use.
With this modification, it works:
http://jsfiddle.net/vxxfxvpL/1/
pop2 = function(arr) {
if(arr.length>0){
for(var w=undefined,x=[],y=0,z=arr.length;y<=z;y++){
if(y+1<z) {
(x[y]=arr[y]);
} else {
(w=arr[y],arr=x);
break;
}
}
}
return w;
}
Arr = [-1,0,1,2];
// Testing...
console.log(pop2(Arr)); // 2
The problem now is to remove the last element. You should construct the original array again without last element. You will have problems with this because you can't modify the original array. That's why this tasks are maded with prototype (Array.prototype.pop2 maybe can help you)
I have the below code causing Internet Explorer to freeze. It's a project that involves processing student grades as an assignment:
var array1 = StudentGradeAreadHugeList();
var nextArrayItem = function() {
var grade = array1.pop();
if (grade) {
nextArrayItem();
}
};
i hope you can help me with this.
You could show more info about the application you're trying to do. But I believe it's a matter of stack overflow (maybe you're using a big list). So, to overcome that you should modify the "nextArrayItem":
window.setTimeout (nextArrayItem, 0)
The freeze incurring mainly from the big data, but now the Event Loop will handle the Recursion process and not your Call Stack.
This is likely caused by an endless recursion. Be aware of proper handling of return values in IE:
var array1 = StudentGradeAreadHugeList();
var nextArrayItem = function() {
var grade = array1.pop();
if ( grade !== null && typeof(grade) !== "undefined" ) {
nextArrayItem();
}
};
pop() on an empty array will not return boolean false but a typeless "undefined".
There's two problems here:
You might be exceeding the call stack limit
Your if-conditional is set-up incorrectly
For the first issue:
As one of the previous responders mentioned, if you have a very large list you can exceed the limit of the call stack since you need to do a recursive call for each element. While doing setTimeout might work, it feels like a hack-y solution. I think the real issue is that your function is handling the array recursively rather than iteratively. I would recommend re-writing your function using a for-loop.
For the second issue:
Let's say in this case your array was set to [100, 90, 80]. When you invoke nextArrayItem() it will work properly the first two time, but the third time you call nextArrayItem() you are popping off the last remaining item (in this case 100) and your grade will be set to 100 which is a truthy value. Therefore, your if-conditional will pass and your function erroneously try to invoke itself again despite the fact that your array is now empty and the program should now exit the call stack.
I tried testing your code using my example in Chrome and what happens is that it will recurse one too many times and invoke pop on an empty array, which will return undefined.
You can fix this issue by changing the if conditional to check for the last element in the array after you have popped the array.
See revised code:
var nextArrayItem = function() {
var grade = array1.pop();
if (array1[array1.length-1]) {
nextArrayItem();
}
};
In Javascript what is the best way to handle scenarios when you have a set of arrays to perform tasks on sets of data and sometimes you do not want to include all of the arrays but instead a combination.
My arrays are labeled in this small snippet L,C,H,V,B,A,S and to put things into perspective the code is around 2500 lines like this. (I have removed code notes from this post)
if(C[0].length>0){
L=L[1].concat(+(MIN.apply(this,L[0])).toFixed(7));
C=C[1].concat(C[0][0]);
H=H[1].concat(+(MAX.apply(this,H[0])).toFixed(7));
V=V[1].concat((V[0].reduce(function(a,b){return a+b}))/(V[0].length));
B=B[1].concat((MAX.apply(this,B[0])-MIN.apply(this,B[0]))/2);
A=A[1].concat((MAX.apply(this,A[0])-MIN.apply(this,A[0]))/2);
D=D[1].concat((D[0].reduce(function(a,b){return a+b}))/(D[0].length));
S=S[1].concat((S[0].reduce(function(a,b){return a+b}))/(S[0].length));
}
It would seem counter-productive in this case to litter the code with tones of bool conditions asking on each loop or code section if an array was included in the task and even more silly to ask inside each loop iteration with say an inline condition as these would also slow down the processing and also make the code look like a maze or rabbit hole.
Is there a logical method / library to ignore instruction or skip if an option was set to false
All I have come up with so far is kind of pointless inline thing
var op=[0,1,1,0,0,0,0,0]; //options
var L=[],C=[],H=[],V=[],B=[],A=[],D=[],S=[];
op[0]&&[L[0]=1];
op[1]&&[C[0]=1,console.log('test, do more than one thing')];
op[2]&&[H[0]=1];
op[3]&&[V[0]=1];
op[4]&&[B[0]=1];
op[5]&&[A[0]=1];
op[6]&&[A[0]=1];
It works in that it sets only C[0] and H[0] to 1 as the options require, but it fails as it needs to ask seven questions per iteration of a loop as it may be done inside a loop. Rather than make seven versions of the the loop or code section, and rather than asking questions inside each loop is there another style / method?
I have also noticed that if I create an array then at some point make it equal to NaN rather than undefined or null the console does not complain
var L=[],C=[],H=[],V=[],B=[],A=[],D=[],S=[];
L=NaN;
L[0]=1;
//1
console.log(L); //NaN
L=undefined;
L[0]=1
//TypeError: Cannot set property '0' of undefined
L=null
L[0]=1
//TypeError: Cannot set property '0' of null
Am I getting warmer? I would assume that if I performed some math on L[0] when isNaN(L)===true that the math is being done but not stored so the line isn't being ignored really..
If I understand what you want I would do something like this.
var op = [...],
opchoice = {
//these can return nothing, no operation, or a new value.
'true': function(val){ /*operation do if true*/ },
'false': function(val){ /*operation do if false*/ },
//add more operations here.
//keys must be strings, or transformed into strings with operation method.
operation: function(val){
//make the boolean a string key.
return this[''+(val == 'something')](val);
}
};
var endop = [];//need this to prevent infinite recursion(loop).
var val;
while(val = op.shift()){
//a queue operation.
endop.push(opchoice.operation(val));
}
I'm sure this is not exactly what you want, but it's close to fulfilling the want of not having a ton of conditions every where.
Your other option is on every line do this.
A = isNaN(A) ? A.concat(...) : A;
Personally I prefer the other method.
It looks like you repeat many of the operations. These operations should be functions so at least you do not redefine the same function over and over again (it is also an optimization to do so).
function get_min(x)
{
return +(MIN.apply(this, a[0])).toFixed(7);
}
function get_max(x)
{
return +(MAX.apply(this, a[0])).toFixed(7);
}
function get_average(x)
{
return (x[0].reduce(function(a, b) {return a + b})) / (x[0].length);
}
function get_mean(x)
{
return (MAX.apply(this, x[0]) - MIN.apply(this, x[0])) / 2;
}
if(C[0].length > 0)
{
L = L[1].concat(get_min(L));
C = C[1].concat(C[0][0]);
H = H[1].concat(get_max(H));
V = V[1].concat(get_average(V));
B = B[1].concat(get_mean(B));
A = A[1].concat(get_mean(A);
D = D[1].concat(get_average(D));
S = S[1].concat(get_average(S));
}
You could also define an object with prototype functions, but it is not clear whether it would be useful (outside of putting those functions in a namespace).
In regard to the idea/concept of having a test, what you've found is probably the best way in JavaScript.
op[0] && S = S[1].concat(get_average(S));
And if you want to apply multiple operators when op[0] is true, use parenthesis and commas:
op[3] && (V = V[1].concat(get_average(V)),
B = B[1].concat(get_mean(B)),
A = A[1].concat(get_mean(A));
op[0] && (D = D[1].concat(get_average(D)),
S = S[1].concat(get_average(S)));
However, this is not any clearer, to a programmer, than an if() block as shown in your question. (Actually, many programmers may have to read it 2 or 3 times before getting it.)
Yet, there is another solution which is to use another function layer. In that last example, you would do something like this:
function VBA()
{
V = V[1].concat(get_average(V));
B = B[1].concat(get_mean(B));
A = A[1].concat(get_mean(A));
}
function DS()
{
D = D[1].concat(get_average(D));
S = S[1].concat(get_average(S));
}
op = [DS,null,null,VBA,null,null,...];
for(key in op)
{
// optional: if(op[key].hasOwnProperty(key)) ... -- verify that we defined that key
if(op[key])
{
op[key](); // call function
}
}
So in other words you have an array of functions and can use a for() loop to go through the various items and if defined, call the function.
All of that will very much depend on the number of combinations you have. You mentioned 2,500 lines of code, but the number of permutations may be such that writing it one way or the other will possibly not reduce the total number of lines, but it will make it easier to maintain because many lines are moved to much smaller code snippet making the overall program easier to understand.
P.S. To make it easier to read and debug later, I strongly suggest you put more spaces everywhere, as shown above. If you want to save space, use a compressor (minimizer), Google or Yahoo! both have one that do a really good job. No need to write your code pre-compressed.
Is there an alternative, faster method of returning the position/index of part of an array within another array (where multiple values match)? It's called a lot within my pathfinding algorithm so could do with being as fast as possible.
My current function is:
// Haystack can be e.g. [[0,1,278.9],[4,4,22.1212]]
function coordinate_location_in_array(needle,haystack){
for(n in haystack){
if(haystack[n][0]==needle[0] && haystack[n][1]==needle[1]) return n;
}
return false;
}
// Needle of [0,1]: returns 0
// Needle of [4,4]: returns 1
// Needle of [6,7]: returns false
Edit:
I've been messing around a bit and come up with a (rather ghastly) string manipulation-based method (thereby avoiding the costly for loop). I think it's still slightly slower. Could anybody benchmark these methods?
function coordinate_location_in_array(needle,haystack) {
var str1 = ':' + haystack.join(':');
var str2 = str1.replace(':'+needle[0]+','+needle[1],'*').split('*')[0];
if(str2.length == str1.length) return false;
var preceedingElements = str2.match(/:/g);
return preceedingElements!=null?preceedingElements.length:0;
}
Perhaps with some improvements this second method might provide some performance gain?
Edit 2:
Bench marked all 3 described methods using jsperf.com (initial method is fastest):
http://jsperf.com/finding-matched-array-within-array/3
Edit 3:
Just replaced the for(..in..) loop with a for(..;..;..) loop (since I know that the haystack array will never have "gaps") and performance seems to have significantly improved:
function coordinate_location_in_array(needle,haystack){
for(var n=0;n<haystack.length;n++){
if(haystack[n][0]==needle[0] && haystack[n][1]==needle[1]) return n;
}
return false;
}
I've updated the jsperf page to include this latest method.
If the "haystack" isn't sorted then there isn't a way to make it faster. Not knowing how the elements in a collection are ordered makes finding something out of it linear by nature, because you just have to check each thing.
If you are using this function over the same "haystack" over and over, you could sort the collection, and use the sorting to make it quicker to find the "needle" (look up different sorting and search algorithms to find one that fits your need best, such as using binary search to find the "needle" after haystack is sorted.)
i don't know if its faster, but you can do something like:
[1,2,3,4].slice(0,2).toString() == [1,2].toString()
in your case it would be:
function coordinate_location_in_array(needle,haystack){
for(n in haystack){
if(haystack[n].slice(0,2).toString() == needle.toString()) return n
}
return false;
}
Also found this post, which covers comparison of JS arrays: compare-two-arrays-javascript-associative
Cheers
Laidback
Using a for(..;..;..) loop rather than a for(..in..) loop made the biggest difference.
(See Edit 3 at the end of the question)
Seems to me this is just a substring search but with numbers instead of characters being the components of the string. As such, Boyer-Moore could be applicable, especially if your needles and haystacks get big.
for (var i=a.length-1;i>0;i--) {
if (i!=a.indexOf(a.charAt(i))) {
a=a.substring(0,i)+a.substring(i+1);
}
}
I found this in a web app I'm auditing, it just baffles me why it's there.
I can't seem to see a case where i!=a.indexOf(a.charAt(i)) would be false.
The value the pass to it is:
a = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
There is no comment either //sigh
This would be true for repeated characters, since indexOf finds the first index of a string, and you're searching from the end. Example:
var a = "xyzxyz";
On first iteration, i === 4, a.charAt(4) === "x", and a.indexOf("x") === 0. So 4 !== 0.
It then sets a = a.substring(0, 4) + a.substring(5). Recalling at substring is inclusive in the first index but exclusive in the last index, that means in this case a = "xyz" + "yz", so we have removed the duplicate "x" from the string.
Since the loop traverses backward, this will continue to work even for characters repeated more than once; you can see that the portion a.substring(i + 1) will always have been covered by the algorithm already, i.e. not contain any duplicates.
As always when encountering this type of thing, applying the extract method refactoring would be a great way to make the code clearer. (Even better than commenting it!) So if you just pulled this out into a method, the code could become a = removeDuplicateChars(a), and everyone is much happier.