Hi there I have been researching and trying to learn how to check for the time complexity of certain algorithms. I've seen this video which was very helpful.
That being said I wondered off and started trying to work out the Worsts Case and an average case of certain algorithms.
1
I believe in the following snippet it is O(n) since to ind the value for sin we have to loop the entire array.
function mySin(x, iterNum) {
var mxx = -x*x;
var sin = 1;
var n = 0;
var term = 1;
for (var i = 1; i <= 2*iterNum; i++) {
n = n + 2;
term = term * mxx / ( n*(n+1) );
sin = sin + term
}
sin = x*sin;
console.log(sin + " = my function.");
console.log(Math.sin(x) + " math.sin");
}
Thanks again
2
function calculateFibonacciSum (num) {
if(cachedNumbers[num]) {
return cachedNumbers[num];
}
if(('number' === typeof num) && num <= 0) {
throw new Error ('Fibonnci series starts with 0. Please, enter any interget greater than or equal to 0');
}
else if(('number' === typeof num) && num === 0) {
return 0;
}
else if(('number' === typeof num) && (num === 1 || num === 2)) {
return 1;
}
else {
var value = calculateFibonacciSum(num-1) + calculateFibonacciSum(num-2);
cachedNumbers[num] = value;
return value;
}
}
While for this one I think it is also O(n) since in the first if/else statement the tc is O(1) since its contestant whilst the final else statement we must loop all the numbers and if the number is not calculated then call the function again (aka recurssion).
TIA
Both of these seem correct to me. Here's a bit more explanation:
1.
This is in fact O(n), as there are n iterations of the loop, the rest constant time; and n is proportional to iterNum
2.
This one is also linear time, but only since you cache the results of previous calculations. Otherwise it would be O(2n).
It is linear time since it essentially runs a loop down to the base cases (0 and 1). In fact, you could re-write this one using a loop instead of recursion.
Related
Thanks to Nina I have a code to compare two sentences word by word and return the number of word matches like this:
function includeWords(wanted, seen) {
var wantedMap = wanted.split(/\s+/).reduce((m, s) => m.set(s, (m.get(s) || 0) + 1), new Map),
wantedArray = Array.from(wantedMap.keys()),
count = 0;
seen.split(/\s+/)
.forEach(s => {
var key = wantedArray.find(t => s === t || s.length > 3 && t.length > 3 && (s.startsWith(t) || t.startsWith(s)));
if (!wantedMap.get(key)) return;
console.log(s, key)
++count;
wantedMap.set(key, wantedMap.get(key) - 1);
});
return count;
}
let matches = includeWords('i was sent to earth to protect you introduced', 'they\'re were protecting him i knew that i was aware introducing');
console.log('Matched words: ' + matches);
The code works fine, but there is still one issue:
What if we want to return a match for introduced and introducing too?
If you want the program to consider the words 'introduce' and 'introducing' as a match, it would amount to a "fuzzy" match (non binary logic). One simple way of doing this would require more code, the algorithm of which would possibly resemble
Take 2 words that you wish to match, tokenize into ordered list
of letters
Compare positionally the respective letters, i.e
match a[0]==b[0]? a[1]==b[1] where a[0] represents the first letter
of the first word and b[0] represents the first tokenized
letter/character potential match candidate
KEep a rolling numeric count of such positional matches. In this case it is 8 (introduc).
divide by word length of a = 8/9 call this f
divide by word length of b = 8/11 call this g
Provide a threshold value beyond which the program will consider it a match. eg. if you say anything above 70% in BOTH f and g can be
considered a match - viola, you have your answer!
Please note that there is some normalization also needed to prevent low length words from becoming false positives. you can add a constraint that the aforementioned calculation applies to words with at least 5 letters(or something to that effect!
Hope this helps!!
Regards,
SR
You could calculate similarites for a word pair and get a relation how many characters are similar bei respecting the length of the given word and the wanted pattern.
function getSimilarity(a, b) {
var i = 0;
while (i < a.length) {
if (a[i] !== b[i]) break;
i++;
}
return i / Math.max(a.length, b.length);
}
console.log(getSimilarity('abcdefghij', 'abc')); // 0.3
console.log(getSimilarity('abcdefghij', 'abcdef')); // 0.6
console.log(getSimilarity('abcdefghij', 'abcdefghij')); // 1
console.log(getSimilarity('abcdef', 'abcdefghij')); // 0.6
console.log(getSimilarity('abcdefghij', 'abcdef')); // 0.6
console.log(getSimilarity('abcdefghij', 'xyz')); // 0
console.log(getSimilarity('introduced', 'introducing')); // 0.7272727272727273
Here's a quick fix solution.
It's not intended as a complete solution.
Since the English language has more than a few quirks that would almost require an AI to understand the language.
First add a function that can compare 2 words and returns a boolean.
It'll also make it easier to test for specific words, and adapt to what's really needed.
For example, here's a function that does the simple checks that were already used.
Plus an '...ed' versus '...ing' check.
function compareWords (word1, word2) {
if (word1 === word2)
return true;
if (word1.length > 3 && word2.length > 3) {
if (word1.startsWith(word2) || word2.startsWith(word1))
return true;
if (word1.length > 4 && word2.length > 4) {
if (/(ing|ed)$/.test(word1) && word1.replace(/(ing|ed)$/, 'inged') === word2.replace(/(ing|ed)$/, 'inged'))
return true;
}
}
return false;
}
//
// tests
//
let words = [
["same", "same"],
["different", "unsame"],
["priced", "pricing"],
["price", "priced"],
["producing", "produced"],
["produced", "producing"]
];
words.forEach( (arr, idx) => {
let word1= arr[0];
let word2= arr[1];
let isSame = compareWords(word1, word2);
console.log(`[${word1}] ≈ [${word2}] : ${isSame}`);
});
Then use it in the code you already have.
...
seen.split(/\s+/)
.forEach(s => {
var key = wantedArray.find(t => compareWords(t, s));
...
Regarding string similarity, here's f.e. an older SO post that has some methods to compare strings : Compare Strings Javascript Return %of Likely
I have implemented this, it seems to work fine. any suggestions would be appreciated..
let speechResult = "i was sent to earth to introducing protect yourself introduced seen";
let expectSt = ['they were protecting him knew introducing that you i seen was aware seen introducing'];
// Create arrays of words from above sentences
let speechResultWords = speechResult.split(/\s+/);
let expectStWords = expectSt[0].split(/\s+/);
function includeWords(){
// Declare a variable to hold the count number of matches
let arr = [];
for(let a = 0; a < speechResultWords.length; a++){
for(let b = 0; b < expectStWords.length; b++){
if(similarity(speechResultWords[a], expectStWords[b]) > 69){
arr.push(speechResultWords[a]);
console.log(speechResultWords[a] + ' includes in ' + expectStWords[b]);
}
} // End of first for loop
} // End of second for loop
let uniq = [...new Set(arr)];
return uniq.length;
};
let result = includeWords();
console.log(result)
// The algorithmn
function similarity(s1, s2) {
var longer = s1;
var shorter = s2;
if (s1.length < s2.length) {
longer = s2;
shorter = s1;
}
var longerLength = longer.length;
if (longerLength == 0) {
return 1.0;
}
return (longerLength - editDistance(longer, shorter)) / parseFloat(longerLength)*100;
}
function editDistance(s1, s2) {
s1 = s1.toLowerCase();
s2 = s2.toLowerCase();
var costs = new Array();
for (var i = 0; i <= s1.length; i++) {
var lastValue = i;
for (var j = 0; j <= s2.length; j++) {
if (i == 0)
costs[j] = j;
else {
if (j > 0) {
var newValue = costs[j - 1];
if (s1.charAt(i - 1) != s2.charAt(j - 1))
newValue = Math.min(Math.min(newValue, lastValue),
costs[j]) + 1;
costs[j - 1] = lastValue;
lastValue = newValue;
}
}
}
if (i > 0)
costs[s2.length] = lastValue;
}
return costs[s2.length];
}
I am working Codewars problem' Happy Numbers ' here is the link https://www.codewars.com/kata/happy-numbers-5/train/javascript Here is the problem, when I am running the code when n > 98 the maximum call stack size is reached. How can I make some changes on my code to fix this problem?
function happyNumbers(x){
var res = [];
for (let i = 1; i <= x; i++){
var str = [];
if (helper(str,i)){res.push(i)}
}
return res
}
function helper(str,n){
var num = 0;
if (n === 1){return true}
if (str.indexOf(n) > -1){return false}
str.push(n);
if (n.toString().length === 1){num = Math.pow(n,2).toString()}
if (n.toString().length >= 2){
num = n.toString().split('')
.reduce((a,b) => Math.pow(a,2)+ Math.pow(b,2)).toString();
}
return helper(str,Number(num))
}
Maybe some more simplyfing would help by
using a Set for visited value to prevent circular loop which never ends (Memoization),
taking numerical values at all, only for splitting into single digits, a string is taken,
summing up by using a simple multiplication,
now some exit function:
check if sum is 1, exit the function with true,
check if sum is already visited and if so, exit with false,
return by calling the function again with sum and updated set visited with sum.
function happyNumbers(x, visited = new Set) {
var sum = 0, value;
for (value of String(x)) sum += value * value;
if (sum === 1) return true;
if (visited.has(sum)) return false;
return happyNumbers(sum, visited.add(sum));
}
console.log(happyNumbers(123));
Hi there I have been researching and trying to learn hw to check for the time complexity of certain algorithms.
I've seen this video which was very helpful.
That being said I wondered off and started trying to work out the Worsts Case and average case of certain algorithms.
Program #1: This is a calculator which calculates in RPN format.
function evaluate() {
var input = prompt("Please enter your input string\n\nExamples of input strings:\n\n\t1. 10 4 5 + *\n\t2. 10 4 5 + * 2 +\n\t3. 10 8 *");
var values = input.split(" ");
var array = new Array();
for (i in values) {
if (values[i] != "+" && values[i] != "*" && values[i] != "-" && values[i] != "/") {
array.push(parseInt(values[i]));
} else {
var operator = values[i];
var val2 = array.pop();
var val1 = array.pop();
switch (operator) {
case "+":
array.push(eval("val1 + val2"));
break;
case "*":
array.push(eval("val1 * val2"));
break;
case "-":
array.push(eval("val1 - val2"));
break;
case "/":
array.push(eval("val1 / val2"));
break;
}
}
}
if (input == "" || input == null) {
document.writeln("Oops, based on your input we have nothing to calculate for you!");
} else {
document.writeln("Your RPN calculation is: ");
document.writeln(array + " with a starting input string of: " + input);
}
}
From what I've understood in this case the the for loop in this case is running for a constant number of times and that is the length of the array then performs a O(1) instruction hence the function reduce has a Time Complexity of O(1) no matter the size of the array passed.
Program #2: Check whether the number is Prime or Not
function isPrime(num) {
if(num <= 1) return false;
const limit = Math.floor(Math.sqrt(num));
for(let i = 2; i <= limit; i++) {
if(num % i === 0) return false;
}
return true;
}
Since there is one for loop and the program needs to at least look at every number from 2 to the inputted number hence I think it is O(n).
For now these are the one's I've done with lots of thinking and am hoping they are correct. My question is there an easy way on how to check the time complexity of functions I've never seen before.
P.S I THINK they are correct. Please if they aren't specify which and why.
Thanks for your help :)
Update
Apparently I was wrong and lexicore corrected me the answers are:
#1 No, it is not O(1). Assuming one step takes O(1) and you have n steps where n is the length of the array, time complexity is O(n).
#2 No, it is not O(n). You cycle makes at most sqrt(n) steps, each step
O(1) so time complexity is O(sqrt(n)) here.
That being said I tried further and stumbled upon this in my textbook. It basically calculates an approximate square root.
#3
function sqrt(num) {
guess = num / 3;
do {
lastGuess = guess;
guess = (num / guess + guess) / 2;
while(Math.abs(lastGuess - guess));
return guess; /
Am I correct to say this is O(n) my reasoning is since we loop until an approx is found.
#4 from the user thanksd in the link he calculates the largest number in an array. Is it safe to say that the time complexity in this is O(n) since the largest number may be at the very last, and hence why O(n).
Thanks again
No, it is not O(1). Assuming one step takes O(1) and you have n steps where n is the length of the array, time complexity is O(n).
No, it is not O(n). You cycle makes at most sqrt(n) steps, each step O(1) so time complexity is O(sqrt(n)) here.
I have a long loop that takes maybe 10 mins or more, and I want to set always a new time to avoid it to continue. But it dosen't works.
function problem3(){
var img = document.getElementById('p_3');
img.style.display = img.style.display === 'block' ? 'none' : 'block';
var number=600851475143;
var t = new Date();
for(var i=3;i*i<=number;i+=2){
if(isPrime(i) && number%i==0){
var maxPrime = i;
}
setInterval(function(){time(t)},5000);
}
document.getElementById("p3").innerHTML = 'Il più grande divisiore primo di <span>'+number+"</span> è <span>" + maxPrime+"</span>";
}
function time(t){
return console.log(Date() - t);
}
If I put console.log(Date() - t);in the problem3() function it works, but I can't do Date()-t every 5 seconds, something like setInterval(Date()-t,5000)
This is a case where you might consider using the workers API. Instead of freezing the browser, let the job be done in the background and call back to the main thread when it's done.
https://developer.mozilla.org/en-US/docs/Web/API/Web_Workers_API
JavaScript is not multithreaded. So we think of setInterval() as running a piece of code every n ms (5000 in your example). But that's not quite true. If there's already script running when the interval elapses, the best that can happen is the bit of code gets added to a queue to be executed - but nothing from that queue is going to run until the already-running script finishes.
So in rough terms that's why it's not working, but what to do? Well, if you want anything to happen before problem3() returns, then problem3() is going to have to make it happen in a synchronous way.
For example, you could create a lastOutputTime variable, initialize it to the current time, and on each iteration through the for loop compare the current time to the stored value. If 5 seconds have passed, output to console and update lastOutputTime.
Your algorithm should be improved to something like this:
function maxPrimeFactor(number) {
if (number == 0 || !Number.isInteger(number) ||
number > Number.MAX_SAFE_INTEGER) return NaN;
number = Math.abs(number);
while(number % 2 == 0) number /= 2;
for (var i = 3; i * i <= number; i += 2) {
while(number % i == 0) number /= i;
}
return number;
}
var number = 600851475143;
console.log('maxPrimeFactor(' + number + ') == ' + maxPrimeFactor(number));
If for some numbers you need too much time, then break the loop into smaller chunks and asynchronize. But never use setInterval for this, and especially never use setInterval inside a long loop. setInterval schedules some task to run every n milliseconds, so if you use it in a loop, after i iterations, the task will run i every n milliseconds! And setInterval is so problematic because it can freeze the browser if the task takes more than n milliseconds. You should use setTimeout instead.
However, this would be useless in this case. The algorithm above can detect that 304250263527209 (15 digits) is a prime almost instantly. Given that the maximum safe integer is 9007199254740991 (16 digits), I don't think you will have problems for any number.
If you say the algorithm takes so long, it may be because you are trying it with bigger numbers. But be aware JS numbers are 64-bit floating point numbers, and thus integers can't be represented accurately above Number.MAX_SAFE_INTEGER. You will get a wrong result anyways, so do not even try to calculate that.
In the case of the Project Euler #551, a brute-force approach would be
function sumOfDigits(n) {
var sum = 0;
while(n != 0) {
sum += n % 10;
n = Math.floor(n/10);
}
return sum;
}
function sumDigitsSeq(n) {
return new Promise(function(resolve) {
var i = 1;
var chunkSize = 1e5;
var sum = 1;
(function chunk() {
chunkSize = Math.min(chunkSize, n-i);
for (var j=0; j<chunkSize; ++j, ++i) {
sum += sumOfDigits(sum);
}
if (i >= n) return resolve(sum);
console.log('Please wait. sumDigitsSeq(' + i + ') == ' + sum);
setTimeout(chunk, 60);
})();
});
}
var number = 1e6;
sumDigitsSeq(number).then(function(result) {
console.log('Done! sumDigitsSeq(' + number + ') == ' + result);
});
Of course brute-force is not the appropriate way to solve the problem.
I'm trying to solve a Coderbyte challenge, and I'm still trying to fully understand recursion.
Here's the problem: Using the JavaScript language, have the function AdditivePersistence(num) take the num parameter being passed which will always be a positive integer and return its additive persistence which is the number of times you must add the digits in num until you reach a single digit. For example: if num is 2718 then your program should return 2 because 2 + 7 + 1 + 8 = 18 and 1 + 8 = 9 and you stop at 9.
Here's the solution I put into jsfiddle.net to try out:
function AdditivePersistence(num) {
var count=0;
var sum=0;
var x = num.toString().split('');
for(var i=0; i<x.length; i++) {
sum += parseInt(x[i]);
}
if(sum.length == 1) {
return sum;
}
else {
return AdditivePersistence(sum);
}
}
alert(AdditivePersistence(19));
It tells me that there's too much recursion. Is there another "else" I could put that would basically just re-run the function until the sum was one digit?
One of the problems is that your if statement will never evaluate as 'true'. The reason being is that the sum variable is holding a number, and numbers don't have a length function. Also, as 'Barmar' pointed out, you haven't incremented the count variable, and neither are you returning the count variable.
Here's a solution that works using recursion.
function AdditivePersistence(num) {
var result = recursive(String(num).split('').reduce(function(x,y){return parseInt(x) + parseInt(y)}), 1);
function recursive(n, count){
c = count;
if(n < 10)return c;
else{
count += 1
return recursive(String(n).split('').reduce(function(x,y){return parseInt(x) + parseInt(y)}), count)
}
}
return num < 10 ? 0 : result
}
To fix the 'too much recursion problem',
if(sum.toString().length == 1)
However, as the others have said, your implementation does not return the Additive Persistence. Use James Farrell's answer to solve the Coderbyte challenge.