Have you ever felt like your head wasn't meant for an algorithm?
I tried solving the maximum subarray problem and I came across this solution on Codewars:
var maxSequence = function(arr){
var min = 0, ans = 0, i, sum = 0;
for (i = 0; i < arr.length; ++i) {
sum += arr[i];
min = Math.min(sum, min);
ans = Math.max(ans, sum - min);
}
return ans;
}
console.log(maxSequence([-2, 1, -3, 4, -1, 2, 1, -5, -4]));
I understand how to solve this problem with a linear time complexity using Kadane's algorithm:
var maxSequence = function(arr){
let max = 0;
let localMax = 0;
for (let i = 0; i < arr.length; i++) {
localMax = Math.max(localMax + arr[i], arr[i]);
max = Math.max(localMax, max);
}
return max;
}
console.log(maxSequence([-2, 1, -3, 4, -1, 2, 1, -5, -4]));
But I can't make sense of why the first solution works. I simply can't grasp the idea behind it. I feel like a need a little help getting over the hump.
Edit: Here's a Codepen with some examples
The algorithm you provided is doing the same thing, in a more complex manner though. In order to explain it properly, I will compare the Codewars algorithm you provided with the Kadanes algorithm in various steps of their execution.
Let us consider the array:
[2 -4 3 2 6 -10 -12 20]
Here is the Codewars algorithm you provided:
var maxSequence = function(arr){
var min = 0, ans = 0, i, sum = 0;
for (i = 0; i < arr.length; ++i) {
sum += arr[i];
min = Math.min(sum, min);
ans = Math.max(ans, sum - min);
}
return ans;
}
Here is the implementation of Kadanes algorithm mentioned in Wikipedia:
def max_subarray(numbers):
"""Find the largest sum of any contiguous subarray."""
best_sum = 0 # or: float('-inf')
current_sum = 0
for x in numbers:
current_sum = max(0, current_sum + x)
best_sum = max(best_sum, current_sum)
return best_sum
First step:-
sum changes to 2 and min remains the same. The ans changes to 2.
Second step:-
sum changes to -2 and min changes to -2. The ans is still 2. An interesting thing to notice here, according the implementation of Kadanes algorithm by Wikipedia, there in the second stage the value of current_sum will change to 0 which is the correct way to proceed.
However in the codewars implementation, the value of sum is still -2. If you notice a little more carefully though, you will observe that the value of sum-min is 0 in the codewars implementation. This is a really important point to notice. Instead of changing sum to 0 when its value reaches less than 0. We store the minimum number that must be substracted from sum to make the net sum 0. This value is stored in min and which also explains why it is named so.
Here is a record of the value of variables so far:
sum min ans
2 0 2 //ans = max(0, 2-0)
-2 -2 2 //ans = max(2, -2+2)
Third step:-
The sum changes to 1. min still remains the same. The ans changes to 3 which is the correct. How did this happen though?
In the Kadanes algorithm, you change the value of current_sum to 3 at this stage. In the codewars implementation, instead of changing sum to 3, what they have done is used a min variable which I repeat again stores the number that should be substracted from answer so that we obtain the same value as we do in current_sum. This is more clear from this part of the algorithm.
ans = Math.max(ans, sum - min); //sum-min is current_max
Here when we substract the min from your sum. It neutralizes that extra negative in your answer. In this array A, the extra negative is 2 + (-4) = -2. In each of the following steps, we will observe that here sum is not containing the maximum continuous subarray sum. Maximum continuous subarray sum there is stored in sum - min. This is the key of this algorithm. sum-min is the current_sum here. Here are the following steps:
sum min ans
1 -2 3 //ans = max(2, 1+2)
3 -2 5 //ans = max(3, 3+2)
9 -2 11 //ans = max(5, 9+2)
-1 -2 11 //ans = max(11, -1+2)
It is interesting to observe that even though the value of sum is negative in the last step, the value of min does not change. Why is that? The answer is it does not need to. If you look at sum-min is this case, it is 1 and not less than 0. Hence there is a possibility that if sufficient positive numbers follow after the current index in A, the value of sum-min might exceed the current value of ans. If you dry run Kadanes algorithm till this step, you will notice that even there the value of current_sum will not change to 0 at this stage, it will be 1.
Remaining steps:-
sum min ans
-1 -2 11 //ans = max(11, -1+2)
-13 -13 11 //ans = max(11, -13+13)
7 -13 20 //ans = max(11, 7+13)
The most important point in this implementation, sum-min here is analogous to current_sum of Kadanes algorithm.
I should also mention that the Kadanes algorithm and codewars algorithm you provided will not work if the input array consists of all negative numbers. Both are not meant for it. There is a small implementation difference in the Kadanes algorithm if you want it to work for array consisting of all negative numbers (initialize current_sum to A[0]).
Do comment if you face any problems in understanding my explanation.
I don't think the the Codewars algorithm will work for every test case.
Following are the test cases where this algorithm will fail:
Test Case 1: arr = [-1]
Test Case 2: arr = [-1, -2]
For both the test cases, the algorithm under test gives an output equal to 0, which is not the correct answer.
PS: I have checked the Codewars problem. The test cases for the problem are not yet comprehensive and this problem has issues.
So for the time being, Kadane's algorithm is a good choice to solve the problem in linear time complexity.
Related
well this is problem number 12 on projecteuler website:
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28. The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
and here's my code (I'm new to Javascript)
let num = 1;
let add= 1;
let divisors = [];
while (divisors.length < 500){
divisors = []
for(let i = 1; i <= num; i++){
if(num % i == 0){
divisors.push(i);
}
}
add ++;
num += add;
}
console.log(num - add)
this code run fine when I change the while loop condition to 300 or less.
and this code running on Intel i7 Q740 1.75GHz.
when I try it nothing shows up on console and my question is that's because of my CPU and lack of power or my code has issue? I wait about 20 mins and still nothing as result.
This code is not very efficient as #Vasil Dininski pointed out but you won't reach the max Integer you just have to wait a while for your program to calculate.
I would recommend to optimize your code e.g. by writing a simple function which returns your number of divisors for your current number.
This could look similar to this:
function numberOfDivisors(num) {
var numOfDivisors = 0;
var sqrtOfNum = Math.sqrt(num);
for(var i = 1; i <= sqrtOfNum; i++) {
if(num % i == 0) {
numOfDivisors += 2;
}
}
// if your number is a perfect square you have to reduce it by one
if(sqrtOfNum * sqrtOfNum == num) {
numOfDivisors--;
}
return numOfDivisors;
}
Then you could use this method in your while loop like this:
var maxNumOfDivisors = 500;
var num = 0;
var i = 1;
while(numberOfDivisors(num) < maxNumOfDivisors) {
num += i;
i++;
}
which would return you the correct triangular number.
Please also note that triangular numbers start at 0 which is why my num is 0.
As you might have noticed, this algorithm might be a bit brute-force. A better one would be combine a few things. Let's assume the number we're looking for is "n":
Find all prime numbers in the range [1, square root of n]. You will
be iterating over n, so the sieve of
eratosthenes
will help in terms of efficiency (you can memoize primes you've already found)
Given that any number can be expressed as a prime number to some power, multiplied by a prime number to some power, etc. you can find all the combinations of those primes to a power, which are divisors of n.
This would be a more efficient, albeit a more complicated way to find them. You can take a look at this quora answer for more details: https://www.quora.com/What-is-an-efficient-algorithm-to-find-divisors-of-any-number
If my calculation is not wrong, you add one more to each next divisor from previous divisor and you get final result.
let prev = 0;
for(let i=0; i < 500; i++) {
prev = prev + i;
}
console.log("AA ", prev);
let say it's given 2k+2+3p=n as the test, how to find out the test is true for a number is valid for a number when k>=0, p>=0, n>=0:
example1 : n=24 should result true since k=5 & p=4 => 2(5)+2+3(4)=24
example2 : n=11 should result true since k=0 & p=3 => 2(0)+2+3(3)=11
example3 : n=15 should result true since k=5 & p=1 => 2(5)+2+3(1)=15
i wonder if there is a mathematic solution to this. i solved it like bellow:
//let say 2k+2+3p=n
var accepted = false;
var betterNumber= n-2;
//assume p=0
var kReminder= (betterNumber)%2==0;
//assume k=0
var pReminder= (betterNumber)%3==0;
if (kReminder || pReminder){
accepted=true;
}else{
var biggerChunk= Math.Max(2,3); //max of 2k or 3p, here i try to find the bigger chunk of the
var smallerChunk= Math.Min(2,3);
if ((betterNumber%bigger)%smallerChunk==0){
accepted=true;
}else
{
accepted=false;
}
}
still there are edge cases that i didn't see. so i wonder if it has a better solution or not.
Update
the test above is just an example. the solution should be efficient enough for big numbers or any combination of number like 1000000k+37383993+37326328393p=747437446239902
By inspection, 2 is the smallest valid even number and 5 is the smallest valid odd number:
2 is valid (k=0, p=0)
5 is valid (k=0, p=1)
All even numbers >= 2 and all odd numbers >= 5 are valid.
Even numbers: k=n/2-1, p=0
odd numbers: k=(n-3)/2-1, p=1
What we're doing here is incrementing k to add 2s to the smallest valid even and odd numbers to get all larger even and odd numbers.
All values of n >= 2 are valid except for 3.
Dave already gave a constructive and efficient answer but I'd like to share some math behind it.
For some time I'll ignore the + 2 part as it is of less significance and concentrate on a generic form of this question: given two positive integers a and b check whether number X can be represented as k*a + m*b where k and m are non-negative integers. The Extended Euclidean algorithm essentially guarantees that:
If number X is not divisible by GCD(a,b), it can't be represented as k*a + m*b with integer k and m
If number X is divisible by GCD(a,b) and is greater or equal than a*b, it can be represented as k*a + m*b with non-negative integer k and m. This follows from the fact that d = GCD(a,b) can be represented in such a form (let's call it d = k0*a + m0*b). If X = Y*d then X = (Y*k0)*a + (Y*m0)*b. If one of those two coefficients is negative you can trade one for the other adding and subtracting a*b as many times as required as in X = (Y*k0 + b)*a + (Y*m0 - a)*b. And since X >= a*b you can always get both coefficients to be non-negative in such a way. (Note: this is obviously not the most efficient way to find a suitable pair of those coefficients but since you only ask for whether such coefficients exist it should be sufficient.)
So the only gray area is numbers X divisible by GCD(a,b) that lie between in the (0, a*b) range. I'm not aware of any general rule about this area but you can check it explicitly.
So you can just do pre-calculations described in #3 and then you can answer this question pretty much immediately with simple comparison + possibly checking against pre-calculated array of booleans for the (0, a*b) range.
If you actual question is about k*a + m*b + c form where a, b and c are fixed, it is easily converted to the k*a + m*b question by just subtracting c from X.
Update (Big values of a and b)
If your a and b are big so you can't cache the (0, a*b) range beforehand, the only idea I have is to do the check for values in that range on demand by a reasonably efficient algorithm. The code goes like this:
function egcd(a0, b0) {
let a = a0;
let b = b0;
let ca = [1, 0];
let cb = [0, 1];
while ((a !== b) && (b !== 0)) {
let r = a % b;
let q = (a - r) / b;
let cr = [ca[0] - q * cb[0], ca[1] - q * cb[1]];
a = b;
ca = cb;
b = r;
cb = cr;
}
return {
gcd: a,
coef: ca
};
}
function check(a, b, x) {
let eg = egcd(a, b);
let gcd = eg.gcd;
let c0 = eg.coef;
if (x % gcd !== 0)
return false;
if (x >= a * b)
return true;
let c1a = c0[0] * x / gcd;
let c1b = c0[1] * x / gcd;
if (c1a < 0) {
let fixMul = -Math.floor(c1a / (b / gcd));
let c1bFixed = c1b - fixMul * (a / gcd);
return c1bFixed >= 0;
}
else { //c1b < 0
let fixMul = -Math.floor(c1b / (a / gcd));
let c1aFixed = c1a - fixMul * (b / gcd);
return c1aFixed >= 0;
}
}
The idea behind this code is based on the logic described in the step #2 above:
Calculate GCD and Bézout coefficients using the Extended Euclidean algorithm (if a and b are fixed, this can be cached, but even if not this is fairly fast anyway).
Check for conditions #1 (definitely no) and #2 (definitely yes) from the above
For value in the (0, a*b) range fix some coefficients by just multiplying Bézout coefficients by X/gcd. F
Find which of the two is negative and find the minimum multiplier to fix it by trading one coefficient for another.
Apply this multiplier to the other (initially positive) coefficient and check if it remains positive.
This algorithm works because all the possible solutions for X = k*a + m*b can be obtained from some base solution (k0, m0) using as (k0 + n*b/gcd, m0 + n*a/gcd) for some integer n. So to find out if there is a solution with both k >= 0 and m >= 0, all you need is to find the solution with minimum positive k and check m for it.
Complexity of this algorithm is dominated by the Extended Euclidean algorithm which is logarithmic. If it can be cached, everything else is just constant time.
Theorem: it is possible to represent number 2 and any number >= 4 using this formula.
Answer: the easiest test is to check if the number equals 2 or is greater or equals 4.
Proof: n=2k+2+3p where k>=0, p>=0, n>=0 is the same as n=2m+3p where m>0, p>=0 and m=k+1. Using p=0 one can represent any even number, e.g. with m=10 one can represent n=20. The odd number to the left of this even number can be represented using m'=m-2, p=1, e.g. 19=2*8+3. The odd number to the right can be represented with m'=m-1, p=1, e.g. 21=2*9+3. This rule holds for m greater or equal 3, that is starting from n=5. It is easy to see that for p=0 two additional values are also possible, n=2, n=4.
Hello Every one i want to solve this problem with javascript and es6 the problem is
if i have an array like that [1,2,3,4] so i want to check every probability like the following and return the greatest number
1--> (1*2) + (3*4) = 14
2--> (1*3) + (2*4) = 11
3--> (1*4) + (2*3) = 10
then the greatest number is 14 as a return result ---> how can i do that using function and keep in mind if the array have 100 number how can i do the 99 probabilities dynamically
notes : maybe the array is not sorted and it may be an odd not only even thank u
Multiplying the largest numbers with each other results in a larger sum than multiplying a large number with a low number and then sum them up (compare given a circumference a square is the rectangle with the largest surface area). One therefore only has to calculate [0]*[1] + [2]*[3] + [4]*[5] + ...
(assuming that the array length is even):
const array = [4,2,1,3].sort();
let sum = 0;
for(let i = 1; i < array.length; i += 2)
sum += array[i - 1] * array[i];
Edit: For that calculation with consecutive numbers starting from 1 one doesn't even need JS, the closed form for the sum is 1/3 * (n - 1)(4*n² - 5*n) (with n being the largest (even) number).
I agree with Stephen, Multiplying the largest numbers with each other results in a larger sum than multiplying a large number with a low number and then sum them up. And also you should be multiplying consecutive numbers in the array like (1*2)+(3*4)+(5*6)+...+(99*100)etc.
You don't need all the combinations.
Mathematically, you will always yield the highest values by multiplying the highest available numbers.
So, it would be pointless to multiply an high number with anything other then the second highest number.
const arr = [4, 3, 2, 5, 6, 7]
let sum = arr.sort().reduce((sum, val, idx) => sum += (idx % 2 !== 0) ? arr[idx - 1] * arr[idx] : 0 , 0)
I'm trying to solve this MaxCollatzLength kata but I'm struggling to optimise it to run fast enough for really large numbers.
In this kata we will take a look at the length of collatz sequences.
And how they evolve. Write a function that take a positive integer n
and return the number between 1 and n that has the maximum Collatz
sequence length and the maximum length. The output has to take the
form of an array [number, maxLength] For exemple the Collatz sequence
of 4 is [4,2,1], 3 is [3,10,5,16,8,4,2,1], 2 is [2,1], 1 is [ 1 ], so
MaxCollatzLength(4) should return [3,8]. If n is not a positive
integer, the function have to return [].
As you can see, numbers in Collatz sequences may exceed n. The last
tests use random big numbers so you may consider some optimisation in
your code:
You may get very unlucky and get only hard numbers: try submitting 2-3
times if it times out; if it still does, probably you need to optimize
your code more;
Optimisation 1: when calculating the length of a
sequence, if n is odd, what 3n+1 will be ?
Optimisation 2: when looping through 1 to n, take i such that i < n/2, what
will be the length of the sequence for 2i ?
A recursive solution quickly blows the stack, so I'm using a while loop. I think I've understood and applied the first optimisation. I also spotted that for n that is a power of 2, the max length will be (log2 of n) + 1 (that only shaves off a very small amount of time for an arbirtarily large number). Finally I have memoised the collatz lengths computed so far to avoid recalculations.
I don't understand what is meant by the second optimisation, however. I've tried to notice a pattern with a few random samples and loops and I've plotted the max collatz lengths for n < 50000. I noticed it seems to roughly follow a curve but I don't know how to proceed - is this a red herring?
I'm ideally looking for a hints in the right direction so I can work towards the solution myself.
function collatz(n) {
let result = [];
while (n !== 1) {
result.push(n);
if (n % 2 === 0) n /= 2;
else {
n = n * 3 + 1;
result.push(n);
n = n / 2;
}
}
result.push(1);
return result;
}
function collatzLength(n) {
if (n <= 1) return 1;
if (!collatzLength.precomputed.hasOwnProperty(n)) {
// powers of 2 are logarithm2 + 1 long
if ((n & (n - 1)) === 0) {
collatzLength.precomputed[n] = Math.log2(n) + 1;
} else {
collatzLength.precomputed[n] = collatz(n).length;
}
}
return collatzLength.precomputed[n];
}
collatzLength.precomputed = {};
function MaxCollatzLength(n) {
if (typeof n !== 'number' || n === 0) return [];
let maxLen = 0;
let numeralWithMaxLen = Infinity;
while (n !== 0) {
let lengthOfN = collatzLength(n);
if (lengthOfN > maxLen) {
maxLen = lengthOfN;
numeralWithMaxLen = n;
}
n--;
}
return [numeralWithMaxLen, maxLen];
}
Memoization is the key to good performance here. You memoize the end results of the function that calculates the Collatz sequence. This will help you on repeated calls to maxCollatzLength, but not when you determine the length of the sequence for the first time.
Also, as #j_random_hacker mentioned, there is no need to actually create the sequence as list; it is enough to store its length. An integer result is light-weight enough to be memoized easily.
You can make use of precalculated results already when you determine the length of a Collatz sequence. Instead of following the sequence all the way down, follow it until you hit a number for which the length is known.
The other optimizations you make are micro-optimizations. I'm not sure that calculating the log for powers of two really buys you anything. It rather burdens you with an extra test.
The memoized implementation below even forgoes the check for 1 by putting 1 in the dictionary of precalculated values initially.
var precomp = {1: 1};
function collatz(n) {
var orig = n;
var len = 0;
while (!(n in precomp)) {
n = (n % 2) ? 3*n + 1 : n / 2;
len++;
}
return (precomp[orig] = len + precomp[n]);
}
function maxCollatz(n) {
var res = [1, 1];
for (var k = 2; k <= n; k++) {
var c = collatz(k);
if (c > res[1]) {
res[0] = k;
res[1] = c;
}
}
return res;
}
I haven't used node.js, but the JavaScript in my Firefox. It gives reasonable performance. I first had collatz as a recursive function, which made the implementation only slightly faster than yours.
The second optimization mentioned in the question means that if you know C(n), you also know that C(2*n) == C(n) + 1. You could use that knowledge to precalculate the values for all even n in a bottom-up approach.
It would be nice if the lengths of the Collatz sequences could be calculated from the bottom up, a bit like the sieve of Erathostenes. You have to know where you come from instead of where you go to, but it is hard to know ehen to stop, because for finding the longest sequence for n < N, you will have to calculate many sequences out of bound with n > N. As is, the memoization is a good way to avoid repetition in an otherwise straightforwad iterative approach.
In this task you are required to write a Python function,
maxLength, that returns two integers:
• First returned value: for each integer k, 1 ≤ k ≤ m, the
length of Collatz sequence for each k is computed and the
largest of these numbers is returned.
• Second returned value is the integer k, 1 ≤ k ≤ m, whose
Collatz sequence has the largest length. In case there are
several such numbers, return the first one (the smallest).
For example, maxLength(10) returns numbers
20 and 9
Which means that among the numbers 1, 2, 3,…, 10, nine has the
longest Collatz sequence, and its length is equal to 20.
In your program you may define other (auxiliary) functions with
arbitrary names, however, the solution function of this task
should be named maxLength(m).
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Generating random numbers in Javascript in a specific range?
How can i get a random value between, for example, from -99 to 99, excluding 0?
var num = Math.floor(Math.random()*99) + 1; // this will get a number between 1 and 99;
num *= Math.round(Math.random()) ? 1 : -1; // this will add minus sign in 50% of cases
Altogether
var ranNum = Math.ceil(Math.random() * 99) * (Math.round(Math.random()) ? 1 : -1)
This returns what you want
function getNonZeroRandomNumber(){
var random = Math.floor(Math.random()*199) - 99;
if(random==0) return getNonZeroRandomNumber();
return random;
}
Here's a functional fiddle
EDIT
To contribute for future readers with a little debate happened in the comments which the user #MarkDickinson made a indeed relevant contribution to my first code posted, I've decided to make another fiddle with a fast comparison between using Math.floor() and Math.round() functions to return the value the op wanted.
First Scenario: Using var random = Math.round(Math.random()*198) - 99; (My first suggestion)
function getNonZeroRandomNumberWithMathRound(){
var random = Math.round(Math.random()*198) - 99;
if(random==0) return getNonZeroRandomNumber();
return random;
}
Second scenario: Using var random=Math.floor(Math.random()*199) - 99; (Mark suggestion)
function getNonZeroRandomNumberWithMathFloor(){
var random = Math.floor(Math.random()*199) - 99;
if(random==0) return getNonZeroRandomNumber();
return random;
}
Methodology
Since it's a short debate I've chosen fiddle.net to do the comparison.
The test consists of running the above functions 100.000 times and then retrieving how much times the extreme numbers 99 and -99 would appear against a other number, let's say 33 and -33.
The test will then give a simple output consisting of the percentage of appearances from 99 and -99 and the percentage of appearances of 33 and -33.
It'll be used the Webkit implementation from Safari 6.0.2 to the give the output from this answer but anyone can test with your favourite browser late on fiddle.net
Result from first scenario:
Percentage of normal ocurrences:0.97%
Percentage of extreme ocurrences:0.52%
Percentage of extreme ocurrences relative to normal ocurrences:53.4% // Half the chances indeed
Result from second scenario:
Percentage of normal ocurrences:1.052%
Percentage of extreme ocurrences:0.974%
Percentage of extreme ocurrences relative to normal ocurrences:92% //Closer of a fair result with a minimal standard deviation
The result can be seen here: http://jsfiddle.net/brunovieira/LrXqh/
Here's a generalized solution that will let you set the boundaries, and opt in/out of including the 0.
var pos = 99,
neg = 99,
includeZero = false,
result;
do result = Math.ceil(Math.random() * (pos + neg)) - neg;
while (includeZero === false && result === 0);
The pos and neg values are inclusive.
This way there's no requirement that the positive and negative ranges be balanced.
Or if you're worried about the rerun due to a single excluded value, you can just make the initial range less by one, and add 1 to any result greater than or equal to 0.
var pos = 5,
neg = 5,
result;
result = Math.floor(Math.random() * (pos + neg)) - neg;
result = result < 0 ? result : result + 1;
That last line could be shorter if you prefer:
result += (result >= 0)