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Problem Explanation:
The smallest common multiple between two numbers is the smallest number that both numbers can divide into. This concept can be extended to more than two numbers as well.
We can first start with just finding the smallest common multiple between two numbers. Naively, you can start writing out multiple of each number until you write a multiple that exists from both numbers.
An example would be the numbers 3
and 4
. The multiples of 3
are 3, 6, 9, 12, 15, 18, ...
and the multiples of 4
are 4, 8, 12, 16, 20, ...
. The first smallest number we run into in both lists is 12
so this is the smallest common multiple between 3
and 4
.
This problem can be confusing because most people look for the smallest common multiple of just the two numbers but forget the keyword range. However, this means that if you are given [1,5]
, then you have to check for the smallest common multiple for all the numbers [1,2,3,4,5]
that is evenly divisible by all of them.
Relevant Links
Hint: 1
Create an array with all the numbers that are missing from the original array to make it easier to check when having to check for even division.
try to solve the problem now
Hint: 2
You can use remainder operator (%
) to check if the reminder of a division is 0, which means it is evenly divisible.
try to solve the problem now
Hint: 3
If you sort the array from greatest to smallest, then you can use the first two numbers as a first check for the smallest common multiple. This is because they are more likely to be the smallest common multiple than the lower numbers.
try to solve the problem now
Spoiler Alert!
Solution ahead!
Basic Code Solution:
function smallestCommons(arr) {
// Sort array from greater to lowest
// This line of code was from Adam Doyle (http://github.com/Adoyle2014)
arr.sort(function(a, b) {
return b - a;
});
// Create new array and add all values from greater to smaller from the
// original array.
var newArr = [];
for (var i = arr[0]; i >= arr[1]; i--) {
newArr.push(i);
}
// Variables needed declared outside the loops.
var quot = 0;
var loop = 1;
var n;
// Run code while n is not the same as the array length.
do {
quot = newArr[0] * loop * newArr[1];
for (n = 2; n < newArr.length; n++) {
if (quot % newArr[n] !== 0) {
break;
}
}
loop++;
} while (n !== newArr.length);
return quot;
}
// test here
smallestCommons([1,5]);
Code Explanation:
- Because of the possibility of the smallest common denominator being among the two biggest numbers, it makes sense to check those first, so sort the array.
- Create a new array to sort all the numbers,
newArr
. - Use a descending
for
loop (var i = arr[0]; i >= arr[1]; i--
) to add the numbers from the biggest to the smallest in the new array. - Declare the variables for the quotient so we can access them outside the loop:
- the quotient that’ll be our smallest common multiple (
quot
) - the loop number we’re checking (
loop
) - the index of the array of numbers (
n
)
- the quotient that’ll be our smallest common multiple (
- Use a
do
while
loop to check what we need whilen
is not the same length as the new array. - In the
do
part, we are going to multiply the very first number, times the number of loops, times the second number (quot = newArr[0] * loop * newArr[1];
). - The
loop
part will allows us to increase the number we’re checking beyond the greatest number we have without having to change the algorithm. - We enter a
for
loop that will go fromn
being 2 and going up by one (loop++
) while it is smaller than the array with all the numbers (n < newArr.length
). - If the quotient does not divide evenly (
quot % newArr[n] !== 0
), then stop the loop (break;
). If it is even, then check for the next elements (n++
) in the array until it is not even or we find our answer. - Outside the loop, increase the value of loop (
loop++
). - At the end of the loop return the quotient (
return quot;
).
Note: If the array only has two elements, then the for
loop never gets used and the return value is the product of said numbers.
Relevant Links
- JS Array Prototype Sort
- JS For Loops Explained
- JS Array Prototype Push
- JS Do While Loop
- String.length
Intermediate Code Solution:
function smallestCommons(arr) {
var range = [];
for (var i = Math.max(arr[0], arr[1]); i >= Math.min(arr[0], arr[1]); i--) {
range.push(i);
}
// can use reduce() in place of this block
var lcm = range[0];
for (i = 1; i < range.length; i++) {
var GCD = gcd(lcm, range[i]);
lcm = (lcm * range[i]) / GCD;
}
return lcm;
function gcd(x, y) { // Implements the Euclidean Algorithm
if (y === 0)
return x;
else
return gcd(y, x%y);
}
}
// test here
smallestCommons([1,5]);
Code Explanation:
- The first, basic solution requires over 2,000 loops to calculate the test case
smallestCommons([1,13])
, and over 4 million loops to calculatesmallestCommons([1,25])
. This solution evaluatessmallestCommons([1,13])
in around 20 loops andsmallestCommons([1,25])
in 40, by using a more efficient algorithm. - Make an empty array range.
- All numbers between the given range are pushed to range using a
for
loop. - The next block of code implements the Euclidean algorithm, which is used for finding smallest common multiples.
Relevant Links
Advanced Code Solution:
function smallestCommons(arr) {
var min = Math.min.apply(null, arr);
var max = Math.max.apply(null, arr);
var grandLCM;
for (var i=min; i<max; i++) {
if(i===min){
grandLCM = (i * (i+1))/gcd(i, i+1);
}else{
grandLCM = (grandLCM * (i+1))/gcd(grandLCM, i+1);
}
}
return grandLCM;
function gcd(x, y) { // Implements The Euclidean Algorithm
if (y === 0)
return x;
else
return gcd(y, x%y);
}
}
// test here
smallestCommons([1,5]);
Code Explanation:
- Get the minimum (min) and maximum (max) in the arr.
- Variable grandLCM will hold our final result.
- In a single loop, iterate from min to max-1.
- In each iteration: if first iteration, compute the lcm of current and next number in range and hold intermediate result in grandLCM else compute the lcm of previous intermediate result and next number in range.
Comparism to code at:
- Unlike the solution at the link (Run Code) above, only a single for loop is used for range iteration and computation.
- The double loop (for and .reduce()) are replaced with just one
for loop
. That is the only difference.
Worthy of note:
- The gcd function uses recursion. Recursion explained.
- Same result can be achieved with a regular loop, e.g a
for loop
. - Here, in each function call in the recursion, a new execution context is created with its own variables
x
andy
. Hence, this recursion approach is more expensive than using a regular loop.
Relevant Links
NOTES FOR CONTRIBUTIONS:
- DO NOT add solutions that are similar to any existing solutions. If you think it is similar but better, then try to merge (or replace) the existing similar solution.
- Add an explanation of your solution.
- Categorize the solution in one of the following categories — Basic, Intermediate and Advanced.
- Please add your username only if you have added any relevant main contents. ( DO NOT remove any existing usernames)
See
Wiki Challenge Solution Template
for reference.