freeCodeCamp Challenge Guide: Smallest Common Multiple

Smallest Common Multiple


Problem Explanation

The smallest common multiple between two numbers is the smallest number that both numbers can divide into evenly. This concept can be extended to more than two numbers as well.

We can first start with finding the smallest common multiple between two numbers. Naively, we can start writing out multiple of each number until we 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.

An faster approach is to check all multiples of 4 to see if they are also multiples of 3, by checking the remainder when we divide the multiple of 4 by 3.

Be careful - do not forget the keyword range. If we are given [1, 5], then we have to check for the smallest common multiple for all the numbers [1, 2, 3, 4, 5], which is the smallest number that is evenly divisible by all of them.

Relevant Links


Hints

Hint 1

You can use remainder operator (%) to check if the reminder of a division is 0, which means it is evenly divisible.


Solutions

Solution 1 - Looping approach (Click to Show/Hide)
function smallestCommons(arr) {
  // Setup
  const [min, max] = arr.sort((a, b) => a - b);
  const numberDivisors = max - min + 1;
  // Largest possible value for SCM
  let upperBound = 1;
  for (let i = min; i <= max; i++) {
    upperBound *= i;
  }
  // Test all multiples of 'max'
  for (let multiple = max; multiple <= upperBound; multiple += max) {
    // Check if every value in range divides 'multiple'
    let divisorCount = 0;
    for (let i = min; i <= max; i++) {
      // Count divisors
      if (multiple % i === 0) {
        divisorCount += 1;
      }
    }
    if (divisorCount === numberDivisors) {
      return multiple;
    }
  }
}

smallestCommons([1, 5]);

In this solution, we check every multiple of the largest value in the range until we find a value that is divisible by every value in the range.

The upper bound for this loop is the product of all values in the provided range, because this number will be divisible by every value in the range.

Solution 2 - ES6 looping (Click to Show/Hide)
function smallestCommons(arr) {
  // Setup
  const [min, max] = arr.sort((a, b) => a - b);
  const range = Array(max - min + 1)
    .fill(0)
    .map((_, i) => i + min);
  // Largest possible value for SCM
  const upperBound = range.reduce((prod, curr) => prod * curr);
  // Test all multiples of 'max'
  for (let multiple = max; multiple <= upperBound; multiple += max) {
    // Check if every value in range divides 'multiple'
    const divisible = range.every((value) => multiple % value === 0);
    if (divisible) {
      return multiple;
    }
  }
}

smallestCommons([1, 5]);

This solution uses ES6 syntax to condense the logic in Solution 1.

Solution 3 - GCD and LCM (Click to Show/Hide)
function smallestCommons(arr) {
  // Setup
  const [min, max] = arr.sort((a, b) => a - b);
  const range = Array(max - min + 1)
    .fill(0)
    .map((_, i) => i + min);
  // GCD of two numbers
  // https://en.wikipedia.org/wiki/Greatest_common_divisor#Euclid's_algorithm
  const gcd = (a, b) => (b === 0) ? a : gcd(b, a % b);
  // LCM of two numbers
  // https://en.wikipedia.org/wiki/Least_common_multiple#Using_the_greatest_common_divisor
  const lcm = (a, b) => a * b / gcd(a, b);
  // LCM of multiple numbers
  // https://en.wikipedia.org/wiki/Least_common_multiple#Other
  return range.reduce((multiple, curr) => lcm(multiple, curr));
}

smallestCommons([1, 5]);

This solution uses formulae and algorithms for the Greatest Common Divisor and Least Common Multiple off of Wikipedia to compactly and quickly compute the Smallest Common Multiple.

Solution 4 - Prime factorization (Click to Show/Hide)
// Find the SCM of a range of numbers
function smallestCommons(arr) {
  let primeFactors = {};
  const [min, max] = arr.sort((a, b) => a - b);
  for (let i = min; i <= max; i++) {
    // Factorize number in range
    let primes = getPrimeFactors(i);
    for (let j in primes) {
      // Add factor to set or update number of occurrences
      if (!primeFactors[j] || primes[j] > primeFactors[j]) {
        primeFactors[j] = primes[j]
      }
    }
  }
  // Build SCM from factorization
  let multiple = 1;
  for (let i in primeFactors) {
    multiple *= i ** primeFactors[i]
  }
  return multiple;
}

// Compute prime factors of a number
function getPrimeFactors(num) {
  const factors = {};
  for (let prime = 2; prime <= num; prime++) {
    // Count occurances of factor
    // Note that composite values will not divide num
    while ((num % prime) === 0) {
      factors[prime] = (factors[prime]) ? factors[prime] + 1 : 1;
      num /= prime;
    }
  }
  return factors;
}

smallestCommons([1, 5]);

This solution uses a prime factorization of the numbers in the range to compute the smallest common multiple. In general, Solution 3 is much faster, but with a very large range or very large values, sometimes Solution 3 may trigger a recursion limit in some browsers.

41 Likes

You forgot the indentation on the intermediate and advanced solution, making them harder to read.
Can I correct this myselft or should I ask you to update it?
Thx again.

3 Likes

You can do it yourself.

4 Likes
function smallestCommons(arr) {
  function isValidMultiple(m, min, max) {
    for (var i = min; i < max; i++) {
      if (m % i !== 0) {
        return false;
      }
    }
    
    return true;
  }
  
  var max = Math.max(arr[0], arr[1]);
  var min = Math.min(arr[0], arr[1]);
  var multiple = max;
  
  while (!isValidMultiple(multiple, min, max)) {
    multiple += max;
  }
  
  return multiple;
}
29 Likes

Alternative basic solution:

function smallestCommons(arr) {
  var min = Math.min.apply(null, arr);
  var max = Math.max.apply(null, arr);
  var listOfNum =[];
  while (min<=max){
    
      listOfNum.push(min);
      min++;
  }
  
  var result = listOfNum[0];
  
for (var i = 1; i<listOfNum.length; i++)
    result = findLCM(result, listOfNum[i]);
  
  return result;
}

function gcd(a, b){
  
  while (a !== b){
    
    if(a > b)
      a = a - b;
    else
      b = b - a;
    
  }

  return a; // GCD of two numbers
}

function findLCM(a, b){
  
  return a * (b / gcd (a,b));
  
}
smallestCommons([1,5]);
2 Likes

Dang, after reading all these solution. I just realized I tackled this problem the hardest way possible. LOL

22 Likes

Came up with a short basic solution:

function smallestCommons(arr) {

  var max = Math.max(arr[0], arr[1]);
  var min = Math.min(arr[0], arr[1]);
  var mltple = max;

  for(var i = max; i >= min; i--){
    if(mltple % i !== 0){
      mltple += max; 
      i = max;
    } 
  }

  return mltple;  
}
71 Likes
var tester = 0;

function smallestCommons(arr) {
var ultimate = 0;
  arr = arr.sort(function(a, b) {
    return a - b;
  });

  console.log(arr);

  var ranges = Range(arr[0], arr[1]);

  console.log(ranges);

  var first = arr[1];

  var final = first;

  ///function will run here
  recurse(first);

  function recurse(current) {

    var temp = 0;
    var tester = 0;
    for (i = arr[0]; i <= arr[1]; i++) {

      if (current % i === 0) {
        console.log(current + " % " + i);
        tester++;
        console.log(i + " works");
        console.log(tester + " tester");

      } else {

        console.log(i + " broke it");

        var counter = current + first;
        
        console.log(temp + " next run");
        recurse(counter);

        break;
      } //else

      console.log(ranges.length + " is length");

      if (tester === ranges.length) {
        console.log(current);
         ultimate = current;
      }

    } //for

  } //recurse
return ultimate;
} //smallest commons

function Range(a, b) {
  var temrange = [];

  for (i = a; i <= b; i++) {

    temrange.push(i);
  }
  return temrange;
}

smallestCommons([5, 1]);

heres my long and messy code that crashes my browser if the numbers are too big, it works for 1,5 but crashes for 1,13.

2 Likes

function smallestCommons(arr) {

var maxPrimes = {};

var min = Math.min(arr[0], arr[1]);
var max = Math.max(arr[0], arr[1]);

var primeFactors = function (num) {
    var primeProd = {};
    var factor = 2;

    if (num < 2) {
      return {};
    } 

    while(factor <= num) {
      if(num % factor === 0){
        primeProd[factor] = primeProd[factor] + 1 || 1;
        num /= factor;
      } else {
        factor++;
      }
    }

    return primeProd;
};

//loop through numbers and save the most instance of the prime factors in the main hash 
for(var i = min; i <= max; i++) {
  var temp = primeFactors(i);
  for(var key in temp) {
    if(!(key in maxPrimes) || maxPrimes[key] < temp[key]) {
      maxPrimes[key] = temp[key];
    }
  }
}

var prod = 1;
//loop through main hash to perform multiplication of all prime factors

for(var num in maxPrimes) {
  prod *= Math.pow(parseInt(num), maxPrimes[num]);
}

return prod;

}

Here’s what I ended up. Not optimized. The brute force method that is available right now was inefficient and broke down at [1, 23]. Works by making a hash of prime factors for each number and calculating the SCM through a main hash.

1 Like

I factored all the numbers in the range and then multiplied them all together.

function smallestCommons(arr) {
  
  var allFactors = { };
  
  var tmpArr = [];
  
  var curNum;
  
  var factorKeys = {};
  
  var count = 0;
  
  var theProduct = 1;
  
  var tmpNum;
  
  if (arr[1] < arr[0]) {
    tmpNum = arr[1];
    arr[1] = arr[0];
    arr[0] = tmpNum;
  }
  
  for (var i = arr[0] > 1 ? arr[0] : 2, l = arr[1]; i <= l; i++) {
    tmpArr = factorMe(i);
    for (var j = 0, l2 = tmpArr.length; j < l2; j++) {
      if (!curNum || curNum != tmpArr[j]) {
        curNum = tmpArr[j];
        count = countNumbers(curNum, tmpArr);
        if (!allFactors[curNum] || allFactors[curNum] < count) {
          allFactors[curNum] = count;
        }
      }
    }
  }
  
  factorKeys = Object.keys(allFactors);

  for (var k = 0, l3 = factorKeys.length; k < l3; k++) {
    theProduct *= Math.pow( factorKeys[k], allFactors[factorKeys[k]] );
  }
 
  return theProduct;
}

function countNumbers(num, arr) {
  var count = 0;
  for (var i = 0, l = arr.length; i < l; i++) {
    if (num == arr[i]) count++;
  }
  return count;
}

function factorMe(num) {
  var factors = [];
  
  var divisor = 2;
  
  while(divisor != num) {
    if ( num % divisor === 0) {
      factors.push(divisor);
      num /= divisor;
    } else {
    divisor++;
    }
  }
  
  factors.push(num);

  
  return factors;
}


smallestCommons([1,5]);

I have done it by getting the highest common multiple first then simplify it.

function smallestCommons(arr) {
  var sorted = arr.sort(),
      res = sorted[1],
      multiplier = sorted[1] - 1,
      simplifier = [2, 3, 5, 7],
      leastSearch = true;
 
 // Get the highest common multiple
  while(leastSearch){
    for(var i = sorted[0]; i <= sorted[1]; i++){
      
      if((res % i) !== 0){
        leastSearch = true;
        break;
      } 
      
      leastSearch = false;
    }
    
    if(leastSearch){
      res *= multiplier;
      multiplier--;
    }
    
  }
  
// Simplify the value
  for(var j = 0; j < simplifier.length; j++){
    var canBeSimplified = true;
    var tempRes;
    
    do {
      tempRes = res / simplifier[j];
      for(var x = sorted[0]; x <= sorted[1]; x++ ){
        if(tempRes % x !== 0){
          canBeSimplified = false;
          tempRes = tempRes * simplifier[j];
          break;
        }
      }
      
      res = tempRes;
    } while (canBeSimplified);
    
  }
  
  
  return res;
}


smallestCommons([1, 13]);

Simple and elegant. Well done

Below is how I pass this challenge, fast and simple:

function smallestCommons(arr) {
  var min=arr[0], max=arr[1], temp;
  if (min>max) {temp=min;min=max;max=temp;}
  var a, b, lim=min;
  for(var i=min;i<max;i++) {
    a=lim;
    b=i+1;
    for (lim; lim%b!==0 ;lim+=a);
  }
  
  return lim;
}


smallestCommons([1,5]);

Run Code

9 Likes

I was able to pass this challenge with this piece of code. Not as efficient as some others I’ve seen here but glad that it works. I’m still new to this whole javascript thing and always looking for ways to make my code more efficient so any feedback is welcome :).

function smallestCommons(arr) {

var newArr = []
var acc = Math.min(…arr);
var firstVal = arr[0];
var secondVal = arr[1];
var boo = 0;

// Sequential Array

 while (acc< Math.max(...arr)+1) {

    newArr.push(acc);
    acc += 1;
  } 

while (boo == 0){
while (firstVal != secondVal)
{
if (firstVal>secondVal) {
secondVal += arr[1];
}

        else if (firstVal<secondVal) {    
          firstVal += arr[0];
        }

    }   

  for (i=1;i<newArr.length-1;i++) {

    if (firstVal % newArr[i] != 0) {  
      
      firstVal += arr[0];
      break;  
    }
    else if (i <newArr.length -2){
      continue; 
    }   
 
    else if (i = newArr.length-2) {
      var boo = 1; 
      continue;
    }
} 

}

return firstVal;

}
smallestCommons([1,13]);

Advanced solution.
This code is quite long, and firstly I thought I tackled this problem the hardest way possible, but after some testing I notices it can astonishingly compute the Smallest Common Multiple (SMC) of [1, 5000] in a few seconds…

// Check if value is a prime number
function isPrime (value) {
  
  if(value%2 === 0 && value > 2)
    return false;
  
  var halfValue = (value-1)/2;
  
  for(var i = 3; i < halfValue; i+=2)
    if(value%i === 0) 
      return false;
  
  return true;
}

// Give the prime factorization of a number
function primeFactors (value) {
  var i = 2;
  var factors = {};
  
  while( value >= i ) 
  {
    if( isPrime(i) ) 
    {
      while( value % i === 0 ) 
      {
        value /= i;
        factors[i] = factors[i] ? factors[i] + 1 : 1; 
      }
      i++; 
    }
    else 
      i++; 
  }
  return factors;
}

// Give the highest value of each factor of 2 primes factorization
function addFactors (f1, f2) {
  Object.keys(f2).forEach 
  (
    function(key) 
    {
      if( ( f1[key] || 0 ) < f2[key] )
        f1[key] = f2[key]; 
    } 
  );
  return f1;
}

// Calculate the smallest common multiple
function smallestCommons(arr) {
  var scm = 1;
  var factors = {};
  
  var min = Math.max( Math.min( arr[0], arr[1] ), 2 );
  var max = Math.max( arr[0], arr[1] );
  
  for(var i = min; i <= max; i++) 
  {
    factors = addFactors(factors,primeFactors(i)); 
  }
  
  Object.keys(factors).forEach 
  (
    function(key) {
      scm *= Math.pow(key, factors[key]); 
    } 
  );
  
  return scm; 
}

console.log(smallestCommons([1,5000]));

So how does it works ?

This code is based on the Prime Factorization (PF) of numbers
e.g : 720 = 2x2x2x2 x3x3 x5

Let’s see the PF of the SMC of [1, 9] :

  • 2520 = 2 x 2 x 2 x 3 x 3 x 5 x 7
    This can be computed with the function primeFactors(2520), which returns
    = > { 2: 3 , 3: 2 , 5: 1 , 7: 1 }

The we do the PF of all numbers from 1 to 9 :

n      2   3   5   7  
---------------------
2    = 1   |   |   |  
3    = |   1   |   |      
4    = 2   |   |   |     
5    = |   |   1   |  
6    = 1   1   |   |  
7    = |   |   |   1   
8    = 3   |   |   |  
9    = |   2   |   |  
---------------------
2520 = 3   2   1   1

We can notice that each value of PF(2520) is the highest value possible in all PF from 1 to 9

The function addFactors return the PF with all the highest possible values.
Then all that is left to do is the inverse of PF.

Fell free to correct / improve my english, it’s quite difficult to explain in another langage.

3 Likes

I think the product should instand of quot in Basic Code Solution

I wrote a similar code, but I was only able to calculate up to the hundreds before it give me a null object.

I copied your code to repl.it online coding environment to see if it works as well as you said, but it gave me infinity. I want to know if this is a JavaScript problem or a machine problem that I couldn’t calculate scm(1,5000)

I kinda like this:

function smallestCommons(arr) {
  const min = Math.min(arr[0],arr[1]), 
        max = Math.max(arr[0],arr[1]),
        range = [...Array(max+1).keys()].slice(min); 

  return range.reduce(function(a,b){
    for (let i = a; i<=a*b; i+=a){
      if (i % b===0 ) return i;
    } 
  });
}
5 Likes