Locate the product of n numerals within a specified interval

Question

Locate the product of n numerals within a specified interval

Is there a way to determine the number of multiples for N unique numbers (provided as an array input) within the range of 1 to K, where 1 < K < 10⁸ and 3 ≤ N < 25?

function findNumberOfMultiples(inputArray, maxSize) {   
    var count = 0;
    var tempArray = [];
    for (var i=0; i<maxSize; i++){
        tempArray[i] = 0;
    }

    for (var j=0; j<inputArray.length; j++) {
        for (var i=1; i<=maxSize; i++) {
            if (i % inputArray[j]) {
                tempArray[i-1] = 1;
            }
        }
    }

    for (var i=0; i<maxSize; i++) {
        if (tempArray[i]==1) {
            count++;
        }
    }
 return count;
}

This program may not be efficient when dealing with large values of K. For instance, if you have inputArray = [2,3,4] and the value of maxSize(k) is 5,

Multiples of 2: 2, 4
Multiples of 3: 3
Multiples of 4: 4

Therefore, the total number of multiples of either 2, 3, or 4 in the range of 1 to 5 is 3.

javascript algorithm math

Answer 1

Answer №1

To tackle this problem, you can utilize an algorithm with a time complexity of O(N^2), where N represents the number of elements in your given array.

Imagine having two elements in your array, denoted as [a1,a2], with a specified range of K.

Your resulting formula would be:

  K/a1 + K/a2 - K/lcm(a1,a2) // due to adding them in both a1 and a2

Therefore, if you have a1,...,ax elements, the equation becomes:

K/a1+...K/ax - K/lcm(ai,aj) (substituting i,j with (n*n-1)/2 combinations)

The operation K/lcm(ai,aj) will run O(N^2) times ((n*n-1)/2 times exactly). Hence, the overall complexity is O(N^2) (with a Log(min(ai,aj)) element that doesn't significantly impact the overall performance).

 public int combinations(int K, int[] input){
    int total = 0;
    for(int i=0;i<input.length;i++){
        total  =  total + Math.floor(K/input[i]);
    }
    for(int i=0;i<input.length;i++){
        for(int j=i+1;j<input.length;j++){
            if(i!=j){
                int lcm =lcmFind(input[i], input[j]);
                total = total - Math.floor(K/lcm);
            }
        }
    }
    return total;
}

The specific test case provided:

Answer 2

To tackle this problem, you can utilize an algorithm with a time complexity of O(N^2), where N represents the number of elements in your given array.

Imagine having two elements in your array, denoted as [a1,a2], with a specified range of K.

Your resulting formula would be:

  K/a1 + K/a2 - K/lcm(a1,a2) // due to adding them in both a1 and a2

Therefore, if you have a1,...,ax elements, the equation becomes:

K/a1+...K/ax - K/lcm(ai,aj) (substituting i,j with (n*n-1)/2 combinations)

The operation K/lcm(ai,aj) will run O(N^2) times ((n*n-1)/2 times exactly). Hence, the overall complexity is O(N^2) (with a Log(min(ai,aj)) element that doesn't significantly impact the overall performance).

 public int combinations(int K, int[] input){
    int total = 0;
    for(int i=0;i<input.length;i++){
        total  =  total + Math.floor(K/input[i]);
    }
    for(int i=0;i<input.length;i++){
        for(int j=i+1;j<input.length;j++){
            if(i!=j){
                int lcm =lcmFind(input[i], input[j]);
                total = total - Math.floor(K/lcm);
            }
        }
    }
    return total;
}

The specific test case provided:

Answer 3

Answer №2

Here is a useful function that gets the job done:

const countMultiples = (arr, max) => {
  const multiplesArr = [];
  for (let j = 0; j < arr.length; j++) {
    const num = arr[j];
    const limit = max / num;
    for (let i = 1; i < limit; i++) {
      const multiple = i * num;
      if (!multiplesArr.includes(multiple)) {
        multiplesArr.push(multiple);
      }
    }
  }
  return multiplesArr.length;
};

Note: While this function does not cause any stack errors, using large values for the range might lead to browser freezing issues.

Answer 4

Here is a useful function that gets the job done:

const countMultiples = (arr, max) => {
  const multiplesArr = [];
  for (let j = 0; j < arr.length; j++) {
    const num = arr[j];
    const limit = max / num;
    for (let i = 1; i < limit; i++) {
      const multiple = i * num;
      if (!multiplesArr.includes(multiple)) {
        multiplesArr.push(multiple);
      }
    }
  }
  return multiplesArr.length;
};

Note: While this function does not cause any stack errors, using large values for the range might lead to browser freezing issues.

Locate the product of n numerals within a specified interval

Answer №1

Answer №2

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