Sum of M maximum distinct digit sum from 1 to N that are factors of K

Given an array of natural numbers upto N and two numbers M and K, the task is to find the sum of M maximum distinct digit sum M numbers from N natural numbers which are factors of K.

Examples:

Input: N = 50, M = 4, K = 30
Output: 16
Explanation:
From 1 to 50, factors of 30 = {1, 2, 3, 5, 6, 10, 15, 30}.
Digit sum of every factor = {1, 2, 3, 5, 6, 1, 6, 3}.
4 largest digit sum = 2, 3, 5, 6.
Sum = 16

Input: N = 5, M = 3, K = 74
Output: 3
Explanation:
From 1 to 5 factors of 74 = {1, 2}
Digit sum of every factor = {1, 2}.
3 largest digit sum = 1, 2 (Here only 2 such numbers are present. But it is asked for atmost M. So these 2 will be considered only.)
Sum = 3

Naive Approach: Simple solution is to run the loop from 1 to N and find the list of all numbers that perfectly divides K. Find the digit sum of each factor and sort them in descending order and print M distinct elements from this list from top.



Efficient Approach:

  • Find all the factors of K in the range of 1 to N by iterating from 2 to √K such that the element completely divides the number. For Detailed Explanation of this approach please refer this article and store them in an array.
  • Find the digit sum of the numbers stored in the factors array.
    For Example:

    For the Given Array - {4, 10, 273 }
    
    Digit Sum of the Elements - 
    Element 0 Digit Sum - "4" = 4
    Element 1 Digit Sum - "10" = 1 + 0 = 10
    Element 2 Digit Sum - "273" = 2 + 7 + 3 = 12
    
  • Remove duplicates from the digit sum as we need distinct elements.
  • Sort the distinct digit sums in reverse order and find the sum of first at most M elements out of this digit-sum array.

Below is the implementation of the above approach.

C++

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// C++ implementation to find the sum of
// maximum distinct digit sum of at most
// M numbers from 1 to N that are factors of K
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to find the factors
// of K in N
vector<int> findFactors(int n, int k)
{
  
    // Initialise a vector
    vector<int> factors;
  
    // Find out the factors of
    // K less than N
    for (int i = 1; i <= sqrt(k); i++) {
        if (k % i == 0) {
            if (k / i == i && i <= n)
                factors.push_back(i);
            else {
                if (i <= n)
                    factors.push_back(i);
                if (k / i <= n)
                    factors.push_back(k / i);
            }
        }
    }
  
    return factors;
}
  
// Find the digit sum of each factor
vector<int> findDigitSum(vector<int> a)
{
  
    // Sum of digits for each
    // element in vector
    for (int i = 0; i < a.size(); i++) {
        int c = 0;
        while (a[i] > 0) {
  
            c += a[i] % 10;
            a[i] = a[i] / 10;
        }
        a[i] = c;
    }
  
    return a;
}
  
// Find the largest M distinct digit
// sum from the digitSum vector
int findMMaxDistinctDigitSum(
    vector<int> distinctDigitSum,
    int m)
{
    // Find the sum of last M numbers.
    int sum = 0;
    for (int i = distinctDigitSum.size() - 1;
         i >= 0 && m > 0;
         i--, m--)
        sum += distinctDigitSum[i];
  
    return sum;
}
  
// Find the at most M numbers from N natural
// numbers whose digit sum is distinct
// and those M numbers are factors of K.
int findDistinctMnumbers(int n, int k, int m)
{
  
    // Find out the factors of
    // K less than N
    vector<int> factors = findFactors(n, k);
  
    // Sum of digits for each
    // element in vector
    vector<int> digitSum = findDigitSum(factors);
  
    // Sorting the digitSum vector
    sort(digitSum.begin(), digitSum.end());
  
    // Removing the duplicate elements
    vector<int>::iterator ip;
    ip = unique(digitSum.begin(), digitSum.end());
  
    digitSum.resize(distance(
        digitSum.begin(),
        ip));
  
    // Finding the sum and returning it
    return findMMaxDistinctDigitSum(digitSum, m);
}
  
// Driver Code
int main()
{
    int n = 100, k = 80, m = 4;
  
    // Function Call
    cout
        << findDistinctMnumbers(n, k, m)
        << endl;
  
    return 0;
}

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Java

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// Java implementation to find the sum of
// maximum distinct digit sum of at most
// M numbers from 1 to N that are factors of K
import java.util.*;
  
class GFG
{
  
// Function to find the factors
// of K in N
public static Vector<Integer> findFactors(int n, int k)
{
  
    Vector<Integer> factors = new Vector<Integer>();
    // Initialise a vector
  
    // Find out the factors of
    // K less than N
    for (int i = 1; i <= Math.sqrt(k); i++)
    {
        if (k % i == 0)
        {
            if (k / i == i && i <= n)
                factors.add(i);
            else 
            {
                if (i <= n)
                    factors.add(i);
                if (k / i <= n)
                    factors.add(k / i);
            }
        }
    }
  
    return factors;
}
  
// Find the digit sum of each factor
public static Vector<Integer> findDigitSum(Vector<Integer> a)
{
  
    // Sum of digits for each
    // element in vector
    for (int i = 0; i < a.size(); i++)
    {
        int c = 0;
        while (a.get(i) > 0)
        {
  
            c += (a.get(i) % 10);
            a.set(i,(a.get(i)/10));
        }
        a.set(i,c);
    }
  
    return a;
}
  
// Find the largest M distinct digit
// sum from the digitSum vector
public static int findMMaxDistinctDigitSum(Vector<Integer> distinctDigitSum,int m)
{
    // Find the sum of last M numbers.
    int sum = 0;
    for (int i = distinctDigitSum.size() - 1;
            i >= 0 && m > 0;i--, m--)
        sum += distinctDigitSum.get(i);
  
    return sum;
}
  
// Find the at most M numbers from N natural
// numbers whose digit sum is distinct
// and those M numbers are factors of K.
public static int findDistinctMnumbers(int n, int k, int m)
{
  
    // Find out the factors of
    // K less than N
    Vector<Integer> factors = findFactors(n, k);
  
    // Sum of digits for each
    // element in vector
    Vector<Integer> digitSum = findDigitSum(factors);
  
    // Sorting the digitSum vector
      
    Collections.sort(digitSum);
  
    // Removing the duplicate elements
  
    HashSet<Integer> hs1 = new HashSet<Integer>(digitSum);
  
    //"HashSet" is stores only unique elements
  
    Vector<Integer> vect2 = new Vector<Integer>(hs1);
  
    // Finding the sum and returning it
    return findMMaxDistinctDigitSum(vect2, m);
}
  
// Driver Code
public static void main(String args[])
{
    int n = 100, k = 80, m = 4;
  
    // Function Call
    System.out.println(findDistinctMnumbers(n, k, m));
}
}
  
// This code is contributed by SoumikMondal

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Output:

24

Performance Analysis:

  • Time Complexity: In the given approach, there are mainly two process which is as follows –
    • Time Complexity to find the factors of a number is: O(√(K))
    • Time complexity to sort and store the unique elements: O(√(K)log(√(K)))

    So, Overall complexity of the approach is: O(√(K) + √(K)log(√(K)))

  • Auxiliary Space: In the given approach there is an extra array used to store the factors of the number K, which is: O(√(K))

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