# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// 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 ` `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; ` `} `

## Java

 `// 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 findFactors(``int` `n, ``int` `k) ` `{ ` ` `  `    ``Vector factors = ``new` `Vector(); ` `    ``// 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 findDigitSum(Vector 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 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 factors = findFactors(n, k); ` ` `  `    ``// Sum of digits for each ` `    ``// element in vector ` `    ``Vector digitSum = findDigitSum(factors); ` ` `  `    ``// Sorting the digitSum vector ` `     `  `    ``Collections.sort(digitSum); ` ` `  `    ``// Removing the duplicate elements ` ` `  `    ``HashSet hs1 = ``new` `HashSet(digitSum); ` ` `  `    ``//"HashSet" is stores only unique elements ` ` `  `    ``Vector vect2 = ``new` `Vector(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 `

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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Improved By : SoumikMondal