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Count numbers whose maximum sum of distinct digit-sum is less than or equals M
• Difficulty Level : Expert
• Last Updated : 08 Apr, 2020

Given an array of integers arr[] and a number M, the task is to find the maximum count of the numbers whose sum of distinct digit-sum is less than or equal to the given number M.

Examples:

Input: arr[] = {1, 45, 16, 17, 219, 32, 22}, M = 10
Output: 3
Explanation:
Digit-sum of the Array is – {1, 9, 7, 8, 12, 5, 4}
Max sum of distinct digit-sum whose sum less than M is {1 + 5 + 4}
Hence, the count of the such numbers is 3.

Input: arr[] = {32, 45}, M = 2
Output: 0
Explanation:
Digit-sum of the Array is – {5, 9}
Max sum of distinct digit-sum less than M is 0
Hence, the count of the such numbers is 0.

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

Approach:
The idea is to find the digit-sum of every element in the array and then sort the digit-sum array.

Now the problem boils down to count number of elements from the sorted distinct digit sum array, with sum less than or equal to M.

To do this, take the minimum distinct digit-sums until the sum of such numbers is less than or equal to the given number M and return the count of such numbers.

Explanation with Example:

```Given Array be - arr[] = {1, 45, 17, 32, 22}, M = 10

Then Digit-sum of each number in the array -
Digit-sum(1) = 1
Digit-sum(45) = 4 + 5 = 9
Digit-sum(17) = 1 + 7 = 8
Digit-sum(32) = 3 + 2 = 5
Digit-sum(22) = 2 + 2 = 4

After sorting the digit-sum array -
Digit-sum[] = {1, 4, 5, 8, 9}

Then there are three numbers such that,
there sum is less than or equal to M = 10
which is {1, 4, 5}
Sum = 1 + 4 + 5 = 10 &leq; M
```

Algorithm:

• Find the digit-sum of every element of the array and store it in another array(say digit-sum[])
• Sort the digit-sum[] array in increasing order.
• Remove Duplicate elements from the sorted digit-sum[] array such that there are only unqiue digit-sum.
• Intialize a variable sum to 0 to store the current sum.
• Iterate over the digit-sum[] array and add the elements to the sum untill the value of the sum is less than equal to M and increment the count.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to find the ``// Maximum count of numbers whose``// sum of distinct digit-sum less``// than or equal to the given number`` ` `#include ``using` `namespace` `std;`` ` `// Function to find the ``// digit-sum of a number ``int` `SumofDigits(``int` `digit)``{``    ``int` `sum = 0;``     ` `    ``// Loop to iterate the number``    ``// digit-wise to find digit-sum``    ``while` `(digit != 0) {``         ` `        ``// variable to store last digit``        ``int` `rem = digit % 10;``        ``sum += rem;``        ``digit /= 10;``    ``}``    ``return` `sum;``}`` ` `// Function to find the count of number``int` `findCountofNumbers(``int` `arr[], ``                   ``int` `n, ``int` `M){``     ` `    ``// Vector to store the Sum of Digits``    ``vector<``int``> SumDigits;`` ` `    ``// Sum of digits for each``    ``// element in vector``    ``for` `(``int` `i = 0; i < n; i++) {``        ``int` `s = SumofDigits(arr[i]);``        ``SumDigits.push_back(s);``    ``}`` ` `    ``// Sorting the digitSum vector``    ``sort(SumDigits.begin(), SumDigits.end());`` ` `    ``// Removing the duplicate elements``    ``vector<``int``>::iterator ip;``    ``ip = unique(SumDigits.begin(), ``                 ``SumDigits.end());``    ``SumDigits.resize(distance(``         ``SumDigits.begin(), ip)``         ``);`` ` `    ``// Count variable to store the Count``    ``int` `count = 0;``    ``int` `sum = 0;``    ``// Finding the Count of Numbers``    ``for` `(``int` `i = 0; i < SumDigits.size(); i++) {``        ``if` `(sum > M)``            ``break``;``        ``sum += SumDigits[i];``        ``if` `(sum <= M)``            ``count++;``    ``}``    ``return` `count;``}`` ` `// Driver Code``int` `main()``{`` ` `    ``int` `arr[] = { 1, 45, 16, 17, ``           ``219, 32, 22 }, M = 10;``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`` ` `    ``// Function Call``    ``cout << findCountofNumbers(arr, n, M);``    ``return` `0;``}`

## Python3

 `# Python 3 implementation to find the ``# Maximum count of numbers whose``# sum of distinct digit-sum less``# than or equal to the given number`` ` `# Function to find the ``# digit-sum of a number ``def` `SumofDigits( digit):``     ` `    ``sum` `=` `0``     ` `    ``# Loop to iterate the number``    ``# digit-wise to find digit-sum``    ``while` `(digit !``=` `0``):``         ` `        ``# variable to store last digit``        ``rem ``=` `digit ``%` `10``        ``sum` `+``=` `rem``        ``digit ``/``/``=` `10``     ` `    ``return` `sum`` ` `# Function to find the count of number``def` `findCountofNumbers(arr, n, M):``     ` `    ``# Vector to store the Sum of Digits``    ``SumDigits ``=` `[]`` ` `    ``# Sum of digits for each``    ``# element in vector``    ``for` `i ``in` `range``( n ):``        ``s ``=` `SumofDigits(arr[i])``        ``SumDigits.append(s)`` ` `    ``# Sorting the digitSum vector``    ``SumDigits.sort()`` ` `    ``# Removing the duplicate elements``    ``ip ``=` `list``(``set``(SumDigits))`` ` `    ``# Count variable to store the Count``    ``count ``=` `0``    ``sum` `=` `0``     ` `    ``# Finding the Count of Numbers``    ``for` `i ``in` `range``(``len``(SumDigits)):``        ``if` `(``sum` `> M):``            ``break``        ``sum` `+``=` `SumDigits[i]``        ``if` `(``sum` `<``=` `M):``            ``count``+``=``1``     ` `    ``return` `count`` ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:`` ` `    ``arr ``=` `[ ``1``, ``45``, ``16``, ``17``, ``        ``219``, ``32``, ``22` `]``    ``M ``=` `10``    ``n ``=` `len``(arr)`` ` `    ``# Function Call``    ``print``( findCountofNumbers(arr, n, M))``     ` `# This ccode is contributed by chitranayal`
Output:
```3
```

Performance Analysis:

• Time Complexity: As in the given approach, we are using sorting which takes O(NlogN) in worst-case and the to find the digit-sum of every element takes O(N*K) where K is the maximum number of digits, Hence the time complexity will be O(NlogN + N*k).
• Auxiliary Space: O(N)

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