N digit numbers divisible by 5 formed from the M digits in C++

We have given a number N along with an array of M digits. Our job is to find the number of n digit numbers formed from the given M digits that are divisible by 5.

Let's see some examples to understand the problem inputs and outputs.

In −

N = 2
M = 3
arr = {5, 6, 3}

Out −

2

There are 2 N digit numbers 35 and 65 possible that are divisible by 5. Let's see another example.

Input −

N = 1
M = 7
arr = {2, 3, 4, 5, 6, 7, 8}

Output −

1

There is only 1 number with 1 digit in the given array that is divisible by 5. So, our task is to find the number of numbers that can be formed from the given number with N digits that are divisible by 5.

The number must end with the digits 0 or 5 to be divisible by 5. Let's see the algorithm

Algorithm

• Check for the 0 and 5 in the given array. 2. If there are both 0 and 5 then there are two ways to place a digit in the units place. Otherwise, there will be a single way to place a digit.
  • Initialise the count to 2.
  • Now, the remaining place can have m - 1, m - 2, m - 3, ... n ways to fill them respectively.
  • Write a loop that iterates from 0 to n - 1.
    • Decrement that size of the array.
    • Multiply it with count.
• If there is a single digit 0 or 5 then there is only one way to place a digit in the units place.
  • Initialise the count to 2.
  • Now, the remaining place can have m - 1, m - 2, m - 3, ... n ways to fill them respectively.
  • Write a loop that iterates from 0 to n - 1.
    • Decrement that size of the array.
    • Multiply it with count.
• If there are no digits 0 or 5 than we can form a number that can be divisible by 5. Return -1 in that case.

Implementation

Following is the implementation of the above algorithm in C++

#include <bits/stdc++.h>

using namespace std;

int numbers(int n, int m, int arr[]) {
   bool isZeroPresent = false, isFivePresent = false;
   int numbersCount = 0;
   if (m < n) {
      return -1;
   }
   for (int i = 0; i < m; i++) {
      if (arr[i] == 0) {
         isZeroPresent = true;
      }
      if (arr[i] == 5) {
         isFivePresent = true;
      }
   }
   if (isZeroPresent && isFivePresent) {
      numbersCount = 2;
      for (int i = 0; i < n - 1; i++) {
         m--;
         numbersCount = numbersCount * m;
      }
   } else if (isZeroPresent || isFivePresent) {
      numbersCount = 1;
      for (int i = 0; i < n - 1; i++) {
         m--;
         numbersCount = numbersCount * m;
      }
   } else {
      return -1;
   }
   return numbersCount;
}
int main() {
   int arr[] = {5, 6, 3};
   cout << numbers(2, 3, arr) << endl;
   return 0;
}

Output

If you run the above code, then you will get the following result.

2