Largest Number with Given Set of N Digits Divisible by 2, 3, and 5 in C++

In this tutorial, we are going to write a program that finds the largest number formed from the array that is divisible by 2, 3, and 5.

Let's see the steps to solve the problem.

• Initialise the array.
• The number must end with 0 and the sum of all the numbers should be divisible by 3 to be divisible by 2, 3, and 5.
• Check for the 0 in the array and print not possible if it's not present in the array.
• Sort the array in descending order.
• Find the remainder for sum % 3.
• If the remainder is not 1, then delete all the digits from the end whose remainder for digit % 3 is equal to the above remainder.
• If there are no digits with the same remainder as above, then subtract 3 from the above remainder and delete the last two digits whose remainder is the same as above.
• Print all the digits from the array.

Example

Let's see the code.

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void findLargestDivibleNumber(int n, vector<int>& v){
   int flag = 0;
   long long sum = 0;
   for (int i = 0; i < n; i++) {
      if (v[i] == 0) {
         flag = 1;
      }
      sum += v[i];
   }
   if (!flag) {
      cout << "Not possible" << endl;
   }else {
      sort(v.begin(), v.end(), greater<int>());
      if (v[0] == 0) {
         cout << "0" << endl;
      }else {
         int flag = 0;
         int remainder = sum % 3;
         if (remainder != 0) {
            for (int i = n - 1; i >= 0; i--) {
               if (v[i] % 3 == remainder) {
                  v.erase(v.begin() + i);
                  flag = 1;
                  break;
               }
            }
            if (flag == 0) {
               remainder = 3 - remainder;
               int count = 0;
               for (int i = n - 1; i >= 0; i--) {
                  if (v[i] % 3 == remainder) {
                     v.erase(v.begin() + i);
                     count++;
                     if (count >= 2) {
                        break;
                     }
                  }
               }
            }
         }
         if (*v.begin() == 0) {
            cout << "0" << endl;
         }else {
            for (int i : v) {
               cout << i;
            }
         }
      }
   }
}
int main() {
   int n = 9;
   vector<int> v{ 4, 5, 0, 3, 2, 4, 5, 6, 7 };
   findLargestDivibleNumber(n, v);
   return 0;
}

Output

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

765544320

Conclusion

If you have any queries in the tutorial, mention them in the comment section.