Largest Multiple of Three in C++

C++Server Side ProgrammingProgramming

Suppose we have one array of digits, we have to find the largest multiple of three that can be formed by concatenating some of the given digits in any order as we want. The answer may be very large so make it as string. If there is no answer return an empty string.

So, if the input is like [7,2,8], then the output will be 87

To solve this, we will follow these steps −

  • Define one 2D array d, there will be three rows

  • sort the array digits

  • sum := 0

  • for initialize i := 0, when i < size of digits, update (increase i by 1), do−

    • x := digits[i]

    • insert digits[i] at the end of d[x mod 3]

    • sum := sum + x

    • sum := sum mod 3

  • if sum is non-zero, then −

    • if not size of d[sum], then −

      • rem := 3 - sum

      • if size of d[rem] < 2, then −

        • return empty string

      • delete last element from d[rem] twice

    • Otherwise

      • delete last element from d[sum]

  • ret := empty string

  • for initialize i := 0, when i < 3, update (increase i by 1), do −

    • for initialize j := 0, when j < size of d[i], update (increase j by 1), do−

      • ret := ret concatenate d[i, j] as string

  • sort the array ret

  • if size of ret and ret[0] is same as '0', then −

    • return "0"

  • return "0"

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   string largestMultipleOfThree(vector<int>& digits) {
      vector<vector<int>> d(3);
      sort(digits.begin(), digits.end(), greater<int>());
      int sum = 0;
      for (int i = 0; i < digits.size(); i++) {
         int x = digits[i];
         d[x % 3].push_back(digits[i]);
         sum += x;
         sum %= 3;
      }
      if (sum) {
         if (!d[sum].size()) {
            int rem = 3 - sum;
            if (d[rem].size() < 2)
            return "";
            d[rem].pop_back();
            d[rem].pop_back();
         }
         else {
            d[sum].pop_back();
         }
      }
      string ret = "";
      for (int i = 0; i < 3; i++) {
         for (int j = 0; j < d[i].size(); j++) {
            ret += to_string(d[i][j]);
         }
      }
      sort(ret.begin(), ret.end(), greater<int>());
      if (ret.size() && ret[0] == '0')
      return "0";
      return ret;
   }
};
main(){
   Solution ob;
   vector<int> v = {7,2,8};
   cout << (ob.largestMultipleOfThree(v));
}

Input

{7,2,8}

Output

87
raja
Published on 09-Jun-2020 06:50:51
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