Minimum number of sets with numbers less than Y in C++

C++Server Side ProgrammingProgramming

Problem statement

Given a string of consecutive digits and a number Y, the task is to find the number of minimum sets such that every set follows the below rule −

  • Set should contain consecutive numbers
  • No digit can be used more than once.
  • The number in the set should not be more than Y.

Example

If str = “1234” and Y = 20 then answer is 3 as below sets are created −

{12} {3} and {4}

Algorithm

  • Convert string to number
  • If the number is not greater than Y, then mark f = 1
  • If the number exceeds Y, then increase count if f = 1 and re-initialize f as 0 and also initialise num as s[i]-‘0’ or num as 0
  • After iterating in the string completely, then increase count if f is 1

Example

#include <iostream>
#include <string>
using namespace std;
int getMinSets(string str, int y) {
   int cnt = 0;
   int num = 0;
   int l = str.length();
   int f = 0;
   for (int i = 0; i < l; ++i) {
      num = num * 10 + str[i] - 48;
      if (num <= y) {
         f = 1;
         continue;
      }
      if (f) {
         ++cnt;
      }
      num = str[i] - '0';
      f = 0;
      if (num <= y) {
         f = 1;
      } else {
         num = 0;
      }
   }
   if (f) {
      ++cnt;
   }
   return cnt;
}
int main() {
   string str = "1234";
   int y = 20;
   cout << "Minimum sets = " << getMinSets(str, y) << endl;
   return 0;
}

Output

When you compile and execute above program. It generates following output −

Minimum sets = 3
raja
Published on 22-Nov-2019 15:12:05
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