Maximum Number on Seven Segment Display Using n Segments in C++

Given the task is to find the maximum number that can be displayed using N segment on ant number of seven segment display.

Let’s now understand what we have to do using an example −

Input − N=5

Output − 71

Explanation − The largest number will be displayed as follows on the seven segment display −

Input − N=6

Output − 111

Approach used in the below program as follows

• The following situation can be divided into 3 case −

  • Case 1 −

    If N is 0 or 1, then it is not possible to display any number.

  • Case 2 −

    If N is odd. Then the numbers that can be displayed with odd number of segments are 2, 3, 5, 7 and 8 out of which 7 takes the least number of segments, that is, 3. Therefore 7 is the most significant digit in this case.

  • Case 3 −

    If N is even. Then the numbers that can be displayed with odd number of segments are 0, 1, 4, 6, 9 and 8 out of which 1 takes the least number of segments, that is, 2. Therefore 1 is the most significant digit in this case.

• In function MaxNumber() first check for the base condition using

  if (N == 1 || N == 0)

• Then using another if statement, check if N is even. If so then print “1” as it is the most significant digit in the even case and call MaxNumber(N - 2) as 2 segments are being used here.

• Use another if statement to check if N is odd. If so then print “7” as it is the most significant digit in the odd case and call MaxNumber(N - 3) as 3 segments are being used here.

Example

 Live Demo

#include <iostream>
using namespace std;
void MaxNumber(int N){
   //Condition to check base case
   if (N == 1 || N == 0){
      return;
   }
   //If the number is even
   if (N % 2 == 0){
      cout << "1";
      MaxNumber(N - 2);
   }
   //If the number is odd
   else if (N % 2 == 1){
      cout << "7";
      MaxNumber(N - 3);
   }
}
//Main function
int main(){
   int N;
   N = 5;
   MaxNumber(N);
   return 0;
}

Output

If we run the above code we will get the following output −

71