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There's are 2 positive integers x and y we have to generate a pattern that looks like,
2,3,3,2,3,3,3,2,3,3,3,2,3,3,2,3,3,3,2,3,3,3,2,3,3,3,2,3,3,2....
If we look at the count of 3s inbetween 2s we can see the same pattern is created,
2 | 3 3 | 2 | 3 3 3 | 2 | 3 3 3 | 2 ......

What I have tried:

C++
#include <iostream.h>
#include <conio.h>

void main() {
  clrscr();
  int x = 5, y = 6;
  int u = x, l = 0, k = 0, j = x;

  while(k < 21) {
    cout<<x<<",";
    if(u == x && l > 0) {
      u = y;
      l = 0;
    }

    if(u == y && l == j) {
      if(j == x) {
	j = y;
      } else {
	j = x;
      }

      u = x;
      l = 0;
    }

    for(int i=0; i<u; i++) {
      cout<<y<<",";
    }
    l++;
    k++;

    cout<<endl;
  }

  getch();
}


It's working but I don't think its the optimal way to do it.
Posted
Updated 16-Nov-20 22:17pm
v2
Comments
[no name] 16-Nov-20 16:48pm    
I agree: lousy variable names; no comments; etc.
Rick York 16-Nov-20 20:03pm    
That doesn't look like "the same pattern" to me. Where is there a sameness in the pattern you posted? If it was 2,3,3,2,3,3,2,3,3,2 I can see a pattern but I do not see one in the sequence you posted.

1 solution

You just need two loops. The outer one should print the "2,3,3,2". The inner one needs to count from 2 up which gives the number of groups of "3,3,3" to display after the outer set.
 
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