TTYT stands for "talk to you tomorrow".
Answer:
O(n²)
Explanation:
The worse case time complexity of insertion sort using binary search for positioning of data would be O(n²).
This is due to the fact that there are quite a number of series of swapping operations that are needed to handle each insertion.
Here is a somewhat cryptic solution that works:
#include <algorithm>
#include <cstdlib>
using namespace std;
void q(char c, int count)
{
for (int i = 0; i < count; i++) {
putchar(c);
}
}
void p(int b1, int plusses)
{
q(' ', b1);
q('+', plusses);
}
int main()
{
for (int i = -3; i <= 3; i++)
{
int pl = min(6, (3 - abs(i)) * 2 + 1);
p(6-pl, pl);
i == 0 ? p(0, 6) : p(6, 0);
p(0, pl);
putchar('\n');
}
getchar();
}
