for int i0 ii j cout

for (int i=0; i<n; i++)
for (int j=n; j>=i; j--)
cout << i << “,” << j <<endl;

Solution

i=0 - 1 unit - 1 time - total 1 unit

i<n - 1unit-n times - n time units

i++ - 1 unit - n times - n time units

j=n - 1 unit - n times - total n units

j>=i - 1 unit - max n * n times when i = 0 - total max n * n units

Similarly j-- takes 1 unit time and executes max n*n times hence total max n*n units

Total time = 2n^2 + 3n + 1 units = O(n^2)