[Solved] How many time is the inner loop running? [closed]

Question

Let’s look at the the two loops individually.

The outer loop starts with i=1, and increments it by k in each iteration, until it’s greater than n. In other words it can run n/k times, or, to be accurate, floor(n/k) times, since a loop can’t run a non-whole number of times.

The inner loop is relatively simpler – it starts with j=1 and increments it by one in each iteration until it’s greater than k, for a total of k times.

Put these two together and you’ll get floor(n/k)*k.

EDIT:
As pointed out in the comment, this analysis is true if n>=k. If n<k the outer loop will run exactly once. I.e., the total times run would be: max(1, floor(n/k))*k.

Accepted Answer

Let’s look at the the two loops individually.

The outer loop starts with i=1, and increments it by k in each iteration, until it’s greater than n. In other words it can run n/k times, or, to be accurate, floor(n/k) times, since a loop can’t run a non-whole number of times.

The inner loop is relatively simpler – it starts with j=1 and increments it by one in each iteration until it’s greater than k, for a total of k times.

Put these two together and you’ll get floor(n/k)*k.

EDIT:
As pointed out in the comment, this analysis is true if n>=k. If n<k the outer loop will run exactly once. I.e., the total times run would be: max(1, floor(n/k))*k.