Solution: Big O of Nested Loop With Multiplication
This review provides a detailed analysis of the different ways to solve the Big O of nested loop with multiplication quiz.
We'll cover the following...
Solution #
n = 10 # Can be anythingsum = 0pie = 3.14var = 1while var < n:print(pie)for j in range(var):sum += 1var *= 2print(sum)
Explanation
The answer is . Have a look at the slides below for an in-depth explanation of the answer.
In the slides below, rtc abbreviates the running time complexity.
Time Complexity
The above slides give a detailed, step-by-step analysis of the code. Here, we provide a more summarized version.
The outer loop here runs times. In the first iteration of the outer loop, the body of the inner loop runs once. In the second iteration, it runs twice, and so on. The number of executions of the body of the inner loop increases in powers of 2. So, if is the number of iterations of the outer loop, the body of the inner loop runs a total of times. This geometric series sums to . The inner loop condition requires that in the last time the inner loop runs, it runs at most times. This requires ...