Algorithms for Coding Interviews in Python/

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Solution: Nested Loop With Multiplication (Advanced)

Solution: Nested Loop With Multiplication (Advanced)

This review provides a detailed analysis of the different ways to solve the nested loop with multiplication challenge.

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Solution #

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n = 10  # can be anything
sum = 0
pie = 3.14
for var in range(n):
    j = 1
    while j < var:
        sum += 1
        j *= 2
    print(sum)

In the main function, the outer loop is O(n)O(n) as it iterates over n. The inner while loop iterates over i which is always less than n and j is doubled each time, therefore we can say that it is O(log2(n))O(log_2(n)). Thus, the total time complexity of the program given above becomes:

O(nlog2(n))O(nlog_2(n))

Here’s another way to arrive at the same result. Let’s look at the inner loop once again.

The inner loop depends upon j which is less than i and is multiplied by 2 in each iteration. This means that the complexity of the inner loop is O(log2(i))O(log_2(\text{i})). But, the value of i at each iteration, of the outer loop, is different. The total complexity of the inner loop in terms of n can be calculated as such:

n×log2(i)⇒∑i=1nlog2(i)n \times log_2(i) \Rightarrow\sum_{i=1}^{n} log_2 (i) ...