Question

What is the time complexity \(T(n)\) of the nested loops below? For simplicity, you may assume that \(n\) is a power of \(2 .\) That is, \(n=2^{k}\) for some positive integer \(k\) : \\[ \begin{array}{l} i=n_{i} \\ \text { while }(i>=1)\\{ \\ \qquad \begin{array}{c} j=i \\ \text { while }(j<=n) \end{array} \end{array} \\] < body of the inner while loop> \(\quad\) I/ Needs \(\Theta(1)\) \\[ j=2^{*} j \\] } \\[ i=|1 / 2| \\] }

Short answer

The time complexity of the given nested loops is \(\Theta((\log n)^2)\).

Step-by-step solution

Understanding the Algorithm

The provided code snippet includes two while loops – one nested within the other. The outer loop begins with 'i' being equal to \(n\) and ends when 'i' becomes less than 1. Each iteration of the loop halves the value of 'i'. The inner loop begins with 'j' equal to 'i' and ends when 'j' becomes greater than \(n\). Each iteration of the loop doubles the value of 'j'. The body of the inner loop takes \(\Theta(1)\) time, which means it runs in constant time.

Analyzing the Outer Loop

The outer loop halves the value of 'i' at each iteration. Given that 'i' starts at \(n\), and assuming \(n = 2^k\), the worst case number of iterations for this loop would be \(k\), since \(2^k = n\). Hence the outer loop runs in logarithmic time, specifically in \(\Theta(\log n)\).

Analyzing the Inner Loop

The inner loop doubles the value of 'j' at each iteration. So for a given value of 'i', the inner loop would run until 'j' exceeds \(n\). Therefore, the number of iterations for the inner loop would be proportional to \(\log n\) as well. Thus, the inner loop also runs in logarithmic time, specifically \(\Theta(\log n)\).

Time Complexity of Nested Loops

Since the inner loop is nested within the outer loop, and both run in \(\Theta(\log n)\) time, the total time complexity of the two nested loops would be the product of their individual time complexities. Namely, \(\Theta((\log n)^2)\).