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Find the time complexity for the following function (the basic operation is the innermost loop body's assignment). function f(n) r ← 0 m ← 1 for i ← 1 to n do m ← 3 × m for j ← 1 to m do r ← r + j return r GROUP OF ANSWER(CHOOSE ONE CORRECT): Θ（n） Θ（n^3） Θ（3^n） Θ（nlogn） Θ（n^2）

Question

Find the time complexity for the following function (the basic operation is the innermost loop body's assignment).

function f(n) r ← 0 m ← 1 for i ← 1 to n do m ← 3 × m for j ← 1 to m do r ← r + j return r GROUP OF ANSWER(CHOOSE ONE CORRECT): Θ（n） Θ（n^3） Θ（3^n） Θ（nlogn） Θ（n^2）

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Solution

The correct answer is Θ(3^n).

Here's why:

  1. The outer loop runs n times, so it has a time complexity of O(n).

  2. The inner loop runs m times, where m is tripled in each iteration of the outer loop. This means the inner loop runs 1 time in the first iteration, 3 times in the second iteration, 9 times in the third iteration, and so on.

  3. The total number of times the inner loop runs across all iterations of the outer loop is the sum of a geometric series: 1 + 3 + 3^2 + 3^3 + ... + 3^(n-1).

  4. The sum of this geometric series is (3^n - 1) / 2, which is O(3^n) because the highest power of n is the dominant term.

  5. Because the inner loop is nested inside the outer loop, the total time complexity is the product of the time complexities of the outer and inner loops.

  6. However, because O(3^n) grows much faster than O(n), the overall time complexity is dominated by the inner loop, making the time complexity of the entire function Θ(3^n).

This problem has been solved

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If an algorithm has a time complexity of Θ(n2), which of the following is TRUE

1/3

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