Provide a big oh runtime analysis for each of the following

Provide a \"big oh\" run-time analysis for each of the following. When a value of \"n\" is used, it is the size of the input. You may assume max() and min() are constant-time in-line functions. void problem_2a() { cin >> rows >> cols; n = rows * cols; step = n; while (step > 1) for (i = 0; i

Solution

Solved Q2 complete, post multiple question to get the remaining answers

Q2a)

Since we are equation n= rows * cols

rows = cols = sqrt(n)

now the for loops i=0 to rows and j=0 to cols will execute for sqrt(n) * sqrt(n) = n operations

But the given for loops are inside the while loop, which will itself take n iterations, hence the total order = n*n = n^2

Hence the function will be O(n^2)

Q2b)

The first for loop will run for n^2 iterations

The second there are three for loops of length n in cascade, so total function value will be n*n*n=n^3

Hence the bigoh of the function will be n^3

Hence O(n^3) is the final answer