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Ksju [112]
3 years ago
15

What is the big-O performance estimate of the following function?

Computers and Technology
2 answers:
larisa [96]3 years ago
8 0

Answer:

The time complexity of the code is O(log₇n).

Explanation:

The i is updated by 7*i.On each iteration i is multiplied by 7.So on finding the time complexity of the code given above it will come out to  be log base 7.

When we divide the input by 2 the time complexity is log base 2.

So on dividing it by 7 we get the time complexity of log base 7.

Digiron [165]3 years ago
4 0

Answer:

Since the loop index variable i is initialized to 0, this function will result in infinite execution for all values of n>=0.

If instead the loop index variable is initialized to 1, Big O performance estimate of the code is O(log n) with logarithm base as 7.

Explanation:

Given function:

int f (n) {

int sum = 0;

for (i = 0; i < n; i = 7 * i)

sum += i;

return sum;

} // end f

The complexity is determined by the for loop. Starting value of the index variable i is 0.  At each iteration the value of the index variable i is multiplied by 7 (i=7*i). But multiplication by 7 still leaves the value of the index variable as 0. So loop condition i<n will fail for all n>=0 and the loop will continue indefinitely since the termination condition will never be achieved.

However if the loop index variable i is initialized to 1 instead, the loop will run O(log n) times with 7 as the base of the logarithm due to successive multiplication of index variable by 7 at each iteration.

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Ket [755]

Answer:

#include <iostream>

using namespace std;

// average function

void average(double arr[], int n) {

double total = 0;

double avgTemp;

for (int i = 0; i < n; i++)

{

 // find temperatures total

 total = total + arr[i];

}

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}

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void printArray(double arr[], int n) {

for (int i = 0; i < n; i++)

{

 cout << arr[i] << "   ";

}

}

int main() {

int const k = 5;

double temps[k];

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double tempValue;  

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for (int i = 0; i < k; i++)

{

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return 0;

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Explanation:

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• note the syntax of void average(double arr[], int n). Since we are passing an array to the function so its formal parameters would be arr[] and an int n which specifies the maximum size of our array temps[].

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