Question

**C++ Question**

For all of the following, determine the total operation count and then the Big-O of the given code segments:
a.
for (int j = 0; j < n; j++)
for (int k = 0; k < j; k++)
sum++;

b.
for (int i = 0; i < q*q; i++)
for (int j = 0; j < i; j++)
sum++;

For all of the following, just determine the Big-O of the given code segments:
c.
for (int i = 0; i < n; i++)
for (int j = 0; j < i*i; j++)
for (int k = 0; k < j; k++)
sum++;

d.
for (int i = 0; i < p; i++)
for (int j = 0; j < i*i; j++)
for (int k = 0; k < i; k++)
sum++;

e.
for (int i = 0; i < n; i++)
{
Circ arr[n];
arr[i].setRadius(i);
}

f.
for (int i = 0; i < n; i++)
{
int k = i;
while (k > 1)
{
sum++;
k = k / 2;
}
}

Answer 1

a.
for (int j = 0; j < n; j++) // It runs for n times so its complexity is O(n)
for (int k = 0; k < j; k++) // It runs for n(n+1)/2 times so its complexity is O(n²)
sum++; // It runs for n(n+1)/2 times so its complexity is O(n²)

Therefore its complexity is O(n²)

b.

for (int i = 0; i < q*q; i++) // It runs for n times so its complexity is O(n²)
for (int j = 0; j < i; j++) // It runs for n²(n²-1)/2 times so its complexity is O(n⁴)
sum++;    // It runs for n²(n²-1)/2 times so its complexity is O(n⁴)

Therefore its complexity is O(n⁴)

c.
for (int i = 0; i < n; i++)   // Its complexity is O(n)
for (int j = 0; j < i*i; j++) // Its complexity is O(n³)
for (int k = 0; k < j; k++) // Its complexity is O(n⁶)
sum++; // Its complexity is O(n⁶)

Therefore its complexity is O(n⁶)

d.
for (int i = 0; i < p; i++)   // Its complexity is O(n)
for (int j = 0; j < i*i; j++)   // Its complexity is O(n³)
for (int k = 0; k < i; k++) // Its complexity is O(n⁶)
sum++; // Its complexity is O(n⁶)

Therefore its complexity is O(n⁶)

According to HomeworkLib guidelines i have to solve first four bits only.