Examples of complexity pattern

Posted 2020-08-15

O(1):constant - the operation doesn‘t depend on the size of its input, e.g. adding a node to the tail of a linked list where we always maintain a pointer to the tail node.
int i=0;
i++;
++i;
i+=6;
O(n):linear - the run time complexity is proportionate to the size of n.
int i,n=100,s=0;
for(i=0;i<n;i++)
{
s+=1;
}
O(n2):quadratic - the run time complexity is proportionate to the square of size of n, e.g., bubble sort.
int i,j,n=100,s=0;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
s+=1;
}

}
O(n3):cubic - very rare.
int i,j,k,n=100,s=0;
for(i=0;i<n;i++)
{
for(j=0;j<n;j++)
{
for(k=0;k<n;k++)
{
s+=1;
}
}
}
O(logmn): logarithmic: normally associated with algorithms that break the problem into smaller chunks per each invocation, e.g. searching a binary search tree.
int i,n=100,m=2; /* m could be any number, e.g.,2,10 */
for(i=0;i<n;i++)
{
i=i*m;
}
O(nlogn): just nlogn: usually associated with an algorithm that breaks the problem into smaller chunks per each invocation, and then takes the results of these smaller chunks and stitches them back together, e.g. quick sort.
int i,n=100;

int m_expo(int m)
{
/* an auxilary function that return the value
of m to the mth exponential, not included to the
time consumation*/

int k = m;
for(j=1;j<m;j++)
{
/* 2 could also be other number */
k = k * k;
}
return k;
}

/* this is the part whose consumation is O(nlogn) */
for(i=0;i<m_expo(n);i++)
{
/* 10 could be other number */
i=i*10;
}
O(n1/2): square root.
int i,n=100;
while(i*i<n)
{
i++;
}
O(2n):exponential - incredibly rare.
int i,n=100;
int expo(int m)
{
/* an auxilary function that return the value
of 2 to the mth exponential, not included to the
time consumation*/

int k =1;
for(j=1;j<m;j++)
{
/* 2 could also be other number */
k = k * 2;
}
return k;
}

/* this is the part whose consumation is O(2n)) */
while(i<expo(n))
{
i++;
}
O(n!):factorial - incredibly rare.
int i,n=100;
int factorial(int m)
{
/* an auxilary function that return the
factorial value of m */

int k =1;
for(j=1;j<=m;j++)
{
/* 2 could also be other number */
k = j * k;
}
return k;
}

/* this is the part whose consumation is O(n!) */
while(i<factorial(n))
{
i++;
}
O(nn):not exist in real life.
int i,n=100;
int mm_expo(int m)
{
/* an auxilary function that return the value
of m to the mth exponential, not included to the
time consumation*/

int k = m;
for(j=1;j<m;j++)
{
k = k * m;
}
return k;
}

/* this is the part whose consumation is O(nn)) */
while(i<mm_expo(n))
{
i++;
}