How to start using Big(O) to understand Algorithms

We’ve all heard of Big(O).
It’s something most of us
learn in college and promptly
forget. We also know it’s
something that Top Coders and
Googlers are good at and many
of us would like to be good
at it too!

I can relate – I find many
algorithms fascinating and
many more intimidating. And
for a long time I struggled
to get my head around the
concept of Big(O). I knew
what it was – vaguely – but I
had no deep understanding –
no intuition for it at all.
And I knew that it was
important in telling me which
algorithms were good and
which weren't.
If you can relate to that
feeling then this article is
for you – you will be able to

*understand*Big(O) and have the beginnings of an intuition about it. As always, this blog post is also a fully executable program - this time you can use it to play with the various Big(O) algorithms and develop a feel for how they react to different inputs.
What

*is*Big(O) anyway?
So what is Big(O)? The
easiest way to understand it
is - Big(O) just describes
how any algorithm scales up.
How does it do this? It
simply focuses on the
upper-limit on the algorithm
ignoring all exceptions,
special cases, and complex
details and irrelevant
constants.
So when you want to know how
a bit of code is going to
handle more and more inputs -
just figure out it's Big(O).

How do we find the Big(O)?

Finding the Big(O) is
surprisingly easy! Just
"squint" at your algorithm -
ignore the details - find the
main repetitions (the
loops/recursions) and you've
trapped the Big(O).
Let's try it out:
Example 1:
----------
for(item in haystack) {
if(item == needle) return item;
}
Here is a simple loop that
walks through every single
input once. Therefore for `n`
inputs it performs `n`
repetitions giving us:
O(n)
Example 2:
----------
low = 0;
high = sortedhaystack.size - 1;
while(low <= high) {
mid = (low+high)/2;
item = sortedhaystack[mid];
if(item == needle) return item;
if(item < needle) low = mid+1;
else high = mid - 1;
}
Squinting at this algorithm
we see:
...
while(low <= high) {
...
mid = (low+high)/2;
if() low = mid;
else high= mid;
...
}
We again start of with the
entire range of inputs but
each repetition

*halves*the range it has to travel by narrowing to the mid point. This kind of discarding half a range is terribly common and terribly useful. Such a constant halving means that each repetition shrinks to log(n) items. Therefore we say this algorithm has: O(log(n)) Because this is common it is useful to keep in mind the relation between discarding chunks of input and*log(n)*.
Why does this matter?

So now we've found the
Big(O)! Big deal! How does
that help us know which
algorithm is better?
Well remember, Big(O) gives
us useful information as the
algorithm scales. If we have
a small haystack then it
really doesn’t matter – even
if each loop takes a second
then a haystack of a few
hundred items will finish
quickly enough in both cases.
But if we have a million
items things get interesting:
Algorithm 1: (1 second per loop)
O(n) = O(1,000,000)
~> 1,000,000 seconds
= 11 days to complete
Algorithm 2: (1 second per loop)
O(log(n)) = O(log(1,000,000))
= O(19.93)
~> 19.93 seconds
= 20 seconds to complete
We can now see there is a

*phenomenal*difference between the two. Just knowing the Big(O) can really help us!
Meet the Big(O) classes

What really makes Big(O)
useful is that most
algorithms fall within a few
Big(O) clases. Once we know
them and how they scale, we
can quickly estimate how
almost any algorithm scales.
Let's look at the important
Big(O) classes now.

Taking each in turn, let's
look at the important
characteristics and try and
build an inutition for each
along with an actual working
algorithm below:

/* (Some setup) */
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include<math.h>
enum OClass {
O1,
O_logn,
O_sqrtn,
O_n,
O_nlogn,
O_n_power_2,
O_2_power_n,
O_n_permut,
O_n_power_n,
};
#define UNUSED(x) (void)(x)
#ifdef DUMP_RESULT
#define RESULT(x) printf("%s\n",#x)
#else
static long _dummy;
#define RESULT(x) (_dummy=1)
#endif
/* The environment ties
* the algorithms, their
* descriptions, and their data.
*/
typedef void (*func)(void* data);
struct environment {
long n;
func algo;
void *data;
enum OClass oclass;
};
struct array {
long sz;
int *vals;
};
struct search {
int needle;
struct array *haystack;
};
struct range_sum {
long *slice_sum;
long root_sz;
long from;
long to;
struct array *array;
};

O(1): Flash - The Fastest O
This is the

**Holy Grail**- an algorithm that always completes in a fixed time irrespective of the size of the input. Completes 1 Million items in: 1 second . Examples: return the head of a list, insert a node into a linked list, pushing/popping a stack, inserting/removing from a queue, deleting from a doubly-linked list,...
void get_first(struct array* array) {
if(array->sz > 0) RESULT(array->vals[0]);
else RESULT(-1);
}

O(log(n)): Shrinking Violet - Divide and Conquer
These algorithms never have
to look at all the input.
They often halve inputs at
each stage and thus have
inverse the performance of
the higher powers (see the
Power-Sisters to contrast).
Completes 1 Million items
in:
20 seconds
.
Examples:
looking up a number in a
phone book, looking up a word
in a dictionary, doing a
binary search, find element
in a binary search tree,...

void binary_jump_search(struct search *s) {
long jump = s->haystack->sz / 2;
long pos = 0;
while(jump > 0) {
while(pos + jump < s->haystack->sz && s->haystack->vals[pos+jump] <= s->needle) pos+=jump;
jump = jump / 2;
}
if(s->haystack->vals[pos] == s->needle) RESULT("Found needle!");
else RESULT("Needle not found!");
}

O(√n): Groot - The Rare O
If we notice:
n
sqrt(n) = ---------
sqrt(n)
in some sense, √n is in the
"middle" of `n`. This type of
algorithm is not very
commonly found.
Completes 1 Million items
in:
16 minutes
.
Examples:
Grover’s algorithm,
the square root trick

void range_sum_query(struct range_sum* rs) {
long sum = 0;
int i = rs->from;
while((i+1)%rs->root_sz != 0 && i <= rs->to) {
sum += rs->array->vals[i++];
}
while(i + rs->root_sz <= rs->to) {
sum += rs->slice_sum[i/rs->root_sz];
i += rs->root_sz;
}
while(i <= rs->to) {
sum += rs->array->vals[i++];
}
RESULT(sum);
}
void setup_slice_sums(struct range_sum* rs) {
int i;
rs->root_sz = (long)sqrt(rs->array->sz);
int max_slice = rs->array->sz/rs->root_sz;
int num_slices = max_slice + 1;
rs->slice_sum = malloc(sizeof(long)*num_slices);
for(i = 0;i < num_slices;i++) {
rs->slice_sum[i] = 0;
}
for(i = 0;i < rs->array->sz;i++) {
rs->slice_sum[i/rs->root_sz] += rs->array->vals[i];
}
}

O(n): Clark Kent - Just a Straight guy
These are

*linear*algorithms which scale directly proportional to the input. This is commonly the case because we often have to access an item at least once. Completes 1 Million items in: 11 days . Examples: finding the maximum/minimum of a collection, finding the max sequential sum, traversing a linked list, deleting from a singly-linked list,...
void linear_search(struct search *s) {
int i;
for(i = 0;i < s->haystack->sz;i++) {
if(s->haystack->vals[i] == s->needle) {
RESULT("Found needle!");
return;
}
}
RESULT("Needle Not Found!");
}

O(nlog(n)): Hisoka - Sorting Cards
Sorting is pretty useful for
many, many, things. When
sorting we need to compare
each of the items with each
other. The cleverest sorting
algorithms compare each item
with an ever reducing set of
other items and are therefore
O( n log(n) )
^ ^
| L with a reducing(log!)
Compare set
each item
It can be shown that
algorithms that need to
compare elements cannot sort
faster than this (Algorithms
like counting sort and radix
sort use other information
and can be faster).
Completes 1 Million items
in:
~ 1 year!
.
Examples:
Merge Sort, Quick Sort, Heap Sort...

long partition_1(int* array, long low, long high) {
long left = low;
long right = high;
long pivot = array[low];
while(left < right) {
while(array[left] <= pivot && left <= high) left++;
while(array[right] > pivot) right--;
if(left < right) {
int tmp;
tmp = array[left];
array[left] = array[right];
array[right] = tmp;
}
}
array[low] = array[right];
array[right] = pivot;
return right;
}
void quick_sort_1(int* array, long low, long high) {
long pivot;
if(low < high) {
pivot = partition_1(array, low, high);
quick_sort_1(array, low, pivot-1);
quick_sort_1(array, pivot+1, high);
}
}
void quick_sort(struct array *array) {
quick_sort_1(array->vals, 0, array->sz-1);
}

O(n^2),O(n^3): The Power Sister - Growing Polynomially
These algorithms grow as a
polynomial of the input.
O(n^2) are known as Quadratic
and are known as Cubic
algorithms. Higher powers
are just known as bad
algorithms :-)
The powers usually reflect
the number of nested loops in
the system.
Completes 1 Million items
in:
O(n^2) ~> 32,000 years
O(n^3) ~> 32,000,000,000 years
.

**NB**: This brings up a**VERY**important point about Big(O) - whenever there are multiple Big-O’s in an algorithm, the biggest class wins out because it dominates the scaling. We can see this by noticing the time that smaller classes take in comparison with larger classes we have seen so far. Examples: O(n^2) - multiplying two n-digit numbers by a simple algorithm, adding two n×n matrices, bubble sort, insertion sort, number of handshakes in a room,... O(n^3) - multiplying two n×n matrices by a naive algorithm,...
void find_max_seq_sum(struct array* array) {
double max_sum = 0;
int i,j;
for(i = 0;i < array->sz;i++) {
double curr_sum = 0;
for(j = i;j < array->sz;j++) {
curr_sum += array->vals[j];
if(curr_sum > max_sum) max_sum = curr_sum;
}
}
RESULT(max_sum);
}

O(2^n): Wonder Woman - Combination Loops
These are exponential
algorithms whose growth
doubles with every new
addition to the input.
You can recognize these as
recursive algorithms that
solve a problem of size `n`
by recursively solving two
problems of size `n-1`.
Another such type is one that
iterates over all subsets of
a set. If you find it hard to
understand how iterating over
subsets translates to,
imagine a set of switches,
each of them corresponding to
one element of a set. Now,
each of the switches can be
turned on or off. Think of
"on" as being in the subset
and "off" being not included.
Now it should be obvious that
there are combinations.
Completes 1 Million items
in:
3.2x10^301019 Millennia!!
.
Examples:
Tower of Hanoi, Naive
Finonacci Calculation,...

void solve_hanoi_1(long num, int from_peg, int to_peg, int spare_peg) {
if(num < 1) return;
if(num > 1) solve_hanoi_1(num-1, from_peg, spare_peg, to_peg);
RESULT("move remaining one from_peg -> to_peg");
if(num > 1) solve_hanoi_1(num-1, spare_peg, to_peg, from_peg);
}
void solve_hanoi(int num) {
solve_hanoi_1(num, 1, 2, 3);
}

O(n!): Link - The Traveling Salesman
These algorithms iterate over
all possible combination of
inputs.
Completes 1 Million items
in:
2.7x10^5565698 Millennia
(good grief!!!)
.
Examples:
The traveling salesman
problem,...

O(n^n): The Blackest Panther - The Slowest O
This is just included for
fun. Such an algorithm will
not scale in any useful way
and I don't know of any.
Examples:
Please don't find any!
Did this help you understand
Big-O better? Which is your
favourite Big-O class? Let me
know what you think!

/* (actually run algos and show results) */
char* oclass_1_str(enum OClass oclass) {
switch(oclass) {
case O1: return "O(1)";
case O_logn: return "O(log(n))";
case O_sqrtn: return "O(sqrt(n))";
case O_n: return "O(n)";
case O_nlogn: return "O(n log(n))";
case O_n_power_2: return "O(n^2)";
case O_2_power_n: return "O(2^n)";
case O_n_permut: return "O(n!)";
case O_n_power_n: return "O(n^n)";
default: return "ERROR!";
}
}
void show_time_msg_1(clock_t begin, clock_t end) {
printf("%lf", (double)(end - begin));
}
void show_time_taken(struct environment* environment) {
clock_t begin,end;
if(!environment->data) {
printf("%-12s(%ld items): (Not executed)\n",
oclass_1_str(environment->oclass),
environment->n);
return;
}
printf("%-12s(%ld items): ",
oclass_1_str(environment->oclass),
environment->n);
fflush(stdout);
begin = clock();
environment->algo(environment->data);
end = clock();
show_time_msg_1(begin, end);
printf("\n");
}
/* [=] Show results of running
* all algos in their
* environments.
*/
void show_algo_results(struct environment* environments) {
while(environments->algo) {
show_time_taken(environments);
environments++;
}
}
/* [=] Return a large, random
* array
*/
int* create_int_array(long sz) {
long i;
int *array = malloc(sz*sizeof(int));
for(i = 0;i < sz;i++) {
array[i] = rand();
}
return array;
}
/* [=] dummy function
* for not implemented
* algorithms.
*/
void do_nothing(void* data) {
}
/* [=] Setup the environment for
* various algorithms
*/
struct environment* create_environments(long sz) {
int i = 0;
time_t t;
srand((unsigned) time(&t));
/* Setup a large block of
* enviroments.
* NB: if the number of
* algo's > 100 this should
* be changed
*/
struct environment *environments = malloc(sizeof(struct environment)*100);
struct environment *environment;
/* setup data */
struct array *sorted_array = malloc(sizeof(struct array));
sorted_array->sz = sz;
sorted_array->vals = create_int_array(sorted_array->sz);
quick_sort(sorted_array);
struct array *array = malloc(sizeof(struct array));
array->sz = sz;
array->vals = create_int_array(array->sz);
struct array *mutable_array = malloc(sizeof(struct array));
mutable_array->sz = sz;
mutable_array->vals = create_int_array(mutable_array->sz);
struct search *search = malloc(sizeof(struct search));
search->haystack = sorted_array;
search->needle = sorted_array->vals[rand()%(sorted_array->sz)];
struct range_sum *rs = malloc(sizeof(struct range_sum));
rs->array = array;
setup_slice_sums(rs);
rs->from = rand()%(rs->array->sz);
rs->to = rand()%(rs->array->sz);
while(rs->from < rs->to) {
rs->from = rand()%(rs->array->sz);
rs->to = rand()%(rs->array->sz);
}
/* O(1) */
environment = &(environments[i++]);
environment->n = array->sz;
environment->algo = (func)&get_first;
environment->data = array;
environment->oclass = O1;
/* O(log(n)) */
environment = &(environments[i++]);
environment->n = search->haystack->sz;
environment->algo = (func)&binary_jump_search;
environment->data = search;
environment->oclass = O_logn;
/* O(sqrt(n)) */
environment = &(environments[i++]);
environment->n = rs->array->sz;
environment->algo = (func)&range_sum_query;
environment->data = rs;
environment->oclass = O_sqrtn;
/* O(n) */
environment = &(environments[i++]);
environment->n = search->haystack->sz;
environment->algo = (func)&linear_search;
environment->data = search;
environment->oclass = O_n;
/* O(nlog(n)) */
environment = &(environments[i++]);
environment->n = mutable_array->sz;
environment->algo = (func)&quick_sort;
environment->data = mutable_array;
environment->oclass = O_nlogn;
/* O(n^2) */
environment = &(environments[i++]);
environment->n = array->sz;
environment->algo = (func)&find_max_seq_sum;
environment->data = array;
environment->oclass = O_n_power_2;
/* O(2^n) */
environment = &(environments[i++]);
environment->n = sz;
environment->algo = (func)&solve_hanoi;
environment->data = (void*)sz;
environment->oclass = O_2_power_n;
/* O(n!) */
environment = &(environments[i++]);
environment->n = sz;
environment->algo = (func)do_nothing;
environment->data = NULL;
environment->oclass = O_n_permut;
/* O(n^n) */
environment = &(environments[i++]);
environment->n = sz;
environment->algo = (func)do_nothing;
environment->data = NULL;
environment->oclass = O_n_power_n;
/* - end marker */
environment = &(environments[i++]);
environment->algo = NULL;
return environments;
}
long get_sz(int argc, char* argv[]) {
if(argc < 2) return 0;
return atol(argv[1]);
}
int main(int argc, char* argv[]) {
long sz = get_sz(argc, argv);
if(!sz) printf("Usage: %s <number of items>\n", argv[0]);
else show_algo_results(create_environments(sz));
}

. . . . . . . . . .

. . . . . . . . . .

Patrick says:

x² and x³ are not exponential but polynomial (quadratic and cubic respoctively).
2^n is the exponential one.

theproductiveprogrammer says:

Thanks Patrick. You are right. I've corrected the mistake in the blog.

Someone says:

How is 'deleting from a doubly-linked list' O(lg n)? If you already have a pointer to the node, it's O(1). If you need to find it, it's still O(n).

theproductiveprogrammer says:

That's correct. Deleting from a doubly-linked list is O(1). Fixed!

nancy says:

Is there a pdf version that I can print out?

. . . . . . . . . .