Big O notation in data structures is the most efficient tool to compare the efficiency of algorithms. it represents the upper bound of an algorithm. it tells the asymptotic behavior of a function that how fast a function f(n) grows as input size n becomes large. ## Big O notation - Definition

A function f(n) is order of g(n) if there exists constants c and n0 such that function f(n) <= c*g(n) where the n is greater then n0. Note: here the g(n) is the upper bound of function f(n)

let's understand it with an example

let's say we have a function f(n) = 5n + 4 then how we calculate the order of the function f(n).

f(n) = 5n + 4
the upper bound of function f(n) is n.
so g(n) = n

let's find the constants c and n0

using the definition we can write the function f(n) like this.

f(n) <= c*g(n)

5n + 4 <= c*n for all n >= n0

let's take c = 6 and n0 = 4 then equation is

5n + 4 <= 6n for all n >= 4

so the function f(n) = 5n + 4 is order of n. and in mathematical term.

5n+4 is O(n)

let's take another example

f(n) = 3n+ 4n + 7

so, g(n) = n2

for values c = 5 and n0 = 6 function f(n) is

3n2 + 4n + 7 <= 5n2 for all n >= 6

so the function f(n) = 3n2 + 4n + 7 is order of n2

3n2 + 4n + 7 is O(n2)

### Method to find the Big O notation of function

to find the Big O notation of a function we have some rules.

#### Rule First:

Keep the fastest growing term and discard the lower terms and constants.

for example, if the function f(n) = 3n2 + 4n + 2 then the fastest-growing term is only 3n2. so we will keep only 3n2 and discard the lower terms 4n and constant 2 from the function f(n).

#### Rule Second:

Ignore the coefficients from the equation.

for example, we have function f(n) = 3n2 then we remove the coefficients 3 from equation 3n2.

#### Rule third

If the function f(n) is constant then we say that f(n) is the order of 1.

for example we have function f(n) = 4 then it's an order of 1. because function f(n) is constant.

#### Rule Fourth

The base of the logarithm is not important.

logan = logbn / logba

both values for log a and b are constants. and we don't consider constant values so we ignore them.

for example if the function f(n) = 8log2n. then we can say that function f(n) is the order of log n.

let's take an example to understand how we find the Big O for function.

Example 1:

f(n) = 45
here the function f(n) is constant then it's and order of 1
f(n) is O(n)

Example 2:

f(n) = 6n3 + 27log2n + 2n

here we apply the rules

first, remove the lower terms and constants, then

f(n) = 6n3

second, remove the coefficients, then

f(n) = n3

so the function f(n) is the order of n3.
f(n) is O(n3)

Example 3:

f(n) = log2n + nlog10n

f(n) = nlog10n

so function f(n) is the order of n logn.
f(n) is O(n logn)

### Bound of an algorithm

there are two bounds of the algorithm. one is tight and the other is loose bound.

for a function f(n) there can be many functions g(n) such that f(n) is O(g(n))

let's say we have a function f(n) = 5n2 + 4n + 8 then the order of function f(n) is n2. but it's also the order of n3, n4 and n
so
the least upper bound of an algorithm is called the tight upper bound and all other bounds are called a loose upper bound.

so the tight upper bound of function f(n) is n2 and all other bounds are loose upper bounds.

Note: the loose upper bounds are not informative as much as the tight upper bound. so we only consider the tight upper bound of an algorithm.

### Big O analysis of algorithms

we can also analyze the big O of an algorithm. so to do this first we need to express the running time of an algorithm as a function of input size n.

let's say we have a function T(n) that represents the running time of an algorithm in terms of n. then we need to find the big O of function T(n).

let's say we have 4 algorithms A, B, C, and D. and the function T(n) represents the running time of an algorithm. then to analyze the algorithm that which one is efficient we need to calculate the big O for function T(n) of all 4 algorithms. so for algorithm A is the order of n3, similarly, algorithms B, C, and D are an order of n2, n, and n2. as shown in the above image.

so the algorithm C is efficient for our problem because the algorithm c has the fastest running time rate.