Alen Abraham
Alen Abraham

Alen Abraham

Big O notation explained: DSA-01

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Alen Abraham
·Mar 14, 2022·

2 min read

Big O notation explained: DSA-01

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What is Big O notation?

Big O notation is used to measure how the running time or space requirements for our program grows as input grows.

  • Measuring running time growth is time complexity. (In this post we focus on time complexity)
  • Measuring space growth is space complexity.

Rules to determine time complexity

  1. Keep fastest growing term.
  2. Drop constants.

Big O refers to very large value of 'n' (n = size) where time, t = an^2 + bn + c.
For example, if we have a function like,

t = 5n^2 + 3n + 20

When value of 'n' is very large, bn+c becomes irrelevant.
ie; an^2 is very larger than bn+c.
For example; if n = 1000 then,
t = 5 1000^2 + 31000+20 = 5000000 + 3020; where 3020, a small value becomes irrelevant.

Different time complexities

Here we discuss about a few of them.

1. O(n)

def foo(arr): size(arr) -> 100 -> 0.22ms
def foo(arr): size(arr) -> 1000 -> 2.30ms

Here time increases as array size increases; time proportional to size(arr).
n = size(arr), b= constant
t = a*n + b

  1. Keep fastest growing term => t = a * n (b is constant)
  2. Drop constants => t = n ( a is constant)
    Therefore; t = O(n)

Example program for t = O(n): To get square numbers

def get_squared_numbers(numbers):
    squared_numbers = []
    for n in numbers:
        square_numbers.append(n * n)
    return squared_numbers

numbers = [2, 5, 8, 9]
get_squared_numbers(numbers)
#returns [4, 25, 64, 81]

2. O(1)

def foo(arr): size(arr) -> 100 -> 0.22ms
def foo(arr): size(arr) -> 1000 -> 0.23ms

time, t = a n + b -> t = a n ( b constant) -> t = n (dropping constants(a))
n = 1 => t = 1
Therefore, t = O(1)

Example program for t = O(1)

def find_first_pe(prices, eps, index):
    pe = prices[index]/eps[index]
    return pe

3. O(n^2)

Example Program for O(n^2): To find duplicates from the list

numbers = [3, 6, 2, 4, 3, 6, 8, 9]

for i in range(len(numbers)):
    for j in range(i + 1, len(numbers)):
        if numbers[i] == numbers[j]:
            print(numbers[i] + " is a duplicate.")
            break

Also there is O(log n), O(2^n) time complexities.

Happy learning

#competitiveprogramming #dsa #datastructures #algorithms

 
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