What is O(log n)? Learn Big O Logarithmic Time Complexity
Jared Nielsen ・7 min read
Is there a computer science topic more terrifying than Big O notation? Don’t let the name scare you, Big O notation is not a big deal. It’s very easy to understand and you don’t need to be a math whiz to do so. In this tutorial, you’ll learn the fundamentals of Big O logarithmic time complexity with examples in JavaScript.
This is the fourth in a series on Big O notation. If you want to stay in the loop, sign up for my weekly newsletter, The Solution.
What Problem(s) Does Big O Solve?
Big O notation helps us answer the question, “Will it scale?”
Big O notation equips us with a shared language for discussing performance with other developers (and mathematicians!).
Quick Refresher
If you’re just joining us, you will want to start with that article, What is Big O Notation?
What is Big O?
Big O notation is a system for measuring the rate of growth of an algorithm. Big O notation mathematically describes the complexity of an algorithm in terms of time and space. We don’t measure the speed of an algorithm in seconds (or minutes!). Instead, we measure the number of operations it takes to complete.
The O is short for “Order of”. So, if we’re discussing an algorithm with O(log N), we say its order of, or rate of growth, is “log n”, or logarithmic complexity.
How Does Big O Work?
Big O notation measures the worst-case scenario.
Why?
Because we don’t know what we don’t know.
We need to know just how poorly our algorithm will perform so we can evaluate other solutions.
The worst-case scenario is also known as the upper bound. When we say upper bound, we mean the maximum number of operations performed by an algorithm.
Remember this table?
O | Complexity | Rate of growth |
---|---|---|
O(1) | constant | fast |
O(log n) | logarithmic | |
O(n) | linear time | |
O(n * log n) | log linear | |
O(n^{2)} | quadratic | |
O(n^{3)} | cubic | |
O(2^{n)} | exponential | |
O(n!) | factorial | slow |
It lists common orders by rate of growth, from fastest to slowest.
We learned O(1), or constant time complexity, in What is Big O?, O(n) in Big O Linear Time Complexity, and O(n^{2)} in Big O Quadratic Time Complexity.
We previously skipped O(log n), logarithmic complexity, because it's easier to understand after learning O(n^{2),} quadratic time complexity.
Now it's time!
Math O’Clock 🧮 🕝
What is a logarithm?
Logarithms allow us to reverse engineer a power. (Like Kryptonite!) They are the inverse operation of exponentiation.
We can think of logarithms as the opposite operation of exponentiation. Remember this analogy format from standardized tests?
A is to B as X is to Y.
We can say, “Addition is to subtraction as exponentiation is to logarithm.”
We can also say, “Multiplication is to division as exponentiation is to logarithm.”
With quadratic time complexity, we learned that n * n is n^{2}.
If the value of n is to 2, then n^{2} is equal to 4.
It then follows that 2 to the third power, 2^{3}, is equal to 8.
In “long” form, that looks like:
2 * 2 * 2 = 8
What if we were given this problem?
2^y = 8
How would you describe this problem?
We could say, “To what power do we raise 2 for a product of 8?”
🧐
We could also say, “How many 2’s do we need to multiply together to get 8?”
The notation for the equation is:
log2(8) = y
Where 2
is the base, 8
is the argument and y
is the exponent, or power.
Why is 8
the argument? (See what I did there?)
Because log2() is a function!
🤯
We could describe the product as, “The base 2 logarithm of 8 is 3.”
What do we mean by base?
“In exponentiation, the base is the number b in an expression of the form b^{n.”}
The three most common bases in regards to logarithms are:
- 2
- 10
- e
In digital electronics and computer science, we (almost always) use base-2, or binary numeral system.
In base-2, we count with two symbols: 0 and 1.
Look familiar?
It’s Boolean! 👻
Unless you were born with six fingers on each hand, you count in base-10, or the decimal numeral system.
10^2 = 100
10^3 = 1000
etc.
If you really want to nerd out, you can read up on e, or Euler’s constant: the unique number whose natural logarithm is equal to one.
☝️ Logarithms are not square roots.
The square root of 8 is 2.82842712475.
The log2 of 8 is 3.
Small difference here, but what if our argument increased?
The square root of 2048 is 45.2548339959.
The log2 of 2048 is 11.
What if we chart a logarithm?
Aerodynamic!
If we compare logarithmic time complexity to other time complexities on the ubiquitous Big O cheat sheet, what do we see?
As we can see, logarithmic time complexity is very good!
What if the problem was, “How many 3s, multiplied together, does it take to get 8?”
The answer is 1.8927892607
If you imagined calculating that number by hand is a tedious process, you’d be right.
That’s why we invented slide rulers and fancy calculators.
Lucky for us, we don’t need to calculate the exact log of a function.
With Big O, we abstract away the details.
We’re not concerned with the specific implementation of our algorithm.
We’re interested in the order of our algorithm so we can make comparisons and evaluate alternative solutions.
We consider the base of our log a constant, so we drop it, and simply use the following notation:
O(log n)
O(log n): Binary Search
The classic example used to illustrate O(log n) is binary search. Binary search is an algorithm that finds the location of an argument in a sorted series by dividing the input in half with each iteration.
Let’s say we are given the following array and asked to find the position of the number 512
:
const powers = [1, 2, 4, 8 ,16, 32, 64, 128, 256, 512];
First, let’s review the brute force solution to this problem.
const bruteSearch = (arr, num) => {
for (let i = 0; i < arr.length; i++) {
if (arr[i] === num) {
return `Found ${num} at ${i}`;
}
}
}
Let's map the values of i
and arr[i]
in an array, J4F:
i | arr[i] |
---|---|
0 | 1 |
1 | 2 |
2 | 4 |
3 | 8 |
4 | 16 |
5 | 32 |
6 | 64 |
7 | 128 |
8 | 256 |
9 | 512 |
What’s the Big O of bruteSearch()
?
O(n)
To find num
, we need to check every item in our array, so the time complexity is linear.
Can we do better?
Definitely.
What’s happening in this function?
const powers = [1, 2, 4, 8 ,16, 32, 64, 128, 256, 512];
const binarySearch = (arr, num) => {
let startIndex = 0;
let endIndex = (arr.length)-1;
while (startIndex <= endIndex){
let pivot = Math.floor((startIndex + endIndex)/2);
if (arr[pivot] === num){
return `Found ${num} at ${pivot}`;
} else if (arr[pivot] < num){
startIndex = pivot + 1;
} else {
endIndex = pivot - 1;
}
}
return false;
}
In the first iteration of our while
loop, we create a pivot at the median of our array. We then check num
against the value at the array indexed by pivot
.
If pivot
is equal to num
, we return.
If pivot
is less than num
, we change the value of our startIndex
to the value of our pivot
plus one.
Why?
If the value of arr[pivot]
is less than num
, we don’t need to check any of the elements below our pivot
in the next iteration.
We just halved our input!
n / 2
If pivot
is greater than num
, we change the value of our endIndex
to the value of our pivot
minus one.
If the value of array[pivot]
is greater than num
, we don’t need to check any of the elements above our pivot
in the next iteration.
In the next iteration, we create a new pivot
at the median between our adjusted startIndex
and endIndex
, and again check the value of arr[pivot]
against the value of num
.
In the example above, we again halve our input.
What’s half of half?
A quarter.
n / 4
In the third iteration, we halve the input again: n / 8
In the final iteration, we find our number. There’s also nothing left to halve at that point.
What’s the pattern?
Let’s make a table!
startIndex | endIndex | pivot | # elements | n |
---|---|---|---|---|
0 | 9 | 4 | 10 | n |
5 | 9 | 7 | 5 | n / 2 |
8 | 9 | 8 | 2 | n / 4 |
9 | 9 | 9 | 1 | n / 8 |
What do you notice about the sequence in the # elements column?
We divide it, more or less, by two, until we reach zero.
What do you notice about the sequence in the n column?
2
4
8
Powers of two!
By dividing our input with each iteration, we are inversing an exponent.
📝 When you see this pattern where the input is divided with each iteration, it’s O(log n).
Big O Logarithmic Time Complexity
Does O(log n) scale?
Definitely.
In this tutorial, you learned the fundamentals of Big O logarithmic time complexity with examples in JavaScript. Stay tuned for part five of this series on Big O notation where we’ll look at O(n log n), or log linear time complexity. If you want to stay in the loop, sign up for my weekly newsletter, The Solution.
Meanwhile, don't forget about big-Θ and big-Ω!
What exactly is big O, big Θ, big Ω ?
pontakornth ・ Jan 15 ・ 1 min read
Big theta and big Omega are upper bounds and lower bounds, right? But in software engineering, we don't mind the upper or lower bounds that's why we use big O? At least that's what I remembered from the book cracking the coding interview. Not sure though. lols
Not quite. Big O is the upper-bound (longest possible run time), while big-Ω is the lower bound (shortest possible run time). It's big-Θ that's the combination of the two, although it can only be used if the upper and lower bounds are the same.
While big-O is the most commonly useful, since we most often care about the longest possible running time of a particular scenario (e.g. of worst-case), it still has a precise meaning.
When we say "the function is at least O(n)", we're almost always misappropriating big-O (upper bound) to describe big-Ω (lower bound).
What's further, when we say "the function always has a runtime of O(n²)", we're really meaning "Θ(n²)"; it's not just the upper bound, but both bounds at once.
To put that another way, "lower bound" means "a runtime of at least", and "upper bound" means "a runtime no greater than". This has nothing to do with best-case/worst-case, as we are always speaking about the runtime of worst-case, unless explicitly stated otherwise.
Take a look at my comment on that article I linked to for more details.
// , If it gets really big, it's big-Θ-Lawd: youtube.com/watch?v=zlplCs00wTE
"Your time complexity is so bad, not even the TARDIS can handle it."
Nice explanation.
Can you provide some other common problems which can be solved in O(logn) ?
TIA
O (log n) is extremely common - and in fact the one that will save you the most time if you've ever dealt with a database slowing down. Just plopping an index on a string value will turn it from an O(n) operation (sequential scan) to an O (log n) operation - divide and conquer. Usual caveats here: databases tend to have some tweaks under the covers, but the typical index is BTREE or the like, which is O (log n).
You can quantify the time savings using the math in this post. On 1000 rows a sequential scan has to do 1000 comparisons to find the record(s) you want. With an index in place, it does log2(1000), which is ~10 (or 10 in our case). That's pretty good - and you an see why an index like that would scale really well.
In terms of algorithms you write - you'd be surprised at the optimizations you can get if you're using something like Redis, which doesn't have indexes (unless you create them). Since it's in memory though, you can scan a list for a given value using binary search - pretty damn useful :).
O(log n) is actually not very common.
You have fast exponentiation if you wanna explore.
At best, you'll encounter O(n*log n) in general for algorithms.
Thanks.
A simple example is binary searching a number in an ordered array.
Thanks for sharing such a great Concept in Easy Way. Kudos,Bro👏