Python Programming Practice: LeetCode #1 -- Two Sum

Summary

In this Python programming tutorial, the host tackles LeetCode's problem number one, which involves finding two numbers in an array that sum up to a given target. Two solutions are presented: a brute force approach using nested loops, resulting in an O(N^2) time complexity, and a more efficient method utilizing a dictionary to reduce the time complexity to O(N). The video guides viewers through the coding process, explains the logic behind each approach, and demonstrates the significant performance improvement of the dictionary method over the brute force solution.

Takeaways

• 📝 The video is a tutorial on solving LeetCode problem number one, which involves finding two numbers in an array that sum up to a specific target.
• 🔍 The problem is described as having a difficulty level of 'easy' and assures that there is exactly one solution for each input.
• 🚫 The constraints of the problem prohibit using the same element twice and require unique indices for the solution.
• 🔢 An example is provided where the numbers at indices 0 and 1 (2 and 7) add up to the target of 9, thus the solution would return indices [0, 1].
• 🔄 The first solution approach discussed is a brute force method using a double for-loop, which results in a time complexity of O(N^2).
• ⏱ The double for-loop solution is acknowledged to be inefficient for large inputs but is used for demonstration purposes.
• 🔑 The second solution approach uses a dictionary to store numbers and their indices, allowing for a single pass through the array with a time complexity of O(N).
• 📈 The dictionary method is highlighted as more efficient, especially for larger datasets, due to its linear time complexity.
• 💡 The video demonstrates coding both solutions in Python, with a focus on clarity and understanding over optimization.
• 📊 After submitting both solutions, the video compares their performance, showing that the dictionary method is significantly faster.
• 🎓 The video concludes by encouraging viewers to continue coding and seeking out different methods to solve programming problems.

Q & A

What is the main topic of the video?

The main topic of the video is solving LeetCode problem number one, which involves finding two numbers in an array that sum up to a specific target.

What is the difficulty level of the problem discussed in the video?

The difficulty level of the problem is labeled as 'easy'.

What is the assumption made about the input array for this problem?

The assumption made is that each input array will have exactly one solution where two numbers add up to the target sum.

Why is using the same element twice not a concern in this problem?

Using the same element twice is not a concern because the problem statement guarantees that there will be exactly one solution without needing to use any element more than once.

What is the first solution approach discussed in the video?

The first solution approach discussed is a brute force method using a double for loop to find all combinations of number additions in the array.

What is the time complexity of the double for loop solution?

The time complexity of the double for loop solution is O(N^2), where N is the length of the input array.

What is the second solution approach mentioned in the video?

The second solution approach is using a dictionary to store numbers and their indices as they are encountered, allowing for a single pass through the array.

What is the time complexity of the dictionary-based solution?

The time complexity of the dictionary-based solution is O(N), where N is the length of the input array.

How does the dictionary-based solution improve efficiency compared to the double for loop?

The dictionary-based solution improves efficiency by reducing the number of operations from N^2 to N, thus avoiding the need to check every possible pair of numbers in the array.

What is the trade-off when using the dictionary-based solution?

The trade-off when using the dictionary-based solution is increased memory usage due to storing numbers and their indices in the dictionary.

What is the final verdict on the dictionary-based solution's performance in the video?

The dictionary-based solution performed significantly faster with a runtime of 48 milliseconds and was in the 95th percentile of speed for Python 3 submissions on the website.

How does the video conclude regarding the solutions to the LeetCode problem?

The video concludes that while there may be many ways to solve the problem, the dictionary-based solution provided a reasonably efficient method that was faster than most submissions.

Outlines

00:00

📝 Introduction to LeetCode Problem 1

The video script begins with a warm welcome to a Python programming practice session focused on solving LeetCode problem number one. The problem, titled 'Two Sum', is of easy difficulty and involves finding two numbers in an array that sum up to a specific target value. The script clarifies that there is exactly one solution to the problem, ensuring no duplicate indices or repeated elements. An example is provided to illustrate the solution approach, where the indices of the numbers that add up to the target are returned. The video outlines two different methods to solve the problem: a brute force approach using nested loops and a more efficient method utilizing a dictionary to reduce the time complexity from O(N^2) to O(N).

05:03

🔍 Brute Force Solution with Nested Loops

The first method discussed in the script is a brute force approach that involves using two nested for-loops to iterate through the array and find all possible combinations of numbers that sum up to the target. The script explains the need to avoid using the same index twice by setting up the loops to ensure the first loop goes through all elements except the last, while the second loop starts from the current index of the first loop and goes to the end of the array. The script provides a code snippet that checks if the sum of the numbers at the current indices equals the target, and if so, returns the indices. The solution is acknowledged to be inefficient with a runtime of O(N^2), and the script mentions that this method is generally to be avoided but is used here for educational purposes.

10:04

🚀 Efficient Solution Using a Dictionary

The second method presented in the script is a more efficient solution that uses a dictionary to store numbers and their indices as they are encountered. This approach allows for a single pass through the array (O(N) time complexity) to find the solution. The script details the process of checking if the complement of the current number (target minus the current number) exists in the dictionary. If it does, the indices are immediately returned. If not, the current number and its index are stored in the dictionary. The script emphasizes the importance of not storing duplicate values and provides a code snippet for this solution. The script concludes with the submission of this solution, which significantly outperforms the brute force method in terms of runtime and is in the 95th percentile of speed for Python 3 submissions on the platform.

Keywords

💡LeetCode

LeetCode is an online platform that provides a collection of coding challenges for candidates to practice and prepare for technical interviews. In the context of this video, it is the source of the 'two sum' problem that the presenter is addressing. The video's theme revolves around solving LeetCode problem number one, which is a coding challenge to find two numbers in an array that sum up to a specific target.

💡Two Sum Problem

The 'Two Sum Problem' is a classic algorithmic challenge where the goal is to find two numbers in an array that add up to a given target value. The video script describes this problem as the central focus, with the presenter outlining two different methods to solve it, emphasizing its significance as a common interview question and a foundational problem in coding practice.

💡Array

An 'array' in programming is a data structure that consists of a collection of elements, each identified by an index or key. In the video, the array is the input data type for the 'two sum' problem, where the presenter discusses finding indices within the array whose elements sum up to a target value.

💡Indices

In the context of arrays, 'indices' refer to the position of elements within the array. The video script mentions returning the indices of two numbers in the array that add up to the target sum, making 'indices' a key concept in understanding how to solve the 'two sum' problem.

💡Brute Force Solution

A 'brute force solution' is a problem-solving approach that involves checking all possible cases until the correct one is found. The video describes one of the methods to solve the 'two sum' problem using a double for-loop, which is a brute force method due to its O(N^2) time complexity.

💡Time Complexity

Time complexity in computer science refers to the amount of time an algorithm takes to run as a function of the size of the input. The video script discusses the time complexity of the brute force solution as O(N^2) and contrasts it with a more efficient solution that has a linear time complexity, O(N).

💡Dictionary

In programming, a 'dictionary' is a data structure that stores key-value pairs. The video script describes using a dictionary to store numbers and their indices as they are encountered in the array, which allows for a more efficient solution to the 'two sum' problem by reducing the time complexity to O(N).

💡Enumerate

The 'enumerate' function in Python provides a convenient way to loop over something and have an automatic counter. In the script, 'enumerate' is used to iterate over the array with both the index and value available, which simplifies the process of adding elements to the dictionary.

💡Runtime

Runtime in programming refers to the duration it takes for a program to execute. The video script compares the runtime of the brute force solution, which is slower, to the more efficient dictionary-based solution, highlighting the importance of algorithm efficiency.

💡Memory Usage

Memory usage refers to the amount of memory a program consumes while running. The video script briefly touches on the memory usage of the dictionary-based solution, noting that while it uses more memory than the brute force approach, it is still efficient and less than half of the submissions.

💡Efficiency

In the context of this video, 'efficiency' refers to how well an algorithm performs in terms of time and space complexity. The script discusses the efficiency of two different solutions to the 'two sum' problem, emphasizing the importance of finding a balance between fast execution and low memory consumption.

Highlights

Introduction to coding solutions for LeetCode number one.

Problem description involves finding indices of two numbers that sum up to a specific target.

Assumption that each input has exactly one solution, avoiding the need to worry about duplicates.

Explanation of the example with numbers and a target of nine, resulting in indices [0, 1].

Two different coding approaches to solve the problem: a brute force method and a dictionary-based method.

The first approach involves a double for loop, resulting in a time complexity of O(N^2).

The second approach uses a dictionary to store numbers and their indices, aiming for a linear time complexity of O(N).

Coding the first solution with a nested for loop and checking for the sum equal to the target.

Avoiding using the same index twice by structuring the loops to prevent overlap.

Submitting the first solution and discussing its slow runtime of 6,000 milliseconds.

Moving on to the second solution using a dictionary for more efficient lookups.

Using enumerate to loop through numbers with both index and value.

Checking if the complement of the current number (target - current number) is in the dictionary.

Returning indices immediately if a match is found in the dictionary.

Storing the current number and its index in the dictionary if not already present.

Submitting the second solution and noting the significantly faster runtime of 48 milliseconds.

Comparison of the two solutions, emphasizing the efficiency of the dictionary-based approach.

Encouragement to keep coding and a sign-off for the next video.

Transcripts

hello everyone and welcome to Python

programming practice this video we're

going to be coding up a couple different

solutions to leet code number one so

we're just gonna start by reading

through the problem and then we will

code up a couple different solutions for

it so the problem is called to sum it

has a difficulty level of easy so it

shouldn't be too difficult or too long

let's read the problem description given

an array of integers return indices of

the two numbers such that they add up to

a specific target you may assume that

each input would have exactly one

solution okay you don't have to worry

about doubling up or getting the wrong

indices I guess and you may not use the

same element twice all right that seems

important as we if we're looping through

something multiple times we can't use

the same element at the same index twice

it's giving an example here given these

numbers and a target of nine

because the zero index which is two and

the first index seven those add up to

nine which is the target we would return

those two indices so you turn in X zero

and one which are the two numbers that

add it up to the target of nine okay so

that is the problem so it's not super

complicated but let's go over to the

code and we could start out by making

some notes here on how we're going to

approach the problem now we're gonna

solve this two different ways start off

with we're just going to do a double for

for loop solution which will essentially

will be looping through the list of

numbers and then looping through the

list of numbers a second time so we can

find every single combination of

additions between the numbers and that

should find us the solution it is a

brute force solution though which means

looping through an array twice with a

double for loop means it's going to be

the runtime is going to be N squared

Oh of N squared which means say if we

have an input list of length 100 we will

end up doing 10,000 operations with the

double for loop that's something that's

generally better avoided if you can

manage it but and for our second

solution we're going to use a dictionary

to make some to store numbers that we

see and that should allow us to only

have to loop through the numbers once so

that will allow us to have a runtime of

only O of n instead of N squared so this

is the general plan here let's take our

notes away and think about how we would

start coding up the first solution here

so basically we need to loop through the

data or the number let's see here

so are given the starting code here

where we have an input list of numbers

and a target and we need to output a

list of integers which are the two

indices that add up to the target or

store the numbers to add up to the

target so with for the double for loop

solution we want to start by looping

through the numbers we probably want to

loop through by indices because that's

what we're returning so we could say for

I which is index in range of the numbers

the length of the list and we don't

we're gonna loop through twice but we

but we don't want to duplicate anything

if we remember from the problem

description we couldn't use the same

element twice which means we want to

avoid having the same index in our hands

at the same time

so we'll set up the loop so that

the first loop will go through

everything other than the last element

and then the second loop will start from

I or where the first loop is or I plus

one I guess and then only look at things

from that index till the end so it will

allow us to avoid looking at the same

index at the same time and it will cut

down somewhat on the total amount of run

time it's still on the order of N

squared but it will be a little bit

better so we want to loop through all of

the numbers except the last one and then

we're gonna do another loop for the

other index in a range from where our

current index is I you know we don't

need to look at ones that we've already

seen so we'll start at I plus 1 which is

the next index over the array and then

we'll go to the end of end of the

numbers so go all the way to the end of

length numbers and then here's the code

that has to check whether the sum of

these two numbers is equal to the target

so if the number at index I plus the

number at index J which are different

because we set up the loops that way is

equal to our target input we should

return a list that is the two indices

right so I and J

so this should be a working double for

loop solution it's not going to be the

most efficient but I'm going to click

Submit here

it's saying pending my cutoff for the

video but it is judging right now and

then

I can pull over

okay so the solution says it's it passed

successfully the runtime unsurprisingly

was pretty slow so about 6,000

milliseconds and it was only faster than

6% of other Python 3 solutions so it was

pretty slow but we knew that going in

alright so now let's go back and try

coding up a more efficient solution so

this time we are going to use a

dictionary and we're going to store the

indices and values that we've already

seen and we can use those stored values

to look up and see whether our current

value plus anything that's stored will

equal the target and if it does we can

just immediately return what those

indices are and that should allow us to

go through the number array only once to

find the solution but we will have to

have a little bit more memory usage

because we're gonna have to store some

numbers but see how we do this well

first we need to define just an empty

dictionary to store things in so we'll

say scene which is the numbers that

we've seen so far and now we're gonna

have to loop through the numbers again

so well maybe use enumerate this time

because we want the index but also the

number and we don't need to stop like at

any specific point like we did with the

first example we just want to do all the

numbers all at once so we'll just go

through everything for index and numb in

enumerate nums so enumerate gets you

both the index and the value at the same

time so I is the index num is the value

and then

first we need to check whether the

dictionary seen contains the number we'd

need to add to our current number that

makes it equal the target so the number

we'd have to add to the current number

you get the target would be the target

minus the current number so if the

target minus the current number is in

the seen dictionary then we can exit and

return the two indices so let's see here

but the second the second index is just

going to be I because that's what we're

looking at but the first index would be

the one we already stored in seen but

means we're going to return the value of

the seen dictionary where the key is

equal to this number that we looked up

and that will give us the index we're

gonna have to store that in our else

clause so we're gonna want to list that

we have to return the first element is

going to be the value in the seen

dictionary associated with this number

and the second index is just I and then

if the target - our current number is

not in the scene dictionary we need to

store the current number and index in

scene at least if it's a number that we

haven't seen before so else if our

current number is not in scene so we've

never seen our current number before we

will store it so seen our current number

and this is equals this index

so so this should loop through every

number if that number plus something

that's we've seen before equals the

target then we just return it the two

indices that made that true and if not

then we have to store what we're

currently looking at in scene but we're

only storing it if it doesn't already

exist in scene I don't know if the

problem definition specified whether you

can have repeated terms in it like in

theory that would be possible but we

don't need to store the same values more

than once like if they're giving us an

array that has several Breeze in it or

something we wouldn't need to store that

more than once

so let's submit this one as well and see

how we do

sure how fats is gonna be compared to

other solutions because this website

from run to run sometimes you get pretty

different values for your solution speed

but it should certainly be better than

the row of N squared solution that we

submitted earlier so let's check on that

all right it looks like it ran in 48

milliseconds this time which was a lot

faster I think the first one was around

6000 and it's saying it was about in the

95th percentile of speed for the Python

3 submission so that seems pretty good

and the memory usage isn't really too

much different than before and it's

actually still less than over half of

the submission so that doesn't seem too

bad now of course there are many

different ways you could probably solve

this problem this might not be the most

efficient method but it got the job done

and seemed to be faster than most of the

submissions that have been sent in so

seems like a reasonably good solution

for that first easy problem on leaked

code so keep coding and I'll see you

again next time