Python Programming Practice: LeetCode #1 -- Two Sum
Summary
TLDRIn 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
📝 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).
🔍 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.
🚀 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.
Mindmap
Keywords
💡LeetCode
💡Two Sum Problem
💡Array
💡Indices
💡Brute Force Solution
💡Time Complexity
💡Dictionary
💡Enumerate
💡Runtime
💡Memory Usage
💡Efficiency
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
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