Merge Sorted Array - Leetcode 88 - Python
Summary
TLDRIn this coding tutorial, the presenter explores an efficient way to merge two sorted arrays into one without using additional memory. The focus is on merging 'nums1' and 'nums2' by starting from the end of 'nums1', leveraging the guaranteed space. The method involves using pointers to compare and insert the larger value from either array, then adjusting the pointers accordingly. An edge case is addressed where the remaining elements in 'nums2' are directly copied to 'nums1'. The script concludes with a concise code implementation and a reminder to consider zero-based indexing.
Takeaways
- đ The video discusses a coding challenge to merge two sorted arrays, nums1 and nums2, into a single sorted array within nums1.
- đ A lazy approach is suggested to create a new array and merge the elements by comparing the smallest values, but this requires extra memory.
- đ The optimal solution is to merge the arrays without a temporary array, aiming for a memory complexity of O(1).
- đ The merging process should start from the end of nums1, taking advantage of the guaranteed space for nums2's elements.
- đ Initialize a pointer at the end of nums1 to begin the merge, comparing the last elements of nums1 and nums2 to determine the largest value to insert.
- đ Update pointers for nums1 and nums2 as elements are compared and merged, always choosing the larger value to place in nums1.
- đĄ Handle edge cases where the smallest value is not in the desired position by adjusting the pointers and merging accordingly.
- đ After merging, any remaining elements in nums2 should be directly copied to nums1 since they are already sorted.
- đ The final code provided is intended to be readable and beginner-friendly, with a focus on avoiding extra memory usage.
- đ A potential bug is identified related to index starting at zero, which is corrected in the script to ensure the code works correctly.
- đ The video encourages viewers to like, subscribe, and support the channel for more content.
Q & A
What is the primary task described in the script?
-The primary task is to merge two input arrays, nums1 and nums2, into a single sorted array, with nums1 having enough space to accommodate all elements from nums2.
Why might someone choose to create a new array instead of merging into nums1?
-Creating a new array is a lazy approach to avoid the complexity of merging into nums1. It simplifies the process by starting with an empty array and filling it with the smallest elements from both arrays.
What is the time complexity of the lazy approach mentioned?
-The time complexity of the lazy approach is O(n), where n is the total number of elements in both arrays.
What is the main disadvantage of the lazy approach in terms of memory usage?
-The main disadvantage is the need for extra memory due to the creation of a temporary array to hold the merged elements before copying them back to nums1.
Why is it beneficial to start merging from the end of nums1 instead of the beginning?
-Starting from the end of nums1 is beneficial because nums1 already has enough empty space at the end to accommodate all elements from nums2, making the merging process more efficient.
What is the strategy for finding the largest value to insert at the end of nums1?
-The strategy involves comparing the last elements of nums1 and nums2 and selecting the larger one to insert at the end of nums1, then moving the corresponding pointer in that array.
How does the script handle the case when all elements from nums2 have been merged?
-When all elements from nums2 have been merged, the remaining elements in nums1 that have not been replaced are left as they are, since they are already in the correct position.
What is the edge case that the script mentions and how is it handled?
-The edge case is when the smallest value is not in the desired position initially. In this case, the script suggests taking all remaining elements in nums2 and filling nums1 with them since they are already in sorted order.
What is the condition for filling nums1 with the leftover elements from nums2?
-The condition for filling nums1 with leftover elements from nums2 is when there are leftover elements in nums2 and the pointer n is greater than zero.
How does the script ensure the final result is stored in nums1 without extra memory?
-The script ensures the result is stored in nums1 by merging elements in reverse order and updating pointers accordingly, thus avoiding the need for a temporary array.
What is the potential bug mentioned in the script and how is it fixed?
-The potential bug is related to forgetting that indexes start at zero, which could lead to incorrect pointer decrements. The script fixes this by properly adjusting the index decrements.
Outlines
đ Introduction to Merging Sorted Arrays
The video begins with a warm welcome and an introduction to the coding challenge of merging two sorted arrays, nums1 and nums2, into a single sorted array within nums1. The challenge includes the condition that nums1 has enough space to accommodate all elements from nums2. The presenter humorously suggests a 'lazy' approach to the problem, which involves creating a new array and then copying the merged result back to nums1, but acknowledges the need for an optimal solution with minimal memory usage, aiming for O(1) space complexity.
đ Optimal Merging Strategy Without Extra Memory
The presenter outlines a strategy to merge the arrays without using extra memory. This involves starting from the end of nums1, where there is guaranteed space, and comparing the last elements of nums1 and nums2 to determine the correct order of insertion. The process is described step by step, with pointers moving in reverse to fill in the correct values from both arrays. The presenter also discusses an edge case where the smallest value is not in the desired position and how the solution handles it by filling nums1 with the remaining sorted elements from nums2.
đšâđ» Coding the Merge Function and Addressing a Bug
The presenter proceeds to translate the discussed strategy into code, detailing the function parameters and the logic for merging the arrays in reverse order. The code checks which array has the larger element and places it at the end of nums1, then updates the pointers accordingly. An edge case is handled by filling the remaining space in nums1 with elements from nums2 if any are left. The presenter identifies and corrects a bug related to index handling, ensuring the code is accurate and efficient.
đ Conclusion and Call to Action
In the final paragraph, the presenter thanks the viewers for watching and encourages them to like, subscribe, and support the channel. The presenter expresses hope to see the viewers again soon, indicating the end of the video and a friendly invitation for continued engagement with the content.
Mindmap
Keywords
đĄMerge Sorted Array
đĄLeetCode Problem 88
đĄLazy Approach
đĄTime Complexity
đĄExtra Memory
đĄPointer
đĄIn-Place Merging
đĄReverse Order
đĄEdge Case
đĄBug
Highlights
Introduction to the problem of merging two sorted arrays, nums1 and nums2, into a single sorted array within nums1.
Lazy approach to merge arrays by creating a new empty array and filling it with the smallest elements from nums1 and nums2.
Discussion on the inefficiency of the lazy approach due to the need for extra memory for the temporary array.
Exploration of an optimal solution that avoids using a temporary array, thus reducing memory usage to O(1).
Explanation of starting the merge process from the end of nums1 to take advantage of the guaranteed space.
Initialization of a pointer at the last value in nums1 to begin the merging process in reverse order.
Comparison of the last elements in nums1 and nums2 to determine the largest value to place at the end of nums1.
Iterative process of comparing and replacing elements from nums1 and nums2 in reverse order.
Handling the case where all remaining elements in nums2 need to be copied to nums1 due to no more elements in nums1 being greater.
Coding the merge function with parameters nums1, m, nums2, and n, representing the last index of the last value in each array.
Use of conditional statements to determine which element from nums1 or nums2 should be placed in nums1.
Updating pointers m and n to reflect the position of the last element considered in each array.
Edge case consideration where the smallest value is not in the desired position and needs to be handled.
Filling nums1 with the remaining sorted elements from nums2 if all elements in nums1 are less than those in nums2.
Finalization of the merge process and ensuring the result is stored correctly in nums1.
Identification and correction of a bug related to index handling in the merge process.
Emphasis on readability and beginner-friendliness in the coding approach.
Encouragement for viewers to like, subscribe, and support the channel for more content.
Transcripts
hey everyone welcome back and let's
write some more neat code today
so today let's look at leak code 88
merge sorted array so we're given
two input arrays nums one and nums two
and we wanna take nums 2 and then merge
it
into nums 1 into 1 sorted array
now lucky for us we are guaranteed
that there is enough space in nums 1
for every element in nums 2 to be
inserted
and we want to make sure that they're in
order now what if you just want to kind
of be
lazy let's say you got these input
arrays nums 1 and nums 2.
it seems kind of annoying having to
merge them into nums too what if we want
to be really
lazy we can just create a fresh copy
a fresh not copy but a fresh empty array
with
the same number of elements as nums one
and then just take
and then we can start at the beginning
of each of these
arrays numbers one and nums two take the
smaller element
insert it into the new array and then we
can repeat that process
the smallest element and then we can
just keep doing that
two three five
six but this array is temporary right
because
they want the answer to be
in nums one so what we could do is then
just
copy this into nums one
the only problem with this solution is
while the time complexity
is big o of n we also need
extra memory we need this temporary
array
but is it possible to solve this problem
exactly how they want us to solve it
with without actually
needing this temporary array
because if we don't need the temporary
array we don't need the extra memory
then the memory would also would be
big o of one so let's see how we can
solve this problem
the most optimal way
okay so when you draw it out like this
the first thing you notice
is that there's the empty space
is at the end of nums one so if we're
going to start
merging these values why should we
start at the beginning and start like
filling this way
when it might just be easier to start
filling this way if we want to sort it
because
there's already empty space we know that
there's enough space
over here for these elements to fit in
no matter what right like that's
guaranteed
so i want to start merging from here
so i'm going to initialize a pointer
here this is where we're going to start
merging it's the last
value in nums now what value
do we want to put here since we want it
to be in order we want the
largest value to be over here how do we
get the largest value
well we can start at the last real
value in nums one
and the last real well the last number
in nums two because there's no empty
space in nums two
and we can compare these two values we
see that
six is greater so we don't need this
anymore and we can replace it with a six
and we also don't need this anymore
so we can move our pointer
of nums two over here now looking at the
five
and we're also done with this pointer
we can shift it over here now we compare
three and five we know five is greater
so that's exactly what we can do we can
replace this zero
with a five and again we know we have to
shift our pointers now
so the last pointer is going to be over
here now this is going to be the next
position we insert into
and this is going to be the next value
from nums 2 that we look at
so now we're going to compare 2 with 3.
finally we get a larger value in nums 1.
so we can get rid of this place it with
a
3. once again let's update our pointers
so last pointer is going to be over here
now we got rid of one pointer replaced
it with another
this is going to be the value we look at
in nums one
and this time they're both equal two and
two so it doesn't really matter which
one we replace
i'll use the first one so get rid of
this
get rid of this and get rid of this
replace it with a
two now we got our last two values
1 and 2. 2 is greater
so we have no more values in nums 2 to
look at
we can insert this 2 here and so
we're pretty much done merging because
there's no more elements
left in uh numbs2
but we only did five values right
lucky for us the one that's over here
doesn't really need to be
changed so we can leave it there and
you know our result is correct
regardless
okay so now let's finally code it up
so in this uh function we're given nums
one
we're given m which is the lat which is
the index of the last
value in nums one not the length of nums
one because we know it has
uh some empty space at the end we got
nums two
and we have n which is the index of the
last value in nums 2.
so the first thing we want to do is get
the last
index of nums 1 like basically
the length so i'm going to store it in
last different way you might need to do
m plus n minus 1.
so the next thing we want to do is start
merging them
in reverse order
so now that we have all three of the
pointers we need last
m and n let's start merging we're gonna
keep going
while there are elements left in both
arrays so while
m greater than zero and
n is greater than zero we want to find
this
the largest value so if nums one
of m is greater than num's two of n
then we can go to our last
position replace it
with nums 1 of m
the else condition is if they were
equal or if nums 2 was greater
so in that case we just want to do the
opposite
numbs 1 last it's going to be
nums 2 of n uh one thing we
should definitely not forget about is
updating our pointers so
since we're going in reverse order we're
going to decrement m
[Music]
and decrement n regardless of
which element we insert we want to
decrement last regardless
so now we've merged them and into
inserted order we did it without extra
memory
and the result is stored in nums one but
there's one last
edge case that we forgot about let me
show you
so it was convenient for us that the
smallest value was already in the
position we wanted it to be in but
what if these two values were the
opposite
what if this value was a 2
and this value was a 1
what would we have done in that case
well we see that the
2 is greater so we would get rid of this
and then put a 2 here but notice
now how the empty elements are in
nums 2 over here right so in this case
what are we going to do we're basically
just going to take
all the remaining elements in nums 2 and
then
fill nums 1 with them right so we'll
just put this one
over here in this case there's only one
value but if there were more values
we would just keep going we would keep
taking each value
and filling it because this is already
in in sorted order so we can just
take that sorted portion and then fill
nums one
that's exactly what i'm gonna do so fill
nums1
with the leftover elements in
nums2 and we're only going to do that
if there are leftover elements and the
condition for that
is if n is greater than zero
so in the last position numbers one of
last
is gonna be nums two of n
and then all we need to do is update our
pointer so
n and last are both going to be
decremented by one
and hopefully this is bug free okay so
this is actually not too bad of a bug we
remember that
indexes start at zero so we forgot to
use the minus one
and i think most of these so let me not
forget
and now it should work awesome so
i tried to make this code somewhat more
like readable and beginner friendly
there are definitely
more tricky ways you can do this as
always
thank you so much for watching please
like and subscribe to support the
channel it helps out a lot
and i hope to see you pretty soon
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