Merge Sorted Array - Leetcode 88 - Python

Summary

In 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

• 😀 The video discusses a coding challenge to merge two sorted arrays, nums1 and nums2, into a single sorted array within nums1.
• 📝 A lazy approach is suggested to create a new array and merge the elements by comparing the smallest values, but this requires extra memory.
• 🔍 The optimal solution is to merge the arrays without a temporary array, aiming for a memory complexity of O(1).
• 👉 The merging process should start from the end of nums1, taking advantage of the guaranteed space for nums2's elements.
• 🔑 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.
• 🔄 Update pointers for nums1 and nums2 as elements are compared and merged, always choosing the larger value to place in nums1.
• 💡 Handle edge cases where the smallest value is not in the desired position by adjusting the pointers and merging accordingly.
• 👌 After merging, any remaining elements in nums2 should be directly copied to nums1 since they are already sorted.
• 🛠 The final code provided is intended to be readable and beginner-friendly, with a focus on avoiding extra memory usage.
• 🐞 A potential bug is identified related to index starting at zero, which is corrected in the script to ensure the code works correctly.
• 👍 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

00:00

😀 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.

05:02

🔍 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.

10:03

👨‍💻 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.

🙌 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.

Keywords

💡Merge Sorted Array

Merging sorted arrays is the process of combining two sorted arrays into a single sorted array. In the context of the video, the task is to merge 'nums1' and 'nums2' such that all elements are in order within 'nums1'. The script describes a method to achieve this without the need for additional memory, which is a key focus of the video's theme.

💡LeetCode Problem 88

LeetCode Problem 88 refers to a specific coding challenge on the LeetCode platform, where the goal is to merge two sorted arrays into one. The video script uses this problem as a basis to demonstrate an efficient algorithm for merging arrays, showcasing a practical application of the concept.

💡Lazy Approach

The 'lazy approach' mentioned in the script refers to a simple but less efficient method of solving the problem by creating a new array and then copying the elements from the two input arrays into it in sorted order. This approach is contrasted with a more optimal solution that is presented later in the video.

💡Time Complexity

Time complexity is a measure of how long an algorithm takes in terms of the amount of input. The script discusses the time complexity of the merging process, emphasizing the importance of an efficient solution with a time complexity of O(n), where n is the total number of elements in both arrays.

💡Extra Memory

In the context of the video, 'extra memory' refers to the additional space required by an algorithm to perform its task. The script initially presents a method that requires a temporary array, which is not optimal due to the extra memory usage. The video then aims to find a solution that does not require this extra memory.

💡Pointer

A 'pointer' in programming is a variable that stores the memory address of another variable. In the script, pointers are used to keep track of the current position in the arrays 'nums1' and 'nums2' during the merging process, which is crucial for the in-place merging technique.

💡In-Place Merging

In-place merging is a technique where the merging of two arrays is done within the existing arrays without using additional memory for a new array. The video script describes an in-place merging algorithm that updates 'nums1' directly, which is the main theme of the video.

💡Reverse Order

The term 'reverse order' in the script refers to the strategy of starting the merging process from the end of the arrays, taking advantage of the existing empty space in 'nums1'. This approach is used to optimize the merging process and is a key part of the video's solution.

💡Edge Case

An 'edge case' is a specific scenario that might not be immediately obvious but could cause issues in an algorithm. The script mentions an edge case where the smallest value is not in the expected position, and the solution includes handling such cases to ensure the algorithm's robustness.

💡Bug

A 'bug' in programming is an error or flaw in the code that causes it to behave incorrectly. The script briefly discusses a bug related to index handling and how it was corrected to ensure the merging algorithm works correctly.

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