Minimum Swaps to Group All 1's Together II - Leetcode 2134 - Python

Summary

In this coding tutorial, the presenter tackles the algorithmic problem of 'minimum swaps to group all ones together.' The challenge involves rearranging an array of zeros and ones with the least number of swaps to make all ones contiguous, taking into account the array's circular nature. The video debunks a common misconception about using the minimum length window approach and introduces a more effective strategy of finding a window of the specific size that contains the maximum number of ones. The presenter also explains two methods to address the circular aspect: one using extra space and another with constant space complexity, concluding with a working solution that iterates through the array with pointers and modulus operations.

Takeaways

• 😀 The video discusses solving a coding problem involving minimum swaps to group all ones together in an array.
• 🔍 The problem is to find the minimum number of swaps required to make all ones in an array adjacent, considering the array as circular.
• 💡 A naive approach of finding the minimum length window containing all ones is introduced but later shown to be incorrect.
• 🔧 A counterexample is provided to demonstrate the flaw in the initial greedy solution approach.
• 📚 The correct approach is to find a window of size equal to the total number of ones that contains the maximum number of ones.
• 🔄 Two methods to address the circular aspect of the problem are discussed: one using extra space and another with constant space complexity.
• 📈 The video explains using a sliding window technique to find the optimal window without creating a new array.
• 👉 The 'count' function in Python is used to determine the total number of ones in the input array.
• 🔄 The sliding window technique involves adjusting a left and right pointer to find the window with the maximum ones.
• 📝 The solution involves keeping track of the number of ones in the current window and updating it as the window slides.
• 🎯 The final result is calculated by subtracting the maximum number of ones found in any window from the total number of ones.

Q & A

What is the main problem being discussed in the video?

The main problem discussed is finding the minimum number of swaps required to group all ones together in a circular array containing only zeros and ones.

What is the significance of considering the array as circular in this problem?

Considering the array as circular allows for the possibility that elements at the end of the array are adjacent to elements at the beginning, which affects how swaps are calculated.

What is the initial approach the speaker tried to solve the problem, and why did it fail?

The initial approach was to find the minimum length window containing all ones. It failed because it didn't account for the possibility of making fewer swaps by rearranging ones in a way that skips holes, as demonstrated in the counterexample provided.

What alternative approach does the speaker suggest to solve the problem?

The speaker suggests finding a window of size equal to the total number of ones in the input that contains the maximum number of ones, which can be more efficient in terms of swaps.

How does the speaker propose to handle the circular aspect of the problem?

The speaker proposes two methods: one that requires extra space by concatenating a copy of the array, and another that uses constant space by employing a sliding window technique with modulo operation to simulate the circular nature without extra memory.

What is the sliding window technique mentioned in the script, and how is it applied here?

The sliding window technique involves maintaining a window of a fixed size and sliding it across the data structure. In this script, it is applied by using two pointers, one representing the start and the other the end of the window, and adjusting them to find the optimal window with the maximum number of ones.

Why is the window size in the sliding window technique set to the total number of ones in the input array?

The window size is set to the total number of ones because the goal is to find a contiguous group of ones that matches this count, which will minimize the number of swaps needed.

How does the speaker calculate the number of swaps needed using the sliding window technique?

The speaker calculates the number of swaps needed by determining the difference between the window size (total number of ones) and the maximum number of ones found in any window of that size during the sliding window process.

What is the 'mod' operation used for in the context of this problem?

The 'mod' operation is used to handle the circular nature of the array without creating a new array. It maps indices that would be out of bounds in a normal array to valid indices in the original array, considering the array's circular property.

What is the time complexity of the solution presented in the video?

The time complexity of the solution is linear, as it involves a single loop that iterates over the array twice (once for counting ones and once for applying the sliding window technique).

How does the speaker ensure that the sliding window does not exceed the total number of ones?

The speaker ensures this by checking if the length of the current window exceeds the total number of ones and, if so, adjusting the left pointer and updating the number of window ones accordingly.

Outlines

00:00

🔍 Introduction to Minimum Swaps Problem

The video script begins with an introduction to a coding challenge focused on minimizing swaps to group all ones in an array. The problem is described as potentially difficult, especially at the start of the month, implying it's a monthly coding series. The basic idea is to arrange all the ones in an array next to each other using the minimum number of swaps. An example is given where three ones are grouped together with one swap. The script also mentions a circular array consideration, where the ends of the array are adjacent, introducing a trick to handle this aspect. The presenter initially attempts a greedy solution by finding the minimum length window containing all ones but realizes it's not always effective, providing a counterexample to illustrate why it fails.

05:01

🔄 Revisiting the Problem with a Sliding Window Approach

The script continues by exploring a different approach to the problem, suggesting finding a window of size equal to the total number of ones in the input array that contains the maximum number of ones. This method is explained with an example, where the goal is to fill zeros with ones using the minimum number of swaps. The presenter discusses two ways to address the circular aspect of the problem: one requiring extra space by concatenating the array with itself, and another using a sliding window technique with two pointers that avoids extra space by using modulo operations to simulate the circular array. The detailed steps of the sliding window algorithm are explained, including counting the total number of ones, maintaining the count of ones in the current window, and adjusting the window size to ensure it doesn't exceed the total number of ones.

10:01

📊 Conclusion and Solution Validation

In the final paragraph, the presenter demonstrates the solution by running the algorithm and showing that it works efficiently. The script emphasizes that the solution is a linear time solution, with one loop for counting ones and another for applying the sliding window technique. The presenter also mentions that the runtime of the code can be variable but confirms its efficiency. The video concludes with an invitation for viewers to check out more content on the NE code platform and thanks the audience for watching, promising to see them in the next video.

Keywords

💡Array

An array is a data structure that stores a collection of elements, typically of the same type, in a contiguous block of memory. In the context of the video, an array is used to represent a sequence of zeros and ones, which is the basis for the problem-solving scenario presented. The script discusses how to manipulate elements within an array to solve a specific algorithmic challenge.

💡Swap

Swapping refers to the process of exchanging the positions of two elements in an array. The video script discusses the concept of swapping as a method to rearrange elements within an array to group all ones together, which is the core of the problem being addressed.

💡Minimum Swaps

Minimum swaps is a term used to describe the least number of operations needed to achieve a desired arrangement of elements in an array. The video's theme revolves around finding the minimum number of swaps required to group all ones together in an array, which is a key aspect of the problem-solving process.

💡Circular Array

A circular array is an array that connects its end with its beginning, forming a circle. The script mentions this concept to explain that the array being considered has a circular nature, which affects how elements are grouped and swapped.

💡Contiguous

Contiguous refers to elements that are adjacent to each other without any interruptions. The video aims to find a way to make all ones in the array contiguous through a series of swaps.

💡Greedy Solution

A greedy solution is an algorithmic paradigm that makes the most optimal choice at each step with the hope of finding the global optimum. The script initially considers a greedy approach to solve the problem by finding the minimum length window containing all ones but later points out its inadequacy.

💡Window

In the context of the video, a window refers to a subset or subarray of a given size within the larger array. The script discusses finding a window of a specific size that either contains all ones or the maximum number of ones as part of the solution strategy.

💡Sliding Window

A sliding window is a technique used in algorithms to move a window of fixed size over an array or string to process elements within that window. The script describes using a sliding window to find the optimal window size that contains the maximum number of ones.

💡Space Complexity

Space complexity refers to the amount of memory space required by an algorithm as a function of the input size. The script discusses two approaches to the problem, one with extra space for a copy of the array and another with constant space by using modulo operations to simulate the circular array.

💡Algorithm

An algorithm is a step-by-step process to solve a problem. The video script is centered around developing an algorithm to determine the minimum number of swaps needed to group all ones in an array, considering both linear and circular aspects.

💡Index

In an array, the index is the position of an element within the array, usually starting from zero. The script uses the concept of index to explain how elements are accessed and manipulated during the swapping process and to handle the circular nature of the array.

Highlights

Introduction to the problem of minimizing swaps to group all ones together in an array.

Explanation of the circular nature of the array and its impact on solving the problem.

Initial approach of finding the minimum length window containing all ones and its limitations.

Counterexample provided to demonstrate the failure of the initial approach.

Introduction of a new approach focusing on finding a window with the maximum number of ones.

Clarification on how to handle the circular aspect by considering the array as concatenated with itself.

Description of a space-efficient method to handle the circular array without extra memory.

Explanation of using two pointers to create a sliding window for the array.

Technique of using modulo operation to map indices for the circular array.

Algorithm to count the total number of ones and calculate the window size based on the number of ones.

Process of updating the window ones count and the maximum window ones found.

Strategy for adjusting the sliding window and maintaining the count of window ones.

Final calculation of the minimum number of swaps based on the difference between the total ones and the maximum window ones.

Demonstration of the solution's efficiency with a linear time complexity.

Conclusion and invitation to check out more content on the NE code platform.

Transcripts

hey everyone welcome back and let's

write some more neat code today so today

let's solve the problem minimum swaps to

group all ones together part two I don't

think we've actually solved the first

one but honestly this is a pretty

difficult problem especially for the

beginning of the month I think it's

going to be a long month the problem is

pretty simple to understand the idea is

we have an array like this one and in

this array we're only going to have

zeros and ones so in this example we

have three ones and we want to group

them all together so that they're all

adjacent so one way for us to do that is

by swapping elements well that's

actually the only way we can do that so

we're going to swap these two together

so this will be a zero this will be a

one now obviously these are all

continuous so it took us one swap to do

this we want to figure out what is the

minimum number of swaps in order for us

to make them all continuous this example

was pretty simple imagine though if we

had a one over here well in that case we

would just take this guy swap it here

and then take this guy and swap it there

and so then we'd have all four of the

ones in the middle this is a bit more

interesting so here you might think we

need to take this guy move it there and

probably take this guy and move it there

and then we'll have all four of them in

the middle but actually this already

counts as grouping them together because

the idea is that this array is

considered to be circular so these are

technically adjacent to these so that's

going to be something we need to deal

with and I'll show you kind of a trick

that we can use to do that that actually

will be probably the easiest part of the

problem one thing I tried to solve this

problem with was find the minimum length

window such that it contains all of the

ones and forgetting about the whole

circular aspect of the problem it seems

like it would work because we could do

something like this for the original

example this would be the minimum length

window that contains all of the ones and

so the length of it is four and if we

knew the count of all the ones in the

input we have three so if you take the

difference between those 4 minus 3 we

get one that's what I thought is a

reasonable way to approach this problem

it seems kind of like a greedy solution

but it's not going to work let me

actually give you a counter example of

why that is so this is a pretty long

string well I guess it's an array but

you see a couple ones here a couple ones

here and a couple ones here here are

kind of the holes that we need to fill

so if I took my original approach I'd

say well this is of length 10 there are

six one on so we take 10 - 6 and then we

get four so of course it's going to take

four operations four swaps to get this

right we'd have to kind of push these

two over here and then these two can go

over here to fill that hole and then

we're good right no it actually takes

less than four we can actually do this

in two swaps I don't want to spoil it

for you so you try to figure it out

yourself for a second the idea is that

by taking these ones and actually

filling them in over here we skipped

this hole this isn't even a hole anymore

because now our ones are all over here

and they're already contiguous there's

six of them or we could have obviously

taken these two and moved them over

there and then they'd be continuous here

so our previous solution doesn't account

for that so we have to get a little more

clever I was just kind of staring at

this for a second and thinking huh well

what's another way we could think about

this problem and then I realized it if

we have all of our ones and they're all

contiguous of course that's going to be

of length six in this problem right cuz

we have six ones in the input so we need

a window of length six so a similar but

slightly different approach would be

find the window of size six that

contains the maximum number of ones so

here we have four ones that tells us

we're trying to fill these two zeros

right cuz 6 - 4 is 2 we're trying to

fill those two zeros it doesn't really

matter where the rest of the ones are

they could be contiguous they could be

scattered wherever they want to be but

it would only take us two swaps to fill

those holes so that's the approach that

will work in this case in this example

and it'll work in the previous examples

find the window that has the maximum

number of ones that is of size six in

this case because six is the number of

ones that were given to us in the input

that's how you kind of handle that part

of it now to address the circular part

of the problem I'll show you two ways to

do it one of them is going to require

extra space and one of them is going to

require constant space but it's going to

be a little more tricky I guess for

Simplicity consider we're given this

input array now obviously you can tell

that the ones are already contiguous but

the way to approach this and considering

the circular aspect is that any kind of

postfix of this could be connected to a

prefix of the beginning portion a better

way to approach it is just to take a

copy of the array and concatenate it

like this now if you see we can kind of

consider any subarrays from here those

are valid we can consider any subarrays

from here but they're going to be

identical to the ones over here but now

if we take a subray like this it is kind

of the circular aspect we have elements

from the end of the array connected to

elements to the beginning of the array

now the only thing is we can't have

something like this like that's not a

valid array cuz we can't just create

elements how are we ever going to get a

subarray of length nine it can't be

longer than of length five that's what

I'm getting at because five was the

original length so no matter how we run

it through the middle it cannot be

larger than five but that doesn't really

matter in our case because remember

we're only looking at windows in this

example there's two ones so we're only

going to be looking at Windows of length

two so that's going to be perfectly fine

for us we would pretty much just do a

sliding window like this sliding window

like this this this and then this would

be the one that we found that has the

maximum number of ones two of them so we

take the size of our window 2 - 2 we get

zero zero swaps is going to be the

result obviously L this approach

requires extra memory cuz we're going to

have to create a new array by

concatenating one to the other is there

a way to do this without that yes we can

kind of Imagine like this is there the

way I'm going to do that is by running

my Loop I'm going to have two pointers

that's how sliding window Works we're

going to have a left pointer here and a

right pointer here and the right pointer

isn't going to go up until the end of

the length of this array consider the

length is n it's actually going to go up

until 2 * n so we're kind of imagining

like this part of the array is there but

what happens when our right pointer is

somewhere over here well let me quickly

label the indexes of everything right

now our right pointer is technically out

of bounds if we try doing nums of right

we're going to get an index out of

bounds error but recall that the length

of the array n is equal to 5 so if we

want the actual value that should be

here the trick is just take the index R

and mod it by n which in this case is

five if we do that this portion is going

to be 6 modded by 5 it's going to be 1

so that tells us that the element over

here is going to be the same element

over here and that does make sense cuz

there is a one toone mapping between

these elements because they're just

copies like these arrays are just copies

of each other so that's how you make the

Math Workout and how you can reduce the

space complexity from oen down to

constant so we won't actually have this

in memory we'll just iterate our pointer

up until 2 * n and then this is how

we'll get what value should be in each

spot so I'm going to get the length of

the input array cuz we're going to need

that a couple times I'm going to count

the total number of ones in the input

and thankfully we can do that with a

built-in function called count in Python

and I'm going to have a left pointer

initially at zero and I'm going to have

my right pointer go not up until n but

actually 2 * n I'm also going to keep

track of two things what are the number

of ones in our window I'm going to call

that window ones I'm also to keep track

of the max window ones because that's

going to tell us what was the window

that had the maximum number of ones and

how many ones could we find in a window

of length this because then we can

calculate the result like this we know

our window has to be of length total

ones and the max number of ones that we

were able to get is going to be

subtracted from this so if they were the

same that means we had all ones in that

window if there were a few zeros then

this difference would be POS positive

remember we're doing a sliding window

where the sliding window is always going

to be of this length or at least that's

you know the one that we're going to be

considering so um just like this we're

going to say if nums of right is equal

to one or we can just do this in Python

then we're going to increment the number

of window 1es and we're trying to

maximize the number of window ones so

we'll set it to the max of itself and

the current window ones just like this

this but remember we don't want our

window to ever be larger than the total

number of ones if it's smaller that's

not really going to change the result so

we don't really care about it but if it

becomes larger then we have to do

something so we'll check if the length

of our window which we can calculate

like this right minus left + 1 is

greater than the total number of ones

then we need to shift our left pointer

but at the same time we should probably

update the number of window ones in our

current window that can actually easily

be updated without using an if statement

we can do this number of window ones is

going to be decremented by this nums of

left so if the number at the index left

was a one we're going to subtract from

this if it was a zero we don't want to

do anything in which case this equation

isn't going to do anything that's the

whole idea behind this problem the only

thing I haven't shown is the modding so

remember that these two indices could

technically be out of bounds so let's

mod this by n and let's mod this guy by

n as well whoops okay so that's the

entire solution let's just run it and as

you can see it works I think it's pretty

efficient but I know that the leak code

run times are pretty random as you can

see though it's definitely a linear time

solution technically we have one Loop

here counting the total number of ones

and we have a loop here if you found

this helpful check out NE code. for a

lot more thanks for watching and I'll

see you soon