Two Sum - Leetcode 1 - HashMap - Python

Summary

The script discusses an efficient algorithm to find two values in an array that sum to a target number, such as 9. Initially, it describes a brute force method with a time complexity of O(N^2), then introduces a hash map approach to reduce the complexity to O(N). The hash map stores the indices of array values, allowing for a single pass check to find the complementary value to reach the target sum. The script explains the process and assures that this method will always find a solution due to the guaranteed presence of two summing elements in the array.

Takeaways

• 🔍 The script discusses a coding problem where the goal is to find two values in an array that sum up to a given target, in this case, 9.
• 🔑 The problem guarantees exactly one solution, so there's no need to handle multiple solutions or the absence of a solution.
• 🛠️ The initial approach described is a brute force method, checking every combination of two values, which has a time complexity of O(N^2).
• 🚀 An alternative, more efficient method is introduced using a hash map to reduce the time complexity to O(N).
• 📚 The hash map stores each value from the array with its corresponding index, allowing for constant time lookups.
• 🔄 The script explains that by the time the second value that sums to the target is encountered, the first value must already be in the hash map.
• 💡 It's highlighted that the hash map should not be initialized with the entire array to avoid reusing the same value.
• 🔎 The script provides a step-by-step explanation of how to iterate through the array and use the hash map to find the two values.
• 👩‍💻 The solution involves checking if the complement of the current value (target minus the value) exists in the hash map before adding the current value to it.
• 📝 The script concludes with a Python code snippet implementing the efficient solution using a hash map.
• 🎯 The final code is demonstrated to work correctly, providing a neat trick to solve the problem in a single pass through the array.

Q & A

What is the primary goal of the algorithm discussed in the script?

The primary goal of the algorithm is to find two values in an input array that sum up to a given target value, in this case, 9, and return their indices.

What is the initial brute force approach to solving the problem?

The initial brute force approach involves checking every possible combination of two values in the array to see if they sum up to the target, which results in a time complexity of O(N^2).

Why is the brute force method considered inefficient?

The brute force method is inefficient because it requires multiple passes through the array for each element, resulting in a worst-case time complexity of O(N^2), where N is the length of the array.

What alternative method is suggested to improve efficiency?

The alternative method suggested is using a hash map to store the values and their indices from the input array, allowing for constant time checks and a single pass through the array.

How does the hash map help in reducing the time complexity of the algorithm?

The hash map allows for constant time operations to check if a value exists and to add a new value, reducing the time complexity to O(N) since the array is iterated only once.

What is the memory complexity of using the hash map approach?

The memory complexity of using the hash map approach is O(N) because it may potentially store every value from the input array.

Why can't the same value be used twice in the solution?

The same value cannot be used twice because the problem statement guarantees exactly one solution, and reusing the same value would imply multiple solutions or a different combination.

What is the significance of the 'previous map' in the algorithm?

The 'previous map' stores every element that comes before the current element in the iteration, allowing for the efficient checking of the required difference against the target sum.

How does the algorithm ensure that it finds the correct pair of indices?

The algorithm ensures the correct pair is found by checking if the difference between the target and the current element exists in the 'previous map'. If it does, it means the pair that sums to the target has been found.

What is the final step in the algorithm once the correct pair is found?

The final step is to return the indices of the two values that sum to the target. If the difference is not found in the 'previous map', the current value and its index are added to the map, and the iteration continues.

Outlines

00:00

🔍 Finding Pairs in Arrays

This paragraph introduces a coding problem where the goal is to find two values in an array that sum up to a given target, in this case, 9. The example given is an array with the values [2, 1, 5, 3], where 2 and 7 sum to 9. The solution involves returning the indices of these values, which are 0 and 1 respectively. It is mentioned that there's exactly one solution, so there's no need to handle multiple or no solutions. The paragraph then discusses the brute force approach of checking every combination of two values, which is inefficient with a time complexity of O(N^2). It hints at a more efficient method using a hash map to store values and their indices for quick lookup, which is the focus of the next paragraph.

05:03

🚀 Optimizing with a Hash Map

The second paragraph delves into the more efficient solution using a hash map to solve the array sum problem. It explains that instead of checking every combination of two numbers, one can use a hash map to store each value's index as it's encountered. The process involves iterating through the array and for each element, calculating the difference needed to reach the target sum and checking if this difference exists in the hash map. If it does, the indices of the two numbers are returned as the solution. The paragraph clarifies that the hash map should be checked and updated in a single pass through the array, ensuring that the first element needed to reach the target sum will always be in the hash map by the time the second element is encountered. The time complexity of this approach is O(N), with an additional memory complexity of O(N) due to the hash map. The paragraph concludes with a brief mention of coding the solution, emphasizing the simplicity and efficiency of this method.

Keywords

💡Leak Code

In the context of the video, 'Leak Code' refers to a problem-solving scenario where a solution is sought for a given challenge, specifically finding two values in an array that sum to a target value. The term is not about leaking actual code but rather solving a coding problem. The script uses the phrase to introduce the algorithmic problem of finding two numbers in an array that add up to a specified target.

💡Input Array

An 'Input Array' is a collection of data elements that serve as the starting point for the algorithm discussed in the video. It is the array within which the algorithm searches for two values that sum to the target number. The video script uses the input array to demonstrate the process of finding the two indices that meet the condition.

💡Target

The 'Target' in the script represents the specific sum that the algorithm is trying to achieve by adding two values from the input array. It is the goal value that the algorithm must reach by selecting appropriate elements from the array. The script uses the number 9 as the target to illustrate the problem-solving process.

💡Indices

'Indices' are the positions or the numerical identifiers of elements within an array. In the video, the indices are the key output, as the algorithm's goal is to find the positions of the two values in the array that sum to the target. The script mentions indices 0 and 1 as the solution to the example problem.

💡Brute Force

In the context of the video, 'Brute Force' refers to a method of solving a problem by trying every possible combination or solution until the correct one is found. The script initially describes a brute force approach to find two numbers in the array that sum to the target, which involves checking every combination of two array elements.

💡Hash Map

A 'Hash Map' is a data structure that allows for efficient storage and retrieval of key-value pairs. In the video, the hash map is used as an efficient way to solve the problem by storing the elements of the array and their indices, allowing for quick lookup to find if the complement of the current element (target minus the element) exists in the map.

💡Time Complexity

'Time Complexity' is a measure of the amount of time an algorithm takes to run as a function of the size of the input. The script discusses the inefficiency of the brute force approach with a time complexity of O(N^2) and then presents a more efficient solution using a hash map, which has a time complexity of O(N).

💡Algorithm

An 'Algorithm' is a set of rules or steps used to solve a problem. The video script describes two different algorithms for finding two numbers in an array that sum to a target. The first is a brute force method, and the second is a more efficient method using a hash map.

💡Runtime

'Runtime' refers to the duration of time that a program or algorithm takes to execute. The script discusses the inefficiency of the brute force method in terms of runtime, stating that it is not very efficient due to its O(N^2) time complexity.

💡Memory Complexity

'Memory Complexity' is the amount of memory space required by an algorithm to solve a problem. The script mentions that while the hash map solution improves time efficiency, it comes at the cost of increased memory usage, as it requires storing all elements and their indices, resulting in O(N) memory complexity.

Highlights

The problem is to find two values in an array that sum to a given target, with the guarantee of a single solution.

A brute force approach checks every combination of two values to find the sum, with a time complexity of O(N^2).

A more efficient method involves using a hash map to store values and their indices, reducing the time complexity to O(N).

The hash map allows for immediate checking of the existence of a required value to complement the current element to the target sum.

The algorithm can iterate through the array in a single pass, improving efficiency over the brute force method.

The hash map is initially empty and is populated as the array is iterated, avoiding the need to pre-populate it with the entire array.

The algorithm cleverly avoids reusing the same value by checking the index of the current element against the index stored in the hash map.

The solution involves returning the indices of the two values that sum to the target, demonstrated with an example using the numbers 2 and 7.

The transcript explains the process of adding elements to the hash map and checking for the required complementing value.

The use of a hash map eliminates the need for multiple passes through the array, streamlining the process.

The algorithm ensures that once the second element that sums to the target is visited, the first one must already be in the hash map.

The time complexity of the hash map approach is O(N) due to the single pass through the array and constant time operations for the hash map.

The memory complexity is also O(N), as every value in the array could potentially be stored in the hash map.

The transcript provides a step-by-step explanation of how to implement the hash map solution in code.

The final code snippet demonstrates the practical application of the hash map to solve the problem efficiently.

The transcript concludes with a successful demonstration of the algorithm's effectiveness in finding the solution in a single pass.

Transcripts

let's all leak code one to some the most

popular leak code question so we're

given an input array and some target in

this case 9 and we want to find the two

values in this input array that sum to 9

so in this case it's 2 & 7 now we want

to return the indices of these two

values so the index of zero of the index

of two is zero the index of 7 is 1 so we

return 0 & 1

we're guaranteed that there's exactly

one solution so we don't have to worry

about not finding a solution and we

don't have to worry about multiple

solutions now the most intuitive way to

solve this problem is basically just

check every combination of two values

and see if they can sum up to our target

so we start at 2 we check every

combination we can make that includes 2

so we scan through the remainder of the

array 1 5 3 and check if any of those

numbers added to 2 some store target for

in this case none of them do so next we

can repeat the process

let's check every combination including

one that sums up the target for so we

scan through every element that comes

after at 5 & 3 and we find that one

added with 3 sums up to our target 4

notice that we didn't have to check the

values that came before 1 because we

already checked the combination 2 & 1

when we were up over here remember when

we checked every combination with 2 so

we didn't have to repeat that work down

here we only had to check the numbers

that came after 1

so the runtime of this algorithm isn't

super efficient this is basically brute

force we're going through the entire

array of length and and we're gonna do

that worst case n times for each number

this means that over all worst case time

complexity will be O of N squared so can

we do better now the thing to notice is

that for each number for example 1 the

value we're looking for is the

difference between the target and this

value 1 so we're looking for 4 minus 1

which is equal to 3 so that means this

is the only value we can add to one

that'll equal the target so we don't

have to check every number we just want

to know if

resist s-- now the easiest way we can do

this the most efficient is by making a

hash map of every value in our input

array so we can instantly check if the

value 3 exists now let's try the same

problem except let's use a hash map this

time now in our hash map we're going to

be mapping each value to the index of

each value so the index of 2 is 0 the

index of 1 is 1 the index of 5 is 2 the

index of 3 is 3 so let's so in our hash

map we're going to be mapping the value

to the index

now we could add every value in this

array into the hash map before we start

iterating through it but there's

actually an easier way to do it if we

added the entire array into the hash map

initially then we would get to the value

2 first right we would want to checked

as the difference between target 4 minus

this value 2 which is equal to 2 exists

in our hash map and we would find that 2

does exist in our hash map but we're not

allowed to reuse the same one right

because they're both at the same index

we can't use the same value twice so we

would have to compare the index of our

current 2 with the index of the 2 that's

in our hash map there's actually an

easier way to do this though and it's a

little clever and let me show you how to

do it that way so doing it this clever

way initially we say our hash map is

empty so we get to the value 2 first of

all right and we want to look for the

difference 4 minus 2 in our hash map our

hash map is empty so we don't find 2 so

then after we visited this element then

we can add it to our hash map so now

that I'm done visiting it I'm gonna move

to the second element 1 and before I do

that I'm gonna add this value 2 to our

hash map and the index of this value is

gonna be 0 now I'm at 1 I'm looking for

4 minus 1 which is 3

I see 3 isn't in our hash map but it

actually is in our array so what's the

problem well for now we're going to say

we don't find our find

three so we add one to our hash map the

index of this one is one and now we move

to the next element five we check does

four minus five it's four minus five

exists in our hash map that's negative

one so no it does not then we add this

five to our hash map and it's index

which is two and we move to the last

value in the array three we checked this

four minus three e

exists in our hash map now that's one so

we see it does exist right right over

here it exists the value exists and it's

index is one so now we found our two

values that sum to the target and we

want to return their indexes their

indices which are going to be one and

three so with this algorithm we don't

have to initialize our hash map it can

be initially empty and then we can just

iterate through this array in one pass

now the reason the algorithm can work in

that way with just one pass is this so

let's say we had a giant array right we

know for sure that there are two

elements in this array that sum to our

target right we don't know where they

are they're at some arbitrary location

when we visit the first one of those

elements our hash map is only going to

be this portion of the array it's only

going to have the values that came

before the first value so we're gonna

we're gonna notice that the second value

that can sum to the target is no is not

going to be in our hash map yet but once

we get to the second value our hash map

is going to be this portion so every

value that comes before this right so

we're gonna be guaranteed that once we

visit the second element that sums up to

the target we're gonna be guaranteed

that the first one is already in our

hash map so we're guaranteed to find the

solution now since we only have to

iterate through the array once and we're

adding each value to our hash map which

is a constant time operation and we're

checking if a value exists in our hash

which is also a constant time operation

the time complexity is going to be Big O

of n we are using extra memory right

that hash map isn't free so the memory

complexity is also going to be O of n

because we could potentially add every

value to the hash map

so now let's code the solution so

remember we need a hash map right I'm

going to call this previous map because

it's basically every element that comes

before the current home that every

previous element is going to be stored

in this map where it can be mapping the

value to the index of that value

so now let's iterate through every value

in this array we need the index as well

as the actual number so let's do it like

this and Python before we add this to

our map let's check if the difference

which is equal to target minus n now

let's check if this difference is

already in the hash map if it is then we

can return the solution which is going

to be a pair of the indices so I can get

the first index like this and the second

index is just AI now if we don't find

the solution then we have to update our

hash map so for this value n I'm gonna

say the index is I and then we're going

to continue since we're guaranteed that

a solution exists we don't have to

return anything out here right but I'll

just put a return for no reason now

let's see if it works and it works

perfectly so with this kind of neat

little trick but just doing it in one

pass you can reduce the amount of code

you have to write and not have to worry

about like edge cases and comparisons

and things like that