Shadowing Practice: Two Sum - Leetcode 1 - HashMap - Python - Learn English Speaking with Video

Let's solve leak code 1 to sum, 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 and 7.

Now we want to return the indices of these two values.

So the index of 2 is 0, the index of 7 is 1.

So we return 0 and 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 two, we check every combination we can make that includes two.

So we scan through the remainder of the array, one, five, three, and check if any of those numbers added to two sums to our target four.

In this case, none of them do.

So next we can repeat the process.

Let's check every combination including one that sums up to target four.

So we scan through every element that comes after it, five and three, and we find that one added with three sums up to our target four.

Notice that we didn't have to check the values

that came before one because we already checked the combination two and one 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 n and we're going to do that worst case n times for each number.

This means that overall 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 1 that'll equal the target.

So we don't have to check every number, we just want to know if 3 exists.

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 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 check does 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

two with the index of the two 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 two 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've visited this element, then we can add it to our hash map.

So now that I'm done visiting it, I'm going to move to the second element 1.

And before I do that, I'm going to add this value 2 to our hash map, and the index of this value is going to 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 a 3.

So we add 1 to our hash map.

The index of this 1 is 1.

And now we move to the next element, 5.

we check does 4 minus 5 uh it's 4 minus 5 exist in our hash map that's negative 1

so no it does not then we add this 5 to our hash map

and it's index which is 2

and we move to the last value in the array 3

we check does 4 minus 3 exist in our hash map now that's 1

so we see it does exist right over here.

The value exists and its 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 gonna be this portion of the array.

It's only gonna have the values that came before the first value.

So we're going to notice that the second value

that can sum to the target 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 going to be guaranteed that once we visit the second element that sums up to the target, we're going to 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 map, 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 element.

Every previous element is going to be stored in this map.

We're going to 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 in Python.

100

Before we add this to our map, let's check if the difference, which is equal to target minus n.

101

Now let's check if this difference is already in the hash map.

102

if it is, then we can return the solution, which is going to be a pair of the indices.

103

So I can get the first index like this, and the second index is just i.

104

Now if we don't find the solution, then we have to update our hash map.

105

So for this value n, I'm going to say the index is i, and then we're going to continue.

106

Since we're guaranteed that a solution exists, we don't have to return anything out here, right?

107

But I'll just put a return for no reason.

108

Now let's see if it works. And it works perfectly.

109

So with this kind of neat little trick with just doing it in one pass, you can reduce the amount of code you have to write

110

and not have to worry about like edge cases and comparisons and things like that.

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