Two Sum - Leetcode 1 - Hashmaps & Sets (Python)

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[GregHogg]

Master Data Structures & Algorithms For FREE at AlgoMap.io!

[harshgamer]

wow, the code was a little hard to understand at first, but when i coded it out myself, i finally got it. thanks greg!. Also, will this solution be accepted in coding interviews?

[HumbleFool-xhq]

def twoSum(self, nums: List[int], target: int) -> List[int]:
hashmap = dict()
for index, num in enumerate(nums):
diff = target-num
if diff in hashmap:
return [index, hashmap[diff]]
hashmap[num]=index

[hamidonai]

You can simply do this with only one for-loop:
```python
def twoSum(self, nums: List[int], target: int) -> List[int]:
indices = {}
for i in range(len(nums)):
diff = target - nums[i]
if diff in indices:
return indices[diff], i
indices[nums[i]] = i
```

[softcodeacademy]

2:05 there is no point of updating this index. If the key already exists we can skip to the next iteration.

[hlubradio]

Hey Greg, the left bracket threw me up off for a bit. LOL. It looked like a reverse 3. Could you just make the middle part a bit more legible? { ... } Thanks!

[cadentoh]

@GregHogg hello, the hash map takes the latest occurrence's index. But what if there was an earlier occurrence of that index? Won't there be an error, not because the logic is incorrect, but the index is not of the earliest one? Might have understood, please be kind thanks.

[baekBlackbeen]

I'm solving this in JAVA. And I didn't know containsKey method takes O(1) time complexity so I thought how is this O(N) time complexity, since it basically looked the same as brute force with using HashMaps.

[nerdyboy]

Make video explanation on Three Sum and Four Sum too...!

[hlubradio]

How are you able to use List instead of list?

[ApnaPakistan]

Why is the time complexity O(n)? Does the hashing solution contain two loops?

[danangyudhatama]

Is this solution obvious for you? As a beginner, i would never get this solution😂

[Sona-ykus]

class Solution:
# Time complexity is O(n) and Space Complexity is O(n)
def twoSum(self, nums: List[int], target: int) -> List[int]:
nums_to_index = {} # Dictionary to store numbers and their indices
for i, n in enumerate(nums):
diff = target - n
if diff in nums_to_index:
return [nums_to_index[diff], i]
nums_to_index[n] = i
return []
Hi Greg, I would like to get to learn the code review comments from you on my code here^. It will be super helpful for my learning journey.
Looking forward to your response.