Given an array of integers, return the indices of the two numbers that add up to the target.
1deftwo_sum(nums, target):
2 seen = {}
3for i, n in enumerate(nums):
4 need = target - n
5if need in seen:
6return [seen[need], i]
7 seen[n] = i
8 # O(n) time · O(n) space
Private Display
Listening
HINT
Think about what value you need to find for each number. Can one lookup replace the inner loop?
Your question
“Can I do this in one pass?”
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01class Solution:
02def maxProfit(self, prices):
03 left = 0
04 right = 1
05while right < len(prices):
06# your approach stays in context
AI has context
Challenge Best Time to Buy and Sell Stock
Current code Two-pointer approach detected
You asked “Why do I move left here?”
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Solution · PythonUpdated from your code
01def two_sum(nums, target):
02 seen = {}
03for i, n in enumerate(nums):
04 need = target - n
05 # fixed: check before storing
06if need in seen:
07return [seen[need], i]
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“Why does a hash map make this O of n instead of O of n squared?”
Because each lookup is constant-time on average. You replace the inner search loop with one hash-table lookup per element.
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for i in range(len(nums)): # focus on your code
HINT
ANSWER
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2
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3
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