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class Solution:
def twoSum(self, nums: list[int], target: int) -> list[int]:
# Hash Map: O(N) Time | O(N) Space
seen = {}
for i, val in enumerate(nums):
complement = target - val
if complement in seen:
return [seen[complement], i]
seen[val] = i
return []
# Test Run: nums = [2, 7, 11, 15], target = 9
# Status: ACCEPTED (Runtime: 38ms, Memory: 17.8MB)Curated Algorithmic Problems & Verified Solutions
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While iterating through the array, we check if the complement (target - nums[i]) exists in our hash map. If present, we return both indices in $O(1)$ lookup time.
seen = {}
for i, val in enumerate(nums):
if target - val in seen:
return [seen[target - val], i]
seen[val] = iCompete in Real-Time Timed Coding Contests
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"Help me write a discussion post explaining how to solve Two Sum in Python with $O(N)$ time complexity and edge-case diagrams."
Summary: Using a single-pass hash map, we eliminate the inner nested loop. Here is the complete intuition, LaTeX complexity proof, and tested Python 3 implementation:
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