Two Sum
Detailed guide and Python implementation for the 'Two Sum' problem.
1. Concept Overview
The 'Two Sum' problem is a key challenge in the Arrays & Hashing section.
This implementation focuses on easy-level logic in Python.
We prioritize technical accuracy and code readability in our provided solutions.
2. Real-World Applications
3. Visual Intuition
Visualizing the logic flow for Two Sum.
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
Read the problem statement for Two Sum carefully.
2. Formulate brute force
Draft a simple iterative solution.
3. Identify inefficiency
Look for redundant calculations.
4. Optimize search path
Use hashing or sorting to speed up the process.
5. Final Implementation
Clean up the code for production standards.
Problem Statement
Given an array of integers nums and an integer target, return the indices of the two numbers such that they add up to target.
You may assume that each input would have exactly one solution, and you may not use the same element twice.
You can return the answer in any order.
Write a function twoSum(nums: List[int], target: int) -> List[int].
- •2 <= len(nums) <= 10^4
- •-10^9 <= nums[i] <= 10^9
- •-10^9 <= target <= 10^9
- •Only one valid answer exists
Examples
nums = [2, 7, 11, 15], target = 9
[0, 1]
nums[0] + nums[1] = 2 + 7 = 9, so we return [0, 1].
nums = [3, 2, 4], target = 6
[1, 2]
nums[1] + nums[2] = 2 + 4 = 6, so we return [1, 2].
nums = [3, 3], target = 6
[0, 1]
nums[0] + nums[1] = 3 + 3 = 6, so we return [0, 1].
Need a Hint?
Edge Cases to Watch
- Empty input structures
- Single element inputs
- Large numerical bounds
Ready to Solve?
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Interview Insights & Variations
Complexity Analysis Breakdown
Why Time: Directly evaluates all possibilities.
Why Space: Uses standard local memory.
Why Time: Optimized paths reduce total operations.
Why Space: May trade memory for speed.
Optimized Solution Python Code
Optimized Solution Python Code
def two_sum_opt(nums, target):
prevMap = {} # val : index
for i, n in enumerate(nums):
diff = target - n
if diff in prevMap:
return [prevMap[diff], i]
prevMap[n] = i
return []Brute Force Code (Spoiler Guarded)
Brute Force Code (Spoiler Guarded)
def two_sum_brute(nums, target):
n = len(nums)
for i in range(n):
for j in range(i + 1, n):
if nums[i] + nums[j] == target:
return [i, j]
return []Algorithm Pattern Checklist
When dealing with Arrays & Hashing data patterns.
Core Prerequisites
Revision Key Notes
Common Mistakes & Pitfalls
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