The Question
Coding

Two Sum

Given an array of integers nums and an integer target, return the indices of the two numbers such that they add up to the target value. You may assume that each input would have exactly one solution, and you may not use the same element twice. The answer can be returned in any order. Constraints: - 2 <= nums.length <= 10^4 - -10^9 <= nums[i] <= 10^9 - -10^9 <= target <= 10^9 - Only one valid answer exists.
Python
Hash Map
Questions & Insights

Clarifying Questions

Input Constraints: What is the maximum size of the input array? (e.g., 10^4, 10^5?)
Value Ranges: Can the integers in the array or the target be negative?
Duplicate Elements: Are we guaranteed that a solution exists, and if there are multiple, should we return any one pair or all of them?
Output Format: Should we return the indices of the numbers or the numbers themselves?

Assumptions:

The array size is between 2 and 10^4.
Each input has exactly one solution.
We must return the indices of the two numbers.
We cannot use the same element twice (e.g., if nums = [3, 2, 4] and target = 6, we cannot use 3 twice).

Thinking Process

Brute Force Approach: Check every possible pair (i, j) and see if their sum equals the target. This results in O(n^2) time complexity, which is inefficient for large inputs.
Optimization via Complementary Lookup: For any element x at index i, we need to find if the value target - x exists elsewhere in the array. A Hash Map (dictionary in Python) allows for O(1) average-time lookups.
One-Pass Strategy: We can iterate through the array once. For each element, check if its "complement" (target - nums[i]) is already in the hash map. If it is, we found our pair. If not, we store the current element's value and its index in the map and continue.
Handling Constraints: Since there is exactly one solution and we can't reuse the same index, the one-pass approach naturally handles this by checking for the complement before adding the current index to the map.
Implementation Breakdown

Problem Set

Functional Requirement: Find two distinct indices i, j such that nums[i] + nums[j] == target.
Time Complexity Constraint: Should be better than O(n^2), preferably O(n).
Space Complexity Constraint: O(n) to store elements in a hash-based structure.

Approach

Algorithm: One-Pass Hash Map lookup.
Data Structure: dict (Hash Map).
Complexity:
Time: O(n) because we traverse the list containing n elements only once.
Space: O(n) as we store at most n elements in the dictionary.

Implementation

Wrap Up

Advanced Topics

Memory vs. Time: If the array is already sorted, we could use a Two-Pointer approach. This would reduce space complexity to O(1) while maintaining O(n) time complexity.
Large Scale Data: If the dataset is too large to fit in memory, we could use a distributed hash table or external sorting followed by a two-pointer scan.
Hash Collisions: While Python's dict handles collisions internally using open addressing, in a low-level system design interview, discussing the impact of hash collisions on the O(1) average lookup time (worst case O(n)) shows depth.