Leetcode #170: Two Sum III - Data structure design
In this guide, we solve Leetcode #170 Two Sum III - Data structure design in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Design a data structure that accepts a stream of integers and checks if it has a pair of integers that sum up to a particular value. Implement the TwoSum class: TwoSum() Initializes the TwoSum object, with an empty array initially.
Quick Facts
- Difficulty: Easy
- Premium: Yes
- Tags: Design, Array, Hash Table, Two Pointers, Data Stream
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input
["TwoSum", "add", "add", "add", "find", "find"]
[[], [1], [3], [5], [4], [7]]
Output
[null, null, null, null, true, false]
Explanation
TwoSum twoSum = new TwoSum();
twoSum.add(1); // [] --> [1]
twoSum.add(3); // [1] --> [1,3]
twoSum.add(5); // [1,3] --> [1,3,5]
twoSum.find(4); // 1 + 3 = 4, return true
twoSum.find(7); // No two integers sum up to 7, return false
Python Solution
class TwoSum:
def __init__(self):
self.cnt = defaultdict(int)
def add(self, number: int) -> None:
self.cnt[number] += 1
def find(self, value: int) -> bool:
for x, v in self.cnt.items():
y = value - x
if y in self.cnt and (x != y or v > 1):
return True
return False
# Your TwoSum object will be instantiated and called as such:
# obj = TwoSum()
# obj.add(number)
# param_2 = obj.find(value)
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.