**Two Sum III – Data structure design** **LeetCode Solution | LeetCode Problems For Beginners | LeetCode Problems & Solutions | Improve Problem Solving Skills | LeetCode Problems Java | LeetCode Solutions in C++**

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** Link for the Problem** – Two Sum III – Data structure design– LeetCode Problem

Two Sum III - Data structure design– LeetCode Problem

**Problem:**

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.`void add(int number)`

Adds`number`

to the data structure.`boolean find(int value)`

Returns`true`

if there exists any pair of numbers whose sum is equal to`value`

, otherwise, it returns`false`

.

#### EXAMPLE 1:

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

**Constraints:**

`-105 <= number <= 105`

`-231 <= value <= 231 - 1`

- At most
`5 * 104`

calls will be made to`add`

and`find`

.

Two Sum III - Data structure design– LeetCode Solutions

Two Sum III - Data structure designSolution in C++:

class TwoSum { public: void add(int number) { ++count[number]; } bool find(int value) { for (const auto& [key, freq] : count) { const int remain = value - key; if (key == remain && freq > 1) return true; if (key != remain && count.count(remain)) return true; } return false; } private: unordered_map<int, int> count; };

Two Sum III - Data structure designSolution in Java:

class TwoSum { public void add(int number) { count.put(number, count.getOrDefault(number, 0) + 1); } public boolean find(int value) { for (Map.Entry<Integer, Integer> entry : count.entrySet()) { final int key = entry.getKey(); final int remain = value - key; if (key == remain && entry.getValue() > 1) return true; if (key != remain && count.containsKey(remain)) return true; } return false; } private HashMap<Integer, Integer> count = new HashMap<>(); }

Two Sum III - Data structure designSolution in Python:

class TwoSum: def __init__(self): self.count = Counter() def add(self, number: int) -> None: self.count[number] += 1 def find(self, value: int) -> bool: for key, freq in self.count.items(): remain = value - key if key == remain and freq > 1: return True if key != remain and remain in self.count: return True return False