Two Sum III – Data structure design LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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Hello Programmers/Coders, Today we are going to share solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python. At Each Problem with Successful submission with all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

In this post, you will find the solution for the Two Sum III – Data structure design in C++, Java & Python-LeetCode problem. We are providing the correct and tested solutions to coding problems present on LeetCode. If you are not able to solve any problem, then you can take help from our Blog/website.

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Link for the ProblemTwo Sum III – Data structure design– LeetCode Problem

Two Sum III - Data structure design– LeetCode 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.


["TwoSum", "add", "add", "add", "find", "find"]
[[], [1], [3], [5], [4], [7]]
[null, null, null, null, true, false]
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


  • -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 design Solution in C++:
class TwoSum {
  void add(int 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;

  unordered_map<int, int> count;
Two Sum III - Data structure design Solution 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 design Solution 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

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