Description
https://leetcode.com/problems/two-sum-iii-data-structure-design/
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 theTwoSum
object, with an empty array initially.void add(int number)
Addsnumber
to the data structure.boolean find(int value)
Returnstrue
if there exists any pair of numbers whose sum is equal tovalue
, otherwise, it returnsfalse
.
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 toadd
andfind
.
Explanation
Use a list to track the elements added and a set to track complements.
Python Solution
class TwoSum:
def __init__(self):
"""
Initialize your data structure here.
"""
self.nums = []
def add(self, number: int) -> None:
"""
Add the number to an internal data structure..
"""
self.nums.append(number)
def find(self, value: int) -> bool:
"""
Find if there exists any pair of numbers which sum is equal to the value.
"""
complements = set()
for num in self.nums:
if value - num in complements:
return True
complements.add(num)
return False
# Your TwoSum object will be instantiated and called as such:
# obj = TwoSum()
# obj.add(number)
# param_2 = obj.find(value)
- Time Complexity: O(N).
- Space Complexity: O(N).