1-TwoSum

Two Sum

Easy

Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

You can return the answer in any order.

Example 1:

Input: nums = [2,7,11,15], target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].

Example 2:

Input: nums = [3,2,4], target = 6
Output: [1,2]

Example 3:

Input: nums = [3,3], target = 6
Output: [0,1]

Constraints:

• $2$ <= nums.length <= $10^4$
• $-10^9$ <= nums[i] <= $10^9$
• $-10^9$ <= target <= $10^9$
• Only one valid answer exists.

Follow-up: Can you come up with an algorithm that is less than $O(n^2)$ time complexity?

Problem can be found in here!

Solution

Basic Solution: Brute Force

# target: int, nums: List[int]

 for i in range(len(nums)):
    number_to_find = target - nums[i]
    for j in range(i + 1, len(nums)):
        if nums[j] == number_to_find:
            return [i, j] # Find indices of the two numbers!

Time Complexity: $O(n^2)$, Space Complexity: $O(1)$

Explanation: Simply iterate the array and find the target value among the array in each iteration.

Improved Solution: Hash Table

memo = {}
for i, j in enumerate(nums):
    number_to_find = target - j
    try:
        return [memo[j], i] # Find indices of the two numbers!
    except KeyError:
        memo[number_to_find] = i

Time Complexity: $O(n)$, Space Complexity: $O(n)$

Explanation: Instead of searching the whole array blindlessly in each iteration, using a hash table can determine whether this element is the target value in $O(1)$ time.