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 <= 104
-109 <= nums[i] <= 109
-109 <= target <= 109
- Only one valid answer exists.
Follow-up: Can you come up with an algorithm that is less than O(n2)
time complexity?
Using hashmap to look up, it can easily to achieve O(n)
class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
check_dict = {}
for index, num in enumerate(nums):
lookup_num = target - num
if lookup_num in check_dict:
return [check_dict[lookup_num], index]
else:
check_dict[num] = index
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