Hi, nice solution you pointed out. I solved a similar problem back at AdventOfCode. I leveraged a HashMap to solve this problem in a time complexity of O(n), but slightly increased space complexity (given the array is already present and not to be computed or serialized from storage).
Given the following input:
Input: nums = {0, -1, 2, -3, 1}, target = -2
Loop through the map and check if target - currentItem (since lookup is possible in constant time) exists also in the map -> sum found
Example (-3 is currentItem):
-2 - (-3) = -2 + 3 = 1
1 exists --> sum found
Looping through whole map (worst case) sums up to a time complexity of O(n) as stated above.
Hi, nice solution you pointed out. I solved a similar problem back at AdventOfCode. I leveraged a HashMap to solve this problem in a time complexity of O(n), but slightly increased space complexity (given the array is already present and not to be computed or serialized from storage).
Given the following input:
Loop through the map and check if
target - currentItem(since lookup is possible in constant time) exists also in the map -> sum foundLooping through whole map (worst case) sums up to a time complexity of O(n) as stated above.
Hey thanks for the above solution. I have posted about this approach in my first blog
alkeshghorpade.me/post/leetcode-tw...