LeetCode sum two: explanations and solutions with Cpp/Java/Python/Ruby
First, we come out with the simplest solution, iterate with array element, check whether there are two value’s sum is target, if true, we get a result, so pseudo code is:
Time complexity: $O(N^2)$ Space complexity: $O(1)$
Optimization with hashmap
Time general idea is use space to optimize time complexity. We can use a hashmap to optimize the second loop from O(N) to O(1).
So we build a hashmap, the key is the element of array, and value is the index of the element. This hashmap can be constructed before hand, but a better way is to construct in the same loop of checking.
Time complexity: $O(N)$ Space complexity: $O(N)$
Java version is same logic with Cpp version:
Why we don’t need a hashmap in Python version, please note the second if test. It’s a pragmatic style for Python check whether a element is in array:
Note we use nums[i+1:] in checking, that means we use part of nums index begin with i+1. According to the document, the time complexity of ‘in’ operator is O(N) for list, so this implementation’s time complexity is: $O(N^2)$
Ruby’s version is short, the logic is same with a Hashmap.
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