en/docs/chapter_searching/replace_linear_by_hashing.md
In algorithm problems, we often reduce the time complexity of algorithms by replacing linear search with hash-based search. Let's use an algorithm problem to deepen our understanding.
!!! question
Given an integer array `nums` and a target element `target`, search for two elements in the array whose "sum" equals `target`, and return their array indices. Any solution will do.
Consider directly traversing all possible combinations. As shown in the figure below, we open a two-layer loop and judge in each round whether the sum of two integers equals target. If so, return their indices.
The code is shown below:
[file]{two_sum}-[class]{}-[func]{two_sum_brute_force}
This method has a time complexity of $O(n^2)$ and a space complexity of $O(1)$, which is very time-consuming with large data volumes.
Consider using a hash table where key-value pairs are array elements and element indices respectively. Loop through the array, performing the steps shown in the figure below in each round:
target - nums[i] is in the hash table. If so, directly return the indices of these two elements.nums[i] and index i to the hash table.=== "<1>"
=== "<2>"
=== "<3>"
The implementation code is shown below, requiring only a single loop:
[file]{two_sum}-[class]{}-[func]{two_sum_hash_table}
This method reduces the time complexity from $O(n^2)$ to $O(n)$ through hash-based search, greatly improving runtime efficiency.
Since an additional hash table needs to be maintained, the space complexity is $O(n)$. Nevertheless, this method achieves a more balanced overall time-space efficiency, making it the optimal solution for this problem.