10.4 Hash optimization strategies¶
In algorithm problems, we often reduce the time complexity of algorithms by replacing linear search with hash search. Let's use an algorithm problem to deepen understanding.
Question
Given an integer array nums and a target element target, please search for two elements in the array whose "sum" equals target, and return their array indices. Any solution is acceptable.
10.4.1 Linear search: trading time for space¶
Consider traversing all possible combinations directly. As shown in Figure 10-9, we initiate a two-layer loop, and in each round, we determine whether the sum of the two integers equals target. If so, we return their indices.
Figure 10-9 Linear search solution for two-sum problem
The code is shown below:
/* Method one: Brute force enumeration */
vector<int> twoSumBruteForce(vector<int> &nums, int target) {
int size = nums.size();
// Two-layer loop, time complexity is O(n^2)
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target)
return {i, j};
}
}
return {};
}
/* Method one: Brute force enumeration */
int[] twoSumBruteForce(int[] nums, int target) {
int size = nums.length;
// Two-layer loop, time complexity is O(n^2)
for (int i = 0; i < size - 1; i++) {
for (int j = i + 1; j < size; j++) {
if (nums[i] + nums[j] == target)
return new int[] { i, j };
}
}
return new int[0];
}
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.
10.4.2 Hash search: trading space for time¶
Consider using a hash table, with key-value pairs being the array elements and their indices, respectively. Loop through the array, performing the steps shown in Figure 10-10 each round.
- Check if the number
target - nums[i]is in the hash table. If so, directly return the indices of these two elements. - Add the key-value pair
nums[i]and indexito the hash table.
Figure 10-10 Help hash table solve two-sum
The implementation code is shown below, requiring only a single loop:
def two_sum_hash_table(nums: list[int], target: int) -> list[int]:
"""Method two: Auxiliary hash table"""
# Auxiliary hash table, space complexity is O(n)
dic = {}
# Single-layer loop, time complexity is O(n)
for i in range(len(nums)):
if target - nums[i] in dic:
return [dic[target - nums[i]], i]
dic[nums[i]] = i
return []
/* Method two: Auxiliary hash table */
vector<int> twoSumHashTable(vector<int> &nums, int target) {
int size = nums.size();
// Auxiliary hash table, space complexity is O(n)
unordered_map<int, int> dic;
// Single-layer loop, time complexity is O(n)
for (int i = 0; i < size; i++) {
if (dic.find(target - nums[i]) != dic.end()) {
return {dic[target - nums[i]], i};
}
dic.emplace(nums[i], i);
}
return {};
}
/* Method two: Auxiliary hash table */
int[] twoSumHashTable(int[] nums, int target) {
int size = nums.length;
// Auxiliary hash table, space complexity is O(n)
Map<Integer, Integer> dic = new HashMap<>();
// Single-layer loop, time complexity is O(n)
for (int i = 0; i < size; i++) {
if (dic.containsKey(target - nums[i])) {
return new int[] { dic.get(target - nums[i]), i };
}
dic.put(nums[i], i);
}
return new int[0];
}
This method reduces the time complexity from \(O(n^2)\) to \(O(n)\) by using hash search, greatly improving the running efficiency.
As it requires maintaining an additional hash table, the space complexity is \(O(n)\). Nevertheless, this method has a more balanced time-space efficiency overall, making it the optimal solution for this problem.