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# 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.
{ class="animation-figure" }
Figure 10-9 Linear search solution for two-sum problem
The code is shown below:
=== "Python"
```python title="two_sum.py"
def two_sum_brute_force(nums: list[int], target: int) -> list[int]:
"""Method one: Brute force enumeration"""
# Two-layer loop, time complexity is O(n^2)
for i in range(len(nums) - 1):
for j in range(i + 1, len(nums)):
if nums[i] + nums[j] == target:
return [i, j]
return []
```
=== "C++"
```cpp title="two_sum.cpp"
/* Method one: Brute force enumeration */
vector twoSumBruteForce(vector &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 {};
}
```
=== "Java"
```java title="two_sum.java"
/* 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];
}
```
=== "C#"
```csharp title="two_sum.cs"
[class]{two_sum}-[func]{TwoSumBruteForce}
```
=== "Go"
```go title="two_sum.go"
[class]{}-[func]{twoSumBruteForce}
```
=== "Swift"
```swift title="two_sum.swift"
[class]{}-[func]{twoSumBruteForce}
```
=== "JS"
```javascript title="two_sum.js"
[class]{}-[func]{twoSumBruteForce}
```
=== "TS"
```typescript title="two_sum.ts"
[class]{}-[func]{twoSumBruteForce}
```
=== "Dart"
```dart title="two_sum.dart"
[class]{}-[func]{twoSumBruteForce}
```
=== "Rust"
```rust title="two_sum.rs"
[class]{}-[func]{two_sum_brute_force}
```
=== "C"
```c title="two_sum.c"
[class]{}-[func]{twoSumBruteForce}
```
=== "Kotlin"
```kotlin title="two_sum.kt"
[class]{}-[func]{twoSumBruteForce}
```
=== "Ruby"
```ruby title="two_sum.rb"
[class]{}-[func]{two_sum_brute_force}
```
=== "Zig"
```zig title="two_sum.zig"
[class]{}-[func]{twoSumBruteForce}
```
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.
1. Check if the number `target - nums[i]` is in the hash table. If so, directly return the indices of these two elements.
2. Add the key-value pair `nums[i]` and index `i` to the hash table.
=== "<1>"
{ class="animation-figure" }
=== "<2>"
{ class="animation-figure" }
=== "<3>"
{ class="animation-figure" }
Figure 10-10 Help hash table solve two-sum
The implementation code is shown below, requiring only a single loop:
=== "Python"
```python title="two_sum.py"
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 []
```
=== "C++"
```cpp title="two_sum.cpp"
/* Method two: Auxiliary hash table */
vector twoSumHashTable(vector &nums, int target) {
int size = nums.size();
// Auxiliary hash table, space complexity is O(n)
unordered_map 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 {};
}
```
=== "Java"
```java title="two_sum.java"
/* Method two: Auxiliary hash table */
int[] twoSumHashTable(int[] nums, int target) {
int size = nums.length;
// Auxiliary hash table, space complexity is O(n)
Map 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];
}
```
=== "C#"
```csharp title="two_sum.cs"
[class]{two_sum}-[func]{TwoSumHashTable}
```
=== "Go"
```go title="two_sum.go"
[class]{}-[func]{twoSumHashTable}
```
=== "Swift"
```swift title="two_sum.swift"
[class]{}-[func]{twoSumHashTable}
```
=== "JS"
```javascript title="two_sum.js"
[class]{}-[func]{twoSumHashTable}
```
=== "TS"
```typescript title="two_sum.ts"
[class]{}-[func]{twoSumHashTable}
```
=== "Dart"
```dart title="two_sum.dart"
[class]{}-[func]{twoSumHashTable}
```
=== "Rust"
```rust title="two_sum.rs"
[class]{}-[func]{two_sum_hash_table}
```
=== "C"
```c title="two_sum.c"
[class]{HashTable}-[func]{}
[class]{}-[func]{twoSumHashTable}
```
=== "Kotlin"
```kotlin title="two_sum.kt"
[class]{}-[func]{twoSumHashTable}
```
=== "Ruby"
```ruby title="two_sum.rb"
[class]{}-[func]{two_sum_hash_table}
```
=== "Zig"
```zig title="two_sum.zig"
[class]{}-[func]{twoSumHashTable}
```
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**.