### **✅ Here is the solution with comments that beats 100%.**
## Approach
Brute Force using HashMap
## Complexity
- Time complexity: O(n)
- Space complexity: O(n)
## Code
```python
def twoSum(self, nums: List[int], target: int) -> List[int]:
# A hash map to store the numbers we've seen and their indices.
# This allows for O(1) average time complexity for lookups.
num_map = {}
# We iterate through the list only once.
# 'enumerate' provides both the index 'i' and the value 'num'.
for i, num in enumerate(nums):
# Calculate the 'complement' needed to reach the target.
complement = target - num
# Check if the complement already exists in our hash map.
# If it does, we've found the two numbers that sum up to the target.
if complement in num_map:
# The problem guarantees a unique solution, so we can return immediately.
# 'num_map[complement]' gives the index of the complement.
return [num_map[complement], i]
# If the complement is not in the map, add the current number and its index.
# This prepares the map for future lookups.
num_map[num] = i
# This part of the code is technically unreachable as the problem statement
# guarantees that a solution always exists.
return []
```
### **✅ Here is the solution with comments that beats 100%.**
## Approach
Brute Force using HashMap
## Complexity
- Time complexity: O(n)
- Space complexity: O(n)
## Code
```python
def twoSum(self, nums: List[int], target: int) -> List[int]:
# A hash map to store the numbers we've seen and their indices.
# This allows for O(1) average time complexity for lookups.
num_map = {}
# We iterate through the list only once.
# 'enumerate' provides both the index 'i' and the value 'num'.
for i, num in enumerate(nums):
# Calculate the 'complement' needed to reach the target.
complement = target - num
# Check if the complement already exists in our hash map.
# If it does, we've found the two numbers that sum up to the target.
if complement in num_map:
# The problem guarantees a unique solution, so we can return immediately.
# 'num_map[complement]' gives the index of the complement.
return [num_map[complement], i]
# If the complement is not in the map, add the current number and its index.
# This prepares the map for future lookups.
num_map[num] = i
# This part of the code is technically unreachable as the problem statement
# guarantees that a solution always exists.
return []
```
