class Solution:
def twoSum(self, nums: List[int], target: int) -> List[int]:
"""
Find two indices in 'nums' such that the sum of their values equals 'target'.
Parameters:
nums (List[int]): A list of integers.
target (int): The target sum for which two numbers in 'nums' should add up.
Returns:
List[int]: A list containing the two indices whose values add up to 'target'.
"""
# Dictionary to hold the numbers and their indices for quick lookup.
num_to_index = {}
# Iterate over the list with index and value.
for i, num in enumerate(nums):
# Calculate the complement needed to reach the target sum.
complement = target - num
# Check if the complement exists in our dictionary.
if complement in num_to_index:
# If found, return the indices of the complement and the current number.
return [num_to_index[complement], i]
# Store the current number and its index in the dictionary.
num_to_index[num] = i
# Since the problem guarantees exactly one solution, no return is needed here.
Summary of Techniques and Approaches:
Hash Map Utilization: The solution leverages a hash map (dictionary) for constant-time look-ups to quickly find if the complement of the current number exists. This is useful when trying to achieve an efficient solution in linear time.
Single-Pass Algorithm: By iterating through the list only once, the approach achieves O(n) time complexity. This method is broadly applicable for scenarios where you need to search for pairs or complements within a collection.
Space-Time Tradeoff: Using extra space (dictionary) is an effective tradeoff to reduce time complexity. This technique can be beneficial when the problem constraints allow additional space usage to optimize speed.
General Applicability: This approach can be applied to other problems involving pair-finding (or k-sum variations) and similar scenarios where identifying complementary elements is key. Always consider leveraging hash-based structures for rapid membership checks to optimize performance.