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This approach leverages a hash map (or dictionary) to solve the problem efficiently. By taking advantage of the average O(1) time complexity for insert and lookup operations in hash maps, we can create a mapping between elements and their indices (or frequencies, depending on the problem requirements). This method not only offers efficient retrieval but also makes it easier to track elements as we iterate through the data structure.
Time Complexity: O(n), where n is the number of elements in the input list.
Space Complexity: O(n), due to the storage of numbers in the hash map.
This C solution uses a simple array-based hash table to achieve a similar effect. It iterates over the numbers, checks for the complement using linear search within the hash table and returns the indices if found.
This approach utilizes a two-pointer technique which is particularly effective when the input is sorted (or can be sorted) without significantly impacting performance. By using two pointers to traverse the array from both ends, we can efficiently find the pair of elements that sum to the target. Note that this approach is based on the assumption that sorting the input is feasible and will not exceed time limits.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(n) because we store tuples of indices and numbers.
1def two_sum(nums, target):
2 nums_sorted = sorted((num, i) for i, num in enumerate(nums))
3 left, right = 0, len(nums) - 1
4 while left < right:
5 current_sum = nums_sorted[left][0] + nums_sorted[right][0]
6 if current_sum == target:
7 return [nums_sorted[left][1], nums_sorted[right][1]]
8 elif current_sum < target:
9 left += 1
10 else:
11 right -= 1
12 return []This Python implementation first sorts the list while retaining the original indices. Two pointers are then used to check for the target sum from the front and back of the sorted list. If the sum is less than the target, the left pointer is moved forward; otherwise, the right pointer is moved backward.