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This approach utilizes a HashSet to keep track of the numbers we have encountered. As we iterate through the list, we check both the presence of a number and its negative in the HashSet. We update the largest number that satisfies the condition.
Time Complexity: O(n), since we iterate through the array once.
Space Complexity: O(1), since the hash set array size is always constant and independent of input size.
1def largest_equal_positive_negative(nums):
2 num_set = set()
3 largest = -1
4 for num in nums:
5 if -num in num_set:
6 largest = max(largest, abs(num))
7 num_set.add(num)
8 return largest
9
10nums = [-1, 2, -3, 3]
11print(largest_equal_positive_negative(nums)) # Output: 3Uses a set to track numbers. For each number, checks if its negative is in the set, updating the largest absolute value if a pair is found.
In this approach, we first sort the array so that we can efficiently find pairs of positive and negative numbers. Once sorted, we use two pointers: one starting from the beginning (for negative numbers) and one from the end (for positive numbers) to find the largest integer pair where both a positive and its negative exist.
Time Complexity: O(n log n) for the sorting operation.
Space Complexity: O(1) additional space beyond input storing.
Implementing a sorting step allows us to use a two-pointer strategy for rapid scanning for valid number pairs, extracting their largest positive k value when found.