45 lines
1.5 KiB
Python
45 lines
1.5 KiB
Python
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"""
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Problem:
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Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to
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target. You can not use the same element twice.
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You can return any one of the answers in any order, if not found return [-1, -1].
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Solution:
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In this solution we check if each element's complement i.e. difference between the target and the element
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(target−nums[i]) exists in the dictionary. If it does exist, we return current element's index and its complement's
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index else we store the complement in the dictionary.
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Beware that the complement must not be nums[i] itself, because we can not use the same element twice.
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Complexity Analysis:
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Time complexity: O(n). We traverse the list containing nn elements only once. Each lookup in the dictionary costs only
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O(1) time.
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Space complexity: O(n). The extra space required depends on the number of items stored in the dictionary,
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which stores at most nn elements.
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"""
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def twoSum(nums, target):
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dictionary = {}
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# using enumerate to iterate through the list keeping track of both index and element
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for i, val in enumerate(nums):
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diff = target - val
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if diff in dictionary:
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return [dictionary[diff], i]
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dictionary[val] = i
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return [-1, -1]
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def main():
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try:
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arr = list(map(int, input("Enter space-seperated integers: ").split()))
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tar = int(input("Enter target: "))
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except ValueError:
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arr = [2, 7, 11, 15]
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tar = 9
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print(twoSum(arr, tar))
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if __name__ == "__main__":
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main()
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