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Two Sum
https://leetcode.com/problems/two-sum
Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target. You may assume that each input would have exactly one solution, and you may not use the same element twice. You can return the answer in any order.
Example 1:
Input: nums = [2,7,11,15],
target = 9
Output: [0,1]
Explanation: Because nums[0] + nums[1] == 9, we return [0, 1].
Problem Overview and Solution
The problem is to identify the indices of two numbers in an array that add up to a given target value. The input to the problem includes an array of integers, referred to as nums, and a single integer called the target. The output should be a pair of indices that point to the two numbers in the array whose sum equals the target value.
To solve this problem efficiently, a hash map is used to store numbers and their indices as the array is traversed. This approach allows for quick lookups to check if the complement of a number (i.e., the value needed to reach the target) has already been seen. By iterating through the array in a single pass, the solution achieves optimal performance while maintaining simplicity.
Algorithm and Complexity
The algorithm begins by initializing a hash map, which is used to store numbers from the array along with their indices. As the algorithm iterates through each number in the array, it calculates the complement of the current number, which is the difference between the target and the number. This complement represents the value needed to reach the target sum.
If the complement is found in the hash map, the algorithm immediately returns the indices of the complement and the current number. If the complement is not found, the algorithm adds the current number and its index to the hash map to be checked against future numbers. This process ensures a solution is found in a single pass through the array.
The time complexity of the algorithm is O(n) , as each element in the array is processed only once. The space complexity is also O(n), because the hash map may store up to N entries, where N is the size of the input array.
Solution
def two_sum(nums, target):
num_to_index = {}
for index, num in enumerate(nums):
complement = target - num
if complement in num_to_index:
return [num_to_index[complement], index]
num_to_index[num] = index
# Example usage
nums = [2, 7, 11, 15]
target = 9
print(two_sum(nums, target)) # Output: [0, 1]