Given an array of integers, return indices of the two numbers such that they add up to a specific target.
You may assume that each input will have at most one solution, and you may not use the same index twice.
In case no solution exists, return [-1, -1]
Example:
Input: nums = [2, 7, 11, 15]
, target = 9
Output: [0, 1]
Explanation: nums[0] + nums[1] = 2 + 7 = 9
Your algorithm should run in O(n log n) time and use O(n) extra space.
The core challenge of this problem is to find two distinct indices in the array such that the sum of the elements at these indices equals the target value. This problem is significant in various applications, such as financial transactions, where you need to find pairs of transactions that sum up to a specific amount.
Potential pitfalls include assuming that there are multiple solutions or using the same index twice, which the problem explicitly forbids.
To solve this problem, we can use a hash map (dictionary in Python) to store the indices of the elements as we iterate through the array. This allows us to check in constant time whether the complement of the current element (i.e., target - current element) has already been seen.
Here is a step-by-step breakdown of the approach:
Let's break down the algorithm step-by-step:
num_to_index
.nums
using a for loop with index i
and value num
.target - num
.num_to_index
.[num_to_index[complement], i]
.num
and its index i
to num_to_index
.[-1, -1]
.def two_sum(nums, target):
# Dictionary to store the index of each number
num_to_index = {}
# Iterate through the list of numbers
for i, num in enumerate(nums):
# Calculate the complement
complement = target - num
# Check if the complement is already in the 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]
# Otherwise, add the current number and its index to the dictionary
num_to_index[num] = i
# If no solution is found, return [-1, -1]
return [-1, -1]
# Example usage
nums = [2, 7, 11, 15]
target = 9
print(two_sum(nums, target)) # Output: [0, 1]
The time complexity of this approach is O(n) because we are iterating through the array once. The space complexity is also O(n) because we are storing each element in the dictionary.
Some potential edge cases include:
[-1, -1]
.[-1, -1]
.[-1, -1]
.Examples:
print(two_sum([], 9)) # Output: [-1, -1]
print(two_sum([1], 9)) # Output: [-1, -1]
print(two_sum([1, 2, 3], 7)) # Output: [-1, -1]
To test the solution comprehensively, we should include a variety of test cases:
Example test cases:
print(two_sum([2, 7, 11, 15], 9)) # Output: [0, 1]
print(two_sum([3, 2, 4], 6)) # Output: [1, 2]
print(two_sum([3, 3], 6)) # Output: [0, 1]
print(two_sum([1, 2, 3], 7)) # Output: [-1, -1]
print(two_sum([], 9)) # Output: [-1, -1]
print(two_sum([1], 9)) # Output: [-1, -1]
When approaching such problems, consider the following tips:
In this blog post, we discussed the Two Sum III problem, explored different approaches to solve it, and provided a detailed explanation of the optimized solution using a hash map. We also covered edge cases, testing, and tips for problem-solving. Understanding and solving such problems is crucial for developing strong algorithmic thinking and coding skills.
For further reading and practice, consider the following resources:
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