Given an array of integers, return a new array containing only the unique values.
The resulting array can be in any order.
Example:
Input: [2, 3, 1, 1, 4, 3, -2, 1] Output: [2, 3, 1, 4, -2]
Your algorithm should run in O(n) time and use O(n) extra space.
Given an array of integers, the task is to return a new array containing only the unique values. The resulting array can be in any order.
An array of integers.
A new array containing only the unique values from the input array.
Input: [2, 3, 1, 1, 4, 3, -2, 1] Output: [2, 3, 1, 4, -2]
The core challenge of this problem is to efficiently remove duplicates from the array while maintaining a time complexity of O(n) and space complexity of O(n). This problem is significant in scenarios where data needs to be cleaned or filtered to remove redundant information.
Common applications include data preprocessing in machine learning, database management, and any system where unique entries are required.
Potential pitfalls include misunderstanding the requirement for O(n) time complexity and using inefficient methods that do not meet the constraints.
To solve this problem, we can use a hash set to keep track of the unique elements we have encountered so far. This approach ensures that we only add unique elements to the result array, and it operates in O(n) time complexity.
A naive solution would involve nested loops to check for duplicates, resulting in O(n^2) time complexity. This is not optimal and does not meet the problem constraints.
We can use a hash set to store unique elements as we iterate through the array. This ensures that each element is processed only once, achieving O(n) time complexity. The hash set provides O(1) average time complexity for insertions and lookups.
1. Initialize an empty hash set to store unique elements.
2. Initialize an empty vector to store the result.
3. Iterate through the input array.
4. For each element, check if it is already in the hash set.
5. If it is not in the hash set, add it to both the hash set and the result vector.
6. Return the result vector.
Here is a step-by-step breakdown of the algorithm:
#include <iostream>
#include <vector>
#include <unordered_set>
std::vector<int> removeDuplicates(const std::vector<int>& nums) {
std::unordered_set<int> seen; // Hash set to store unique elements
std::vector<int> result; // Vector to store the result
for (int num : nums) {
// If the number is not in the hash set, add it to the result
if (seen.find(num) == seen.end()) {
seen.insert(num);
result.push_back(num);
}
}
return result;
}
int main() {
std::vector<int> nums = {2, 3, 1, 1, 4, 3, -2, 1};
std::vector<int> uniqueNums = removeDuplicates(nums);
std::cout << "Unique elements: ";
for (int num : uniqueNums) {
std::cout << num << " ";
}
std::cout << std::endl;
return 0;
}
The time complexity of this approach is O(n) because we iterate through the array once, and each insertion and lookup in the hash set is O(1) on average.
The space complexity is O(n) because we use a hash set to store unique elements, which in the worst case can be the same size as the input array.
Potential edge cases include:
Examples:
Input: [] Output: [] Input: [1, 1, 1, 1] Output: [1] Input: [1, 2, 3, 4] Output: [1, 2, 3, 4]
To test the solution comprehensively, consider the following test cases:
Testing frameworks such as Google Test can be used to automate and validate these test cases.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to remove duplicates from an array in O(n) time and O(n) space using C++. We explored the problem definition, understood the core challenges, and developed an optimized solution using a hash set. We also analyzed the complexity, considered edge cases, and provided tips for effective problem-solving.
Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills. Practice regularly and explore further to enhance your proficiency.
For further reading and practice, consider the following resources: