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.
In this problem, you are given an array of integers and you need to return a new array that contains only the unique values from the original array. The order of the elements in the resulting array does not matter.
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 linear time complexity. This is significant in scenarios where the array is large, and performance is critical.
Common applications include data cleaning, where duplicate entries need to be removed, and in algorithms where unique elements are required for further processing.
Potential pitfalls include using algorithms that do not meet the O(n) time complexity requirement, such as nested loops, which would result in O(n^2) complexity.
To solve this problem, we can use a Set data structure, which inherently does not allow duplicate values. This will help us achieve the desired time complexity of O(n).
A naive solution would involve using nested loops to check for duplicates, but this would result in O(n^2) time complexity, which is not optimal.
An optimized solution involves using a Set to track unique values as we iterate through the array. This approach ensures that each element is processed only once, resulting in O(n) time complexity.
Here is a step-by-step breakdown of the algorithm:
uniqueSet
.uniqueSet
.uniqueSet
, add it to uniqueSet
.uniqueSet
to an array and return it.// Function to remove duplicates from an array
function removeDuplicates(arr) {
// Initialize a Set to store unique values
const uniqueSet = new Set();
// Iterate through the array
for (let num of arr) {
// Add each element to the Set (duplicates will be ignored)
uniqueSet.add(num);
}
// Convert the Set back to an array and return it
return Array.from(uniqueSet);
}
// Example usage
const inputArray = [2, 3, 1, 1, 4, 3, -2, 1];
const uniqueArray = removeDuplicates(inputArray);
console.log(uniqueArray); // Output: [2, 3, 1, 4, -2]
The time complexity of this approach is O(n) because we iterate through the array once. The space complexity is also O(n) because, in the worst case, we store all elements in the Set.
Consider the following edge cases:
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:
Example test cases:
console.log(removeDuplicates([2, 3, 1, 1, 4, 3, -2, 1])); // [2, 3, 1, 4, -2]
console.log(removeDuplicates([])); // []
console.log(removeDuplicates([1, 1, 1, 1])); // [1]
console.log(removeDuplicates([1, 2, 3, 4])); // [1, 2, 3, 4]
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 complexity using JavaScript. We explored the problem definition, understood the core challenges, and walked through an optimized solution using a Set. We also analyzed the complexity, considered edge cases, and provided testing strategies.
Understanding and solving such problems is crucial for improving algorithmic thinking and coding skills. Keep practicing and exploring further to enhance your problem-solving abilities.
For further reading and practice, consider the following resources: