Given an array of integers, count how many distinct values exist in the array.
Example:
Input: [1, 5, -3, 1, -4, 2, -4, 7, 7] Output: 6 Explanation: the distinct values in the array are [1, 5, -3, -4, 2, 7]
Your algorithm should run in O(n) time and use O(n) extra space.
The core challenge of this problem is to efficiently count the number of distinct values in an array. This is a common problem in data processing and analysis, where you need to identify unique elements from a dataset.
Potential pitfalls include not accounting for negative numbers or duplicates properly. Misconceptions might arise from misunderstanding the requirement for distinct values.
To solve this problem, we need to think about how to efficiently track unique values. A naive solution might involve nested loops to compare each element with every other element, but this would result in an O(n^2) time complexity, which is not optimal.
Instead, we can use a JavaScript Set
to store unique values. The Set
data structure automatically handles duplicates and allows for O(1) average time complexity for insertions and lookups.
Using a Set
is optimal because it ensures that each element is processed only once, resulting in O(n) time complexity. The space complexity is also O(n) because, in the worst case, all elements are unique and need to be stored in the Set
.
1. Initialize an empty Set
.
2. Iterate through each element in the array.
3. Add each element to the Set
.
4. The size of the Set
at the end of the iteration is the number of distinct values.
// Function to count distinct values in an array
function countDistinctValues(arr) {
// Initialize a new Set to store unique values
const uniqueValues = new Set();
// Iterate through each element in the array
for (let value of arr) {
// Add the value to the Set
uniqueValues.add(value);
}
// The size of the Set is the number of distinct values
return uniqueValues.size;
}
// Example usage
const inputArray = [1, 5, -3, 1, -4, 2, -4, 7, 7];
console.log(countDistinctValues(inputArray)); // Output: 6
The time complexity of this approach is O(n) because we iterate through the array once. The space complexity is O(n) because, in the worst case, all elements are unique and stored in the Set
.
Consider the following edge cases:
Examples:
// Edge case: empty array
console.log(countDistinctValues([])); // Output: 0
// Edge case: all identical elements
console.log(countDistinctValues([1, 1, 1, 1])); // Output: 1
// Edge case: negative and positive numbers
console.log(countDistinctValues([-1, -1, 2, 2, 3, -3])); // Output: 4
To test the solution comprehensively, include a variety of test cases:
Using a testing framework like Jest can help automate and manage these tests.
When approaching such problems, consider the following tips:
In this blog post, we discussed how to count the number of distinct values in an array using an efficient O(n) time complexity algorithm with JavaScript. We explored the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing strategies. Understanding and solving such problems is crucial for improving your coding and problem-solving skills.