Introduction
Welcome to another post at Art of Coding! Today, we’re delving into the challenge: 2824. Count Pairs Whose Sum is Less than Target.
If you’ve been following our blog, you’re likely familiar with the two-pointer technique from our previous posts. It’s a go-to strategy for solving array and string problems with elegance and efficiency. In this article, we’ll not only tackle the problem “2824. Count Pairs Whose Sum is Less than Target” using this technique but also deepen our understanding of its mechanics and applications.
We’ll break down the problem, walk through solutions in both Java and Python, and analyze the complexities involved.
Let’s get started!
Understanding Problem 2824. Count Pairs Whose Sum is Less than Target
Optimizing Pair Sum Counts: Java & Python Solution
In this section, we dive into the Problem “2824 Count Pairs Whose Sum is Less than Target”. This problem is a common challenge in coding interviews and it goes like this: Given an array of integers and a target sum, count the number of pairs in the array whose sum is less than the target value.
At first glance, this might seem straightforward – just check all possible pairs and count the ones that fit the criteria, right? However, this brute-force approach can be inefficient, especially with larger arrays. This is where the two-pointer technique comes into play, offering a more optimized solution.
The two-pointer technique is an elegant method that utilizes two pointers to traverse the array. In the case, we use this technique to efficiently find and count pairs. The beauty of this method lies in its simplicity and efficiency, significantly reducing the time complexity compared to the brute-force approach.
So, how does it work exactly for this problem?
We will explore this in detail with our Java and Python solutions in the following sections. But in essence, by sorting the array and moving two pointers from opposite ends, we can quickly identify and count the pairs that meet our condition – their sum being less than the target.
Exploring the Algorithm: The Two-Pointer Technique in Action
After delving into the intricacies of Problem 2824. Count Pairs Whose Sum is Less than Target and the two-pointer technique, we’re now ready to translate our understanding into code.
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
public class CountPairs {
public static int countPairs(List<Integer> nums, int target) {
Collections.sort(nums); // Sorting the array
int result = 0;
int left = 0;
int right = nums.size() - 1;
while (left < right) {
if (nums.get(left) + nums.get(right) < target) {
result += right - left; // Adding count of valid pairs
left++; // Moving the left pointer forward
} else {
right--; // Moving the right pointer backward
}
}
return result;
}
public static void main(String[] args) {
List<Integer> nums = Arrays.asList(-6, 2, 5, -2, -7, -1, 3);
System.out.println(countPairs(nums, -2));
}
}
def countPairs(nums, target):
nums.sort() # Sorting the array
left, right = 0, len(nums) - 1
count = 0
while left < right:
if nums[left] + nums[right] < target:
count += right - left
left += 1
else:
right -= 1
return count
The two-pointer technique is a powerful method in algorithm design, particularly useful for solving problems involving arrays or sequences. In the context of Problem “2824. Count Pairs Whose Sum is Less than Target”, this technique allows us to efficiently count the number of pairs whose sum is less than the target value.
Here’s how the algorithm works:
Sort the Array: We start by sorting the array in ascending order. Sorting is crucial because it arranges the elements such that we can exploit their order to optimize our pair counting.
Initialize Pointers: We use two pointers,
leftandright, which initially point to the first (index 0) and last element (indexsize - 1) of the array, respectively.Iterate and Evaluate: We iterate through the array using these pointers with the following logic:
- If the sum of elements at
leftandrightpointers is less than the target, we move to the next step of counting valid pairs (explained further below). - If the sum is equal to or greater than the target, we move the
rightpointer one step to the left (decrementingright), as we need a smaller sum.
- If the sum of elements at
Counting the Pairs: When we find that the sum of
nums[left] + nums[right]is less than the target, it means that not only does this particular pair meet the condition, but also all pairs formed between theleftelement and each element fromlefttoright - 1(inclusive) will satisfy the condition. This is due to the sorted nature of the array.
Key Line Explained:
The line result += right - left in our algorithm is critical. It efficiently calculates the number of valid pairs that can be formed with the left element and all elements up to the right element, excluding the right element itself. This is based on the reasoning that all these elements, when added between right and left, will produce sums less than the target. By using this line, we add the count of all these valid pairs to our result in a single step, showcasing the efficiency of the two-pointer technique.
Complexity Analysis - Evaluating Efficiency: Time and Space Considerations
In our solutions to Problem 2824. Count Pairs Whose Sum is Less than Target using the two-pointer technique, both time and space complexities play significant roles.
Time Complexity:
- Sorting the Array: The initial step in our algorithm involves sorting the array. In both Java and Python, the sort operation typically has a time complexity of O(n log n), where n is the number of elements in the array.
- Iterating with Two Pointers: Once the array is sorted, we use a while loop to iterate through the array with our two pointers. This iteration, in the worst case, goes through the array once, leading to a time complexity of O(n).
- Overall Time Complexity: When combining these steps, the dominating factor is the sorting step, making the overall time complexity of our solution O(n log n).
Space Complexity:
- In both the Java and Python implementations, we only use a fixed number of variables (pointers and counters). There’s no additional space that grows with the size of the input array.
- Therefore, the space complexity of our solution is O(1), also known as constant space complexity. This means the space required by our algorithm does not increase with the size of the input data.
Why This Matters:
This analysis shows that our solution is efficient, especially for large datasets. The O(n log n) time complexity is much more performant than a brute-force approach, which would typically have a time complexity of O(n^2). Additionally, the constant space complexity implies that our solution is memory-efficient, an important consideration in real-world applications.
Conclusion
Using the Two-Pointer Technique not only simplified the problem but also provided an excellent demonstration of algorithmic efficiency in both Java and Python.
Key Takeaways:
- Two-Pointer Technique: An elegant solution for problems involving sequences or arrays, particularly when paired with sorting.
- Efficiency: Our solutions in Java and Python highlight the importance of considering time and space complexities in coding solutions. The two-pointer technique, with its O(n log n) time complexity and O(1) space complexity, proves to be a powerful tool in algorithm design.
- Applicability: This technique is versatile and can be applied to a variety of problems, especially those involving sorted arrays or pairs.
Your Next Steps:
As you continue your journey in coding and algorithm design, we encourage you to:
- Experiment: Try applying the two-pointer technique to other problems.
- Analyze: Always consider the time and space complexity of your solutions.
- Engage: Share your experiences and solutions with the community. If you’ve tried solving Problem 2824 or a similar problem using a different approach, we’d love to hear about it in the comments!
Remember, each problem solved is a step forward in your coding journey. Keep exploring, keep coding, and keep sharing!