Question

Given a list of \(n\) integers and a value \(v\), describe an \(O(n \log (n))\) algorithm to find whether there are two values \(x\) and \(y\) in the list with sum \(v\).

Short answer

Sort the list and use two pointers to check for the sum in \(O(n \log (n))\) time.

Step-by-step solution

Sort the List

First, sort the given list of integers in ascending order. Sorting can be done using an efficient algorithm like quicksort or mergesort, which takes \(O(n \log (n))\) time. This sorted list will allow us to efficiently identify pairs of integers that sum to \(v\).

Initialize Two Pointers

After sorting, set two pointers: the first pointer, \(i\), at the beginning of the list (i.e., the smallest element), and the second pointer, \(j\), at the end of the list (i.e., the largest element).

Iterate with Two Pointers

Check if the sum of the elements at pointers \(i\) and \(j\) equals \(v\). - If so, you've found the pair \((x, y)\) such that \(x + y = v\), and the algorithm returns true.- If the sum is less than \(v\), increment the \(i\) pointer to increase the sum.- If the sum is greater than \(v\), decrement the \(j\) pointer to decrease the sum.Repeat this process until the pointers cross each other.

Conclusion

If the pointers cross without finding a pair that sums to \(v\), then no such pair exists in the list. The algorithm should then return false.

Key concepts

Sorting Algorithms

Sorting algorithms are essential in organizing data, especially when solving problems efficiently. Many everyday computer tasks involve sorted data, like searching or comparing items.
One powerful sorting algorithm is quicksort, which uses the divide-and-conquer technique to sort data. Quicksort has an average complexity of \(O(n \log(n))\), making it ideal for sorting large lists quickly.

• Quicksort picks a 'pivot' and partitions the array into smaller and larger elements than the pivot.
• It then recursively sorts the sub-arrays.

In some cases, mergesort might be used. Mergesort divides the list into smaller lists, sorts them, and then merges them back together. It also runs in \(O(n \log(n))\) time and is favored for its stability in sorting.
Sorting is crucial as it prepares the data, allowing other techniques, like the two-pointer method, to work effectively.

Two-Pointer Technique

The two-pointer technique is a handy tool in problem solving. It involves setting two pointers at different locations in data structures, like arrays, and then moving them under certain conditions to achieve a goal.
This method is particularly effective in sorted lists, allowing for efficient exploration of potential solutions.

• One pointer is placed at the start of the list, and the other at the end.
• The pointers move towards each other until they meet, inspecting or modifying the elements they point to.

This technique helps solve problems like finding pairs of numbers in a list that sum to a specific value. By leveraging sorted data and the two-pointer method, one can check sums efficiently, which is computationally more palatable than comparing all possible pairs.

Problem Solving Approach

A structured problem-solving approach improves the clarity and efficiency of solutions. Initial solution steps often involve understanding the problem requirements and constraints.
With this algorithm, the need to determine two integers' sum matches a given value, benefits from clear steps:

• First, sort the list of integers to arrange them in order.
• Next, set up necessary tools, like two pointers, to explore possible solutions.

Iterate through the list with a defined strategy for pointer movement. For instance, if the current sum is too low, adjust the pointers to explore higher values instead of aimlessly searching.
Finally, have a condition to conclude when no valid pair exists, ensuring the process stops when exhaustive search confirms it's necessary.

Complexity Analysis

Understanding complexity analysis is crucial for evaluating algorithm performance. It determines how the algorithm's resource demands scale with input size.
The algorithm in this example leverages the efficiency of sorting, with an essential merging step to achieve a desired complexity of \(O(n \log(n))\).
The steps include:

• Sorting the list, which forms the bulk of the \(O(n \log(n))\) complexity.
• Utilizing two pointers to check pairs along the list, requiring at most \(n\) comparisons.

This linear check through the list is \(O(n)\), meaning it doesn't change the overall complexity. Thus, by understanding each part, we ensure the solution is both optimal and scalable for larger inputs, which is essential in real-world applications.