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This approach involves sorting the data first and then using a two-pointer technique. By sorting, we can simplify the problem as the elements will be in order. The two-pointer method then efficiently checks possible solutions by limiting the number of checks needed.
Time complexity is O(n log n) due to sorting, and space complexity is O(1).
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This JavaScript code performs sorting utilizing Array.sort(). By maneuvering two pointers from respective ends towards the center, it quickly isolates potential pairs whose sum matches the target.
This approach uses a hash table (or dictionary) to keep track of the elements and their complements needed to reach the target sum. By utilizing this lookup, we can reduce the problem to a linear time complexity.
Time complexity is O(n) assuming uniform distribution of hash function (no collisions), and space complexity is O(n).
1import java.util.HashSet;
2
3public class PairFinder {
4 public static void findPair(int[] arr, int sum) {
5 HashSet<Integer> seen = new HashSet<>();
6
7 for (int number : arr) {
8 int complement = sum - number;
9 if (seen.contains(complement)) {
10 System.out.println("Pair found: (" + number + ", " + complement + ")");
11 return;
12 }
13 seen.add(number);
14 }
15 System.out.println("No pair found");
16 }
17}In this Java implementation, a HashSet stores elements while checking complements efficiently. Each iteration checks if the complement has already been recorded, leading to quick pair identification.