702810. Find pairs with given relation
Here is the detailed Markdown blog post for the GeeksforGeeks problem:
Find Pairs with Given Relation
Summary
Find pairs with given relation is an array-based algorithmic problem that involves finding all pairs in an array that satisfy a specific condition. This problem requires a basic understanding of arrays and hash data structures.
Detailed Explanation
The problem statement is not available, so we'll assume it's a general algorithmic problem based on the title and tags. The goal is to find all pairs in an array where the difference between the pair elements satisfies a certain relation (e.g., less than or equal to a given value).
Here's a step-by-step breakdown of the solution:
- First, sort the array in ascending order.
- Initialize an empty hash data structure (e.g., a Java
HashMapor Pythondict) to store the elements and their counts. - Iterate through the sorted array, and for each element:
- Check if the element is already present in the hash with a count greater than 0.
- If yes, calculate the difference between the current element and its corresponding value in the hash (if it exists).
- If the difference satisfies the given relation, add this pair to the result set.
- Continue iterating through the array until all elements have been processed.
The time complexity of this algorithm is O(n log n) due to the sorting step, where n is the size of the input array. The space complexity is O(n) for storing the hash data structure and result set.
Here's an ASCII art diagram illustrating the problem:
+---------------+
| Array: [3, 4, 5, 7, 8] |
+---------------+
(Sorted array)
+---------------+
| Array: [3, 4, 4, 5, 5, 5, 7, 8] |
+---------------+
(Hash data structure)
+---------------+
| Hash: { 3: 1, 4: 2, 5: 3, 7: 1, 8: 1 } |
+---------------+
(Pairs with given relation)
+---------------+
| Result set: [(4, 3), (5, 0)] |
+---------------+
Optimized Solutions
Java
import java.util.*;
public class FindPairs {
public static List<List<Integer>> findPairs(int[] arr, int k) {
List<List<Integer>> result = new ArrayList<>();
Map<Integer, Integer> countMap = new HashMap<>();
Arrays.sort(arr);
for (int i = 0; i < arr.length; i++) {
if (countMap.containsKey(arr[i] - k)) {
result.add(Arrays.asList(arr[i], arr[i] - k));
}
countMap.put(arr[i], countMap.getOrDefault(arr[i], 0) + 1);
}
return result;
}
}