Pairs with difference k1713

Summary

Given an array nums and an integer k, find all pairs (x, y) in the array where y - x = k. The problem is to write a function that returns these pairs. This problem requires understanding of arrays and data structures.

Detailed Explanation

To solve this problem, we can iterate over the array and check for each pair if their difference is equal to k. We'll store the result in a list and return it at the end.

Here's the step-by-step breakdown:

Initialize an empty list result to store the pairs.
Iterate over the array using two nested loops, the outer loop starting from the first element and the inner loop starting from the second element.
For each pair (x, y), calculate their difference d = y - x.
If d == k, add the pair to the result list.
Return the result list.

The time complexity of this solution is O(n^2), where n is the size of the array, as we're using two nested loops. The space complexity is O(n) for storing the result list.

Here's a diagram representing the pairs with difference k:

  nums = [1, 3, 4, 5, 6]
  k = 2

  (1, 3)
  (3, 5)

  y - x = 2

Optimized Solutions

Java

import java.util.ArrayList; import java.util.List; public List<List<Integer>> pairsWithDifferenceK(int[] nums, int k) { List<List<Integer>> result = new ArrayList<>(); for (int i = 0; i < nums.length; i++) { for (int j = i + 1; j < nums.length; j++) { if (nums[j] - nums[i] == k) { result.add(Arrays.asList(nums[i], nums[j])); } } } return result; }