Min Subsets with Consecutive Numbers

Summary

Given a list of integers, find all subsets that contain consecutive numbers. The problem is to identify these subsets and return them as output.

Detailed Explanation

The problem can be solved using a straightforward approach. First, sort the input array in ascending order. Then, iterate through the sorted array and maintain a set of consecutive numbers found so far. If the current number is one more than the last number in the set, add it to the set. Otherwise, start a new subset.

Here's a step-by-step breakdown:

Sort the input array nums in ascending order.
Initialize an empty list result to store the subsets of consecutive numbers.
Iterate through the sorted array:
- If the current number is one more than the last number in the set, add it to the set.
- Otherwise, start a new subset and add the current number to it.
Add each subset to the result list.
Return the result list.

Time complexity: O(n log n) due to sorting Space complexity: O(1) as we only use a constant amount of space

Here's an ASCII art diagram representing the process:

  +---------------+
  |        nums     |
  |  [1, 2, 3, 5, 6]  |
  +---------------+
           |
           |
           v
  +---------------+
  |    sorted nums   |
  |  [1, 2, 3, 5, 6]  |
  +---------------+
           |
           |
           v
  +---------------+
  |     result      |
  |  [[1, 2], [2, 3], [3, 5], [5, 6]]  |
  +---------------+

Optimized Solutions

Java

import java.util.ArrayList; import java.util.List; public class Main { public static List<List<Integer>> minSubsets(int[] nums) { List<List<Integer>> result = new ArrayList<>(); Arrays.sort(nums); int start = nums[0]; List<Integer> subset = new ArrayList<>(); for (int num : nums) { if (num == start + 1) { subset.add(num); start++; } else { result.add(subset); subset = new ArrayList<>(); subset.add(num); start = num; } } result.add(subset); return result; } }