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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0401_0500.s0491_increasing_subsequences;

// #Medium #Array #Hash_Table #Bit_Manipulation #Backtracking
// #2022_07_21_Time_24_ms_(38.30%)_Space_68.7_MB_(63.52%)

import java.util.ArrayList;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

/**
 * 491 - Increasing Subsequences\.
 *
 * Medium
 *
 * Given an integer array `nums`, return all the different possible increasing subsequences of the given array with **at least two elements**. You may return the answer in **any order**.
 *
 * The given array may contain duplicates, and two equal integers should also be considered a special case of increasing sequence.
 *
 * **Example 1:**
 *
 * **Input:** nums = [4,6,7,7]
 *
 * **Output:** [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
 *
 * **Example 2:**
 *
 * **Input:** nums = [4,4,3,2,1]
 *
 * **Output:** [[4,4]]
 *
 * **Constraints:**
 *
 * *   `1 <= nums.length <= 15`
 * *   `-100 <= nums[i] <= 100`
**/
@SuppressWarnings("java:S5413")
public class Solution {
    public List> findSubsequences(int[] nums) {
        if (nums == null || nums.length == 1) {
            return new ArrayList<>();
        }
        Set> answer = new HashSet<>();
        List list = new ArrayList<>();
        return new ArrayList<>(backtracking(nums, 0, list, answer));
    }

    private Set> backtracking(
            int[] nums, int start, List currList, Set> answer) {
        if (currList.size() >= 2) {
            answer.add(new ArrayList<>(currList));
        }
        for (int i = start; i < nums.length; i++) {
            if (currList.isEmpty() || currList.get(currList.size() - 1) <= nums[i]) {
                currList.add(nums[i]);
                backtracking(nums, i + 1, currList, answer);
                currList.remove(currList.size() - 1);
            }
        }
        return answer;
    }
}