<p>Given an integer array <code>nums</code>, return <em>all the different possible non-decreasing subsequences of the given array with at least two elements</em>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,6,7,7]
<strong>Output:</strong> [[4,6],[4,6,7],[4,6,7,7],[4,7],[4,7,7],[6,7],[6,7,7],[7,7]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,3,2,1]
<strong>Output:</strong> [[4,4]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 15</code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> findSubsequences(vector<int>& nums) {
        set<vector<int>> ans;
        int n = nums.size();
        for (int i = 0; i < (1 << n); i++) {
            int lastindex = -1;
            vector<int> ss;
            bool k = true;
            for (int j = 0; j < n; j++) {
                if ((i >> j) & 1) {
                    if (lastindex == -1) {
                        lastindex = j;
                        ss.push_back(nums[j]);
                    } else {
                        if (nums[lastindex] <= nums[j]) {
                            ss.push_back(nums[j]);
                            lastindex = j;
                        } else {
                            k = false;
                            break;
                        }
                    }
                }
            }
            if (k && ss.size()>1) {

                ans.insert(ss);
            }
        }
        vector<vector<int>>fans;
        for (auto x:ans){
            fans.push_back(x);
        }
        return fans;
    }
};
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