<p>Given a collection of candidate numbers (<code>candidates</code>) and a target number (<code>target</code>), find all unique combinations in <code>candidates</code>&nbsp;where the candidate numbers sum to <code>target</code>.</p>

<p>Each number in <code>candidates</code>&nbsp;may only be used <strong>once</strong> in the combination.</p>

<p><strong>Note:</strong>&nbsp;The solution set must not contain duplicate combinations.</p>

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

<pre>
<strong>Input:</strong> candidates = [10,1,2,7,6,1,5], target = 8
<strong>Output:</strong> 
[
[1,1,6],
[1,2,5],
[1,7],
[2,6]
]
</pre>

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

<pre>
<strong>Input:</strong> candidates = [2,5,2,1,2], target = 5
<strong>Output:</strong> 
[
[1,2,2],
[5]
]
</pre>

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

<ul>
	<li><code>1 &lt;=&nbsp;candidates.length &lt;= 100</code></li>
	<li><code>1 &lt;=&nbsp;candidates[i] &lt;= 50</code></li>
	<li><code>1 &lt;= target &lt;= 30</code></li>
</ul>

class Solution {
public:
    vector<vector<int>> combinationSum2(vector<int>& nums, int target) {
        int n=nums.size(),mx=0;
        sort(nums.begin(),nums.end());
        vector<vector<vector<int>>>prevdp(target+1),curdp(target+1);
        map<vector<int>,int>mp;
        for(int j=0;j<n;j++)
        {
            for(int i=nums[j];i<=target;i++)
            {
               if(i==nums[j]) 
               {
                    if(mp[{i}]) continue;
                    curdp[i].push_back({i});
                    mp[{i}]++;
               }
               else if(i-nums[j]<=target&&prevdp[i-nums[j]].size())
               {
                    for(auto v:prevdp[i-nums[j]]) 
                    {
                        v.push_back(nums[j]);
                        if(mp[v]) continue;
                        curdp[i].push_back(v);
                        mp[v]++;
                    }
               }
            }
            prevdp=curdp;
        }
        return prevdp[target];
    }
};
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