<p>There are&nbsp;<code>n</code>&nbsp;items each&nbsp;belonging to zero or one of&nbsp;<code>m</code>&nbsp;groups where <code>group[i]</code>&nbsp;is the group that the <code>i</code>-th item belongs to and it&#39;s equal to <code>-1</code>&nbsp;if the <code>i</code>-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.</p>

<p>Return a sorted list of the items such that:</p>

<ul>
	<li>The items that belong to the same group are next to each other in the sorted list.</li>
	<li>There are some&nbsp;relations&nbsp;between these items where&nbsp;<code>beforeItems[i]</code>&nbsp;is a list containing all the items that should come before the&nbsp;<code>i</code>-th item in the sorted array (to the left of the&nbsp;<code>i</code>-th item).</li>
</ul>

<p>Return any solution if there is more than one solution and return an <strong>empty list</strong>&nbsp;if there is no solution.</p>

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

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png" style="width: 191px; height: 181px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
<strong>Output:</strong> [6,3,4,1,5,2,0,7]
</pre>

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

<pre>
<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
<strong>Output:</strong> []
<strong>Explanation:</strong>&nbsp;This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
</pre>

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

<ul>
	<li><code>1 &lt;= m &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>group.length == beforeItems.length == n</code></li>
	<li><code>-1 &lt;= group[i] &lt;= m - 1</code></li>
	<li><code>0 &lt;= beforeItems[i].length &lt;= n - 1</code></li>
	<li><code>0 &lt;= beforeItems[i][j] &lt;= n - 1</code></li>
	<li><code>i != beforeItems[i][j]</code></li>
	<li><code>beforeItems[i]&nbsp;</code>does not contain&nbsp;duplicates elements.</li>
</ul>

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        groupAdj={}
        nodeAdj={}
        groupAdjList={}
        dummyGroupNanme=m
        for i in range(n):
            nodeAdj[i]=[]
            if group[i]==-1:
                group[i]=dummyGroupNanme
                dummyGroupNanme+=1
        for j in range(dummyGroupNanme):
            groupAdj[j]=[]
            groupAdjList[j]={}
        for i in range(n):
            for k in beforeItems[i]:
                if group[k]!=group[i]:
                    groupAdj[group[k]].append(group[i])
                else:
                    if k not in groupAdjList[group[k]]:
                        groupAdjList[group[k]][k]=[]
                    
                    groupAdjList[group[k]][k].append(i)
                nodeAdj[k].append(i)
            if i not in groupAdjList[group[i]]:
                groupAdjList[group[i]][i]=[]
        groupSort=[]
        groupVisit=set()
        groupPath=set()
        
        for j in range(dummyGroupNanme):
            if not self.dfs(j,groupVisit,groupPath,groupSort,groupAdj):
                return []
        #groupSort.reverse()
        
        topSort=[]
        
        for j in groupSort:
            nodeVisit=set()
            nodePath=set()
            for key in groupAdjList[j]:
                if not self.dfs(key,nodeVisit,nodePath,topSort,groupAdjList[j]):
                    return []
        
        topSort.reverse()
        
        return topSort
        
    def dfs(self,src,visit,path,sortList,itrDict)->bool:
        if src in visit and src in path:
            return False
        if src in visit:
            return True
        visit.add(src)
        path.add(src)
        for nei in itrDict[src]:
            if not self.dfs(nei,visit,path,sortList,itrDict):
                return False
        sortList.append(src)
        path.remove(src)
        return True    
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