## Microsoft Interview Question for Software Engineer in Tests

Country: United States

Comment hidden because of low score. Click to expand.
4
of 4 vote

Pretty much the same as LCA of a BST, we just need to traverse all children.

``````Node LCA(Node a, Node b, Node root)
{
if(a == root || b == root)
return root;

int count = 0;
Node temp = null;

for(Node iter : root.children)
{
Node res = LCA(a, b, iter);
if(res != null)
{
count++;
temp = res;
}
}

if(count == 2)
return root;

return temp;
}``````

Comment hidden because of low score. Click to expand.
0

This code assumes that both the nodes (a and b) are present in the tree right?
If only one of them is present then it will return that whereas it must return NULL for such a case.

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1
of 1 vote

http{://}community.topcoder.com/tc?module=Static&d1=tutorials&d2=lowestCommonAncestor

Comment hidden because of low score. Click to expand.
1
of 1 vote

``````class FindClosestCommonAncestorNaryTree
{
public Node Solve(Node tree, Node node1, Node node2)
{
List<Node> ancestors1 = new List<Node>();
List<Node> ancestors2 = new List<Node>();

tree.GetAncestorsOf(node1, ref ancestors1);
tree.GetAncestorsOf(node2, ref ancestors2);

foreach (Node n1 in ancestors1)
{
foreach (Node n2 in ancestors2)
{
if (n1 == n2)
{
return n1;
}
}
}

return null;
}
}

public class Node
{
public int data;
public List<Node> children = null;

public Node(int _data, List<Node> _children)
{
data = _data;
children = _children;
}

internal bool GetAncestorsOf(Node node, ref List<Node> ancestors)
{
if (this == node)
{
return true;
}

if (children != null)
{
foreach (Node child in children)
{
if (child.GetAncestorsOf(node, ref ancestors))
{
return true;
}
}
}

return false;
}
}``````

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0

what is the time complexity of this??

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1
of 3 vote

get ancestor list of both nodes n1 and n2.
compare top ancestor of n1 with top ancestor of n2.
if equal, got to the a level below of ancestor for n1 & n2 and compare in iterative manner.
at some point, either they will not be equal or one of the ancestor will be null.
return the previous compared ancestor.

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0

can you show an implementation of this?

Comment hidden because of low score. Click to expand.
0

two stacks. push ancestor's from both the trees in two stacks. pull until different encounter or one of the stacks is empty. last pull is the answer.

can someone help with the complexity?

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0
of 0 vote

Lowest Common Ancestor

{
{
}
{
}
else
}

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0

Its n-ary. Not binary tree not even binary search tree.

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0
of 0 vote

if not having parent pointers: O(n) + O(h) = O(n)
if having parent pointers: O(h)

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````def lca_dfs(self,key1,key2,root):
execut=[]
execut.extend([key1,key2])
lst1=self.dfs(execut[0],root)
lst2=self.dfs(execut[1],root)
intrsctn=(set(lst1)).intersection(set(lst2))
lst=list(intrsctn)
if len(lst)>0:
print "lca is:",lst[len(lst)-1]
else:
print "lca is:",root.val

def dfs(self,key,root):
if len(root.child)==0:
return
for nde in root.child:
if nde.val==key:
return ([])
else:
fnd=self.dfs(key,nde)
if isinstance(fnd,list):
fnd.append(nde.val)
return fnd

return fnd``````

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0
of 0 vote

``````def lca_dfs(self,key1,key2,root):
execut=[]
execut.extend([key1,key2])
lst1=self.dfs(execut[0],root)
lst2=self.dfs(execut[1],root)
intrsctn=(set(lst1)).intersection(set(lst2))
lst=list(intrsctn)
if len(lst)>0:
print "lca is:",lst[len(lst)-1]
else:
print "lca is:",root.val

def dfs(self,key,root):
if len(root.child)==0:
return
for nde in root.child:
if nde.val==key:
return ([])
else:
fnd=self.dfs(key,nde)
if isinstance(fnd,list):
fnd.append(nde.val)
return fnd

return fnd``````

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0
of 0 vote

``````public static N_ary_tree findCommonAncestor(N_ary_tree root,
N_ary_tree node1,N_ary_tree node2){
if(root == null){
return root;
}
if(root==node1 || root==node2){
return root;
}

N_ary_tree res = null;
int count = 0;
for(N_ary_tree ele : root.children){
N_ary_tree temp = findCommonAncestor(ele,node1,node2);
if(temp!=null){
res=temp;
count++;
}
}

if(count>=2){
return root;
}

return res;

}``````

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