Facebook Interview Question for Android Engineers


Country: United States
Interview Type: Phone Interview




Comment hidden because of low score. Click to expand.
2
of 4 vote

public static int getMaxPathSum(Node root) {
		int suml = 0;
		int sumr =0;
		if (root.l == null && root.r == null) return root.val;
		if (root.l != null ) {
			suml = suml + root.val +  getMaxPathSum(root.l);
		}
		if (root.r != null ) {
			sumr = sumr + root.val + getMaxPathSum(root.r);
		}
		
		return Math.max(suml, sumr);

}

- Anonymous May 07, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

// This function returns overall maximum path sum in 'res' 
// And returns max path sum going through root. 
int findMaxUtil(Node* root, int &res) 
{ 
    //Base Case 
    if (root == NULL) 
        return 0; 
  
    // l and r store maximum path sum going through left and 
    // right child of root respectively 
    int l = findMaxUtil(root->left,res); 
    int r = findMaxUtil(root->right,res); 
  
    // Max path for parent call of root. This path must 
    // include at-most one child of root 
    int max_single = max(max(l, r) + root->data, root->data); 
  
    // Max Top represents the sum when the Node under 
    // consideration is the root of the maxsum path and no 
    // ancestors of root are there in max sum path 
    int max_top = max(max_single, l + r + root->data); 
  
    res = max(res, max_top); // Store the Maximum Result. 
  
    return max_single; 
} 
  
// Returns maximum path sum in tree with given root 
int findMaxSum(Node *root) 
{ 
    // Initialize result 
    int res = INT_MIN; 
  
    // Compute and return result 
    findMaxUtil(root, res); 
    return res; 
}

- Ak13 October 23, 2018 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

What if some values were negative ? I think the last row should be :

return Math.max(Math.max(suml, sumr),root.val);

Because the root.val could be larger than its pathes L and R .

- Mathboy January 29, 2019 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int getMaxPathSum(Node root) {
int suml = 0;
int sumr =0;
if (root.l == null && root.r == null) return root.val;
if (root.l != null ) {
suml = suml + root.val + getMaxPathSum(root.l);
}
if (root.r != null ) {
sumr = sumr + root.val + getMaxPathSum(root.r);
}

return Math.max(suml, sumr);
}

- Anonymous May 07, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int getMaxPathSum(Node root) {
		int suml = 0;
		int sumr =0;
		if (root.l == null && root.r == null) return root.val;
		if (root.l != null ) {
			suml = suml + root.val +  getMaxPathSum(root.l);
		}
		if (root.r != null ) {
			sumr = sumr + root.val + getMaxPathSum(root.r);
		}
		
		return Math.max(suml, sumr);

}

- Imad May 07, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

function getMaxPathSum(Tree) {
   const leftNode = Tree.left;
   const rightNode = Tree.right;
   if (leftNode === null && rightNode === null) {
      return Tree.value;
   }
   const sumLeft = Tree.value + getMaxPathSum(leftNode);
   const sumRight = Tree.value + getMaxPathSum(rightNode);
   if (sumLeft >= sumRight) {
      return sumLeft;
   }
   return sumRight;
}

- mgonyan May 30, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Well, I think that you can have a path which doesn't necessarily go through the root...That means that we have to find the path between any two nodes which constitutes the maximum sum...The below code achieves that

def find_height_by_value(node):
    if node is None:
        return 0
    return node.data + max(find_height_by_value(node.left), find_height_by_value(node.right))

def find_max_sum(node):
    if node is None:
        return 0
    lheight = find_height_by_value(node.left)
    rheight = find_height_by_value(node.right)

    lmax = find_max_sum(node.left)
    rmax = find_max_sum(node.right)

    return max(max(lmax, rmax), node.data + lheight + rheight)

root = Node(4)
root.left = Node(3)
root.right = Node(5)
root.left.left = Node(2)
root.right.right = Node(6)
print("MAXIUMUM SUM = ",find_max_sum(root))

The output for the above test case is 20...

- Devansh Singh July 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

def find_height_by_value(node):
    if node is None:
        return 0
    return node.data + max(find_height_by_value(node.left), find_height_by_value(node.right))

def find_max_sum(node):
    if node is None:
        return 0
    lheight = find_height_by_value(node.left)
    rheight = find_height_by_value(node.right)

    lmax = find_max_sum(node.left)
    rmax = find_max_sum(node.right)

    return max(max(lmax, rmax), node.data + lheight + rheight)


root = Node(4)
root.left = Node(3)
root.right = Node(5)
root.left.left = Node(2)
root.right.right = Node(6)
print("MAXIUMUM SUM = ",find_max_sum(root))

The path can be between any two nodes constituting the maximum sum...It doesn't necessarily has to pass through the root node.

- Devansh Singh July 24, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

class Solution(object):
    def maxPathSum(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.res = float('-inf')
        self.dfs(root)
        return self.res

    def dfs(self, node):
        if not node: return 0
        l = self.dfs(node.left)
        if l < 0: l = 0
        r = self.dfs(node.right)
        if r < 0: r = 0
        if l+r+node.val > self.res:
            self.res = l+r+node.val
        return node.val + max(l,r)

- aasimon@uc.cl August 11, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/*
Given a binary tree, where each node represents an integer, 
find the max value of path sum.
*/

/*
Clarification:

Can I have negative values as nodes values? YES
Can I assume that all the numbers are integer? YES
Should the root be on the path? NO

Brute force: For every every node, I'll consider this nodes as the start node
and I'll try with the rest of the nodes as the finish node.

Time complexity is O(N*N)

Better solution is: For every node, has a value that is the max path that start
in that node and goes to the subtree.

Then, we can compare and merge this values to get the final answer.

*/

class BTree{
 public:
	long long value,sum;
	BTree * left, * right;
	BTree(long long _value, BTree * _left, BTree * _right){
		value=_value; 
		left=_left;
		right=_right;
		sum=max({0LL,value,getSum(left)+value,getSum(right)+value});
	}
	static long long maxPath(BTree * tree){
		if(!tree)return 0;
		long long ans=max({tree->sum,maxPath(tree->left),maxPath(tree->right)});
		ans=max(ans,tree->value+getSum(tree->left)+getSum(tree->right));
		return ans;
	}
	static long long getSum(BTree * tree){
		if(!tree)return 0;
		return tree->sum;
	}
};

int main() {FIN;
	BTree f(2,NULL,NULL);
	BTree g(1,NULL,NULL);
	BTree c(3,&f,&g);
	BTree d(4,NULL,NULL);
	BTree e(5,NULL,NULL);
	BTree b(3,&d,&e);
	BTree a(2,&b,&c);
	cout<<BTree::maxPath(&a)<<"\n";
}

- mrz August 29, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/*
Given a binary tree, where each node represents an integer, 
find the max value of path sum.
*/

/*
Clarification:

Can I have negative values as nodes values? YES
Can I assume that all the numbers are integer? YES
Should the root be on the path? NO

Brute force: For every every node, I'll consider this nodes as the start node
and I'll try with the rest of the nodes as the finish node.

Time complexity is O(N*N)

Better solution is: For every node, has a value that is the max path that start
in that node and goes to the subtree.

Then, we can compare and merge this values to get the final answer.

*/

class BTree{
 public:
	long long value,sum;
	BTree * left, * right;
	BTree(long long _value, BTree * _left, BTree * _right){
		value=_value; 
		left=_left;
		right=_right;
		sum=max({0LL,value,getSum(left)+value,getSum(right)+value});
	}
	static long long maxPath(BTree * tree){
		if(!tree)return 0;
		long long ans=max({tree->sum,maxPath(tree->left),maxPath(tree->right)});
		ans=max(ans,tree->value+getSum(tree->left)+getSum(tree->right));
		return ans;
	}
	static long long getSum(BTree * tree){
		if(!tree)return 0;
		return tree->sum;
	}
};

int main() {FIN;
	BTree f(2,NULL,NULL);
	BTree g(1,NULL,NULL);
	BTree c(3,&f,&g);
	BTree d(4,NULL,NULL);
	BTree e(5,NULL,NULL);
	BTree b(3,&d,&e);
	BTree a(2,&b,&c);
	cout<<BTree::maxPath(&a)<<"\n";
}

- mrz August 29, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/*
Given a binary tree, where each node represents an integer, 
find the max value of path sum.
*/

/*
Clarification:

Can I have negative values as nodes values? YES
Can I assume that all the numbers are integer? YES
Should the root be on the path? NO

Brute force: For every every node, I'll consider this nodes as the start node
and I'll try with the rest of the nodes as the finish node.

Time complexity is O(N*N)

Better solution is: For every node, has a value that is the max path that start
in that node and goes to the subtree.

Then, we can compare and merge this values to get the final answer.

*/

class BTree{
 public:
	long long value,sum;
	BTree * left, * right;
	BTree(long long _value, BTree * _left, BTree * _right){
		value=_value; 
		left=_left;
		right=_right;
		sum=max({0LL,value,getSum(left)+value,getSum(right)+value});
	}
	static long long maxPath(BTree * tree){
		if(!tree)return 0;
		long long ans=max({tree->sum,maxPath(tree->left),maxPath(tree->right)});
		ans=max(ans,tree->value+getSum(tree->left)+getSum(tree->right));
		return ans;
	}
	static long long getSum(BTree * tree){
		if(!tree)return 0;
		return tree->sum;
	}
};

- Anonymous August 29, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/*
Given a binary tree, where each node represents an integer, 
find the max value of path sum.
*/

/*
Clarification:

Can I have negative values as nodes values? YES
Can I assume that all the numbers are integer? YES
Should the root be on the path? NO

Brute force: For every every node, I'll consider this nodes as the start node
and I'll try with the rest of the nodes as the finish node.

Time complexity is O(N*N)

Better solution is: For every node, has a value that is the max path that start
in that node and goes to the subtree.

Then, we can compare and merge this values to get the final answer.

*/

class BTree{
 public:
	long long value,sum;
	BTree * left, * right;
	BTree(long long _value, BTree * _left, BTree * _right){
		value=_value; 
		left=_left;
		right=_right;
		sum=max({0LL,value,getSum(left)+value,getSum(right)+value});
	}
	static long long maxPath(BTree * tree){
		if(!tree)return 0;
		long long ans=max({tree->sum,maxPath(tree->left),maxPath(tree->right)});
		ans=max(ans,tree->value+getSum(tree->left)+getSum(tree->right));
		return ans;
	}
	static long long getSum(BTree * tree){
		if(!tree)return 0;
		return tree->sum;
	}
};

- mrz August 29, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int maxPath(Node node) {
if(node==null){
return 0;
}
return Math.max(node.data+maxPath(node.left),node.data+maxPath(node.right));
}

- simplest October 03, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static int maxPath(Node node) {
        if(node==null){
            return 0;
        }
        return Math.max(node.data+maxPath(node.left),node.data+maxPath(node.right));
    }

- simplest October 03, 2018 | Flag Reply


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