Amazon Interview Question for Software Engineers


Country: United States




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Basically you need to find two nodes which are same to start with. And then call the isDuplicate(root1, root2) method on both. This exhaustive traversal is worst case matching of each node with each node and then isIdentical is O(n). Hence the overall complexity is O(n3)
Now any two subtrees can be duplicate only if one is not ancestor of other.

- Yoda May 05, 2018 | Flag Reply
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static class TreeNode {
		@Override
		public int hashCode() {
			final int prime = 31;
			int result = 1;
			result = prime * result + ((children == null) ? 0 : children.hashCode());
			result = prime * result + data;
			return result;
		}

		@Override
		public boolean equals(Object obj) {
			if (this == obj)
				return true;
			if (obj == null)
				return false;
			if (getClass() != obj.getClass())
				return false;
			TreeNode other = (TreeNode) obj;
			if (children == null) {
				if (other.children != null)
					return false;
			} else if (!children.equals(other.children))
				return false;
			if (data != other.data)
				return false;
			return true;
		}

		TreeNode(int data) {
			this.data = data;
		}

		@Override
		public String toString() {
			return this.data + "";
		}

		int data;
		List<TreeNode> children;

		public int getData() {
			return data;
		}

		public void setData(int data) {
			this.data = data;
		}

		public List<TreeNode> getChildren() {
			return children;
		}

		public void setChildren(List<TreeNode> children) {
			this.children = children;
		}
	}


	public static void main(String args[]) throws Exception {
          Node root=new Node(2);
          root.left=new Node(1);
          root.right=new Node(1);
          List<TreeNode> temp=findDuplicateSubtrees(root);
	}

	public static List<TreeNode> findDuplicateSubtrees(Node root) {
		String inOrder = inOrderTraversal(root);
		String preOrder = preOrderTraversal(root);
		List<TreeNode> list = new ArrayList<TreeNode>();
		for (int i = 0; i < inOrder.length(); i++) {
			for (int j = i; j < inOrder.length(); j++) {
				String subStr = inOrder.substring(i, j);
				if (inOrder.indexOf(subStr)  != inOrder.lastIndexOf(subStr)
				 && preOrder.indexOf(subStr) != preOrder.lastIndexOf(subStr) && subStr.length()>0) {
					list.add(new TreeNode(Character.getNumericValue(subStr.charAt(0)))) ; 
				}
			}
		}
		return list;
	}

	private static String preOrderTraversal(Node root) {
		if (root == null)
			return "";
		return root.data + preOrderTraversal(root.left) + preOrderTraversal(root.right);
	}

	private static String inOrderTraversal(Node root) {
		if (root == null)
			return "";
		return inOrderTraversal(root.left) + root.data + inOrderTraversal(root.right);
	}

- koustav.adorable May 06, 2018 | Flag Reply
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using System;
using System.Collections.Generic;
using System.Linq;

namespace test
{
    public class TreeNode
    {
        public int val { get; set; }
        public TreeNode left { get; set; }
        public TreeNode right { get; set; }
        public TreeNode(int x)
        {
            val = x;
        }
    }

    
    public class FindDupNodes
    { 
        public FindDupNodes()
        {
            List<TreeNode> tree = new List<TreeNode>();
            tree = ConstructTree();

            TreeNode node = FindMaxDups(tree);
            Console.WriteLine($"{node.val} -- {node?.left?.val} -- {node?.right?.val}");


        }
        public List<TreeNode> ConstructTree()
        {
            List<TreeNode> nodes = new List<TreeNode>();
            TreeNode t1 = new TreeNode(1);
            TreeNode t2 = new TreeNode(2);
            TreeNode t3 = new TreeNode(3);
            TreeNode t4 = new TreeNode(4);

            t1.left = t2; t1.right = t3;
            t2.left = t4;
            t3.left = t2;
            t3.right = t4;

            nodes.Add(t1);
            nodes.Add(t2);
            nodes.Add(t3);
            nodes.Add(t4);
            return nodes;
        }
        
        public TreeNode FindMaxDups(List<TreeNode> tree)
        {
            //get the root
            TreeNode rootNode = tree.Where(t => t.val == 1).FirstOrDefault();
            //iterate over in recursion and see which one appears more than once
            IterateTree(rootNode, tree);
            int maxValue = nodeCounts.Max(n => n.Value); //.Select(nd => nd.Key);
            List<TreeNode> maxNodes = nodeCounts.Where(n => n.Value == maxValue).Select(n => n.Key).ToList();
            return maxNodes.FirstOrDefault();
        }

        public Dictionary<TreeNode, int> nodeCounts = new Dictionary<TreeNode, int>();
        public void IterateTree(TreeNode currentNode, List<TreeNode> tree)
        {
            if (currentNode == null)
                return;
           
            if (nodeCounts.ContainsKey(currentNode))
                nodeCounts[currentNode] = nodeCounts[currentNode] + 1;
            else
                nodeCounts.Add(currentNode, 1);

            if (currentNode.left == null && currentNode.right == null)
                return;

            IterateTree(currentNode.left, tree);
            IterateTree(currentNode.right, tree);

        }
    }
}

- GeekLion May 13, 2018 | Flag Reply


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