Zoho Interview Question for Senior Software Development Engineers


Team: Theriyathu
Country: India
Interview Type: In-Person




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1
of 1 vote

First you can build a map of relationships. In java it might look like:

Map<String, List<String>> relations = new HashMap<String, List<String>>();

Then you enter your values into the map (i.e. "Syam" would be the key for an ArrayList<String> containing "Ram" and "Akil")

Once we have our map, we can use getDescendantCount("Syam", 2) on the below function to get our desired output.

    private int getDescendantCount(String key, int depth) {
        int descendant  = 0;

        if(depth <= 0) { return 1; }
        if(!relations.containsKey(key)) { return 0; }

        for(String child : tree.get(key)) {
            descendant  += getDescendantCount(child, depth - 1);
        }

        return descendant  ;
    }

- Bradley Wheeler December 19, 2016 | Flag Reply
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0
of 0 vote

Is finding the first level parents is part of the task or it's an input?

- Nikolay December 19, 2016 | Flag Reply
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0
of 0 vote

It is simple if we make father childList Map.
1. Create Map<Father, List<Child>> by iterating over pairs.
2. For level =1 map.get(father) which in turn returns childList (List<Child>) totalChildren = childList.size()
3. For level = 2 , Iterate over childList and each element of childList is now father so map.get(childList(i)) and add the size of return value for totalChildren.

repeat same for the next level.

- Avinash Kumar December 19, 2016 | Flag Reply
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0
of 0 vote

function getChildren (map, name, level) {
	if (level === 1) {
		return map[name]
	}
	var children = []
	for (var i = 0; i < map[name].length; ++i) {
		children = children.concat(getChildren(map, map[name][i], level - 1));
	}
	return children;
}
module.exports = function (A, name, level) {
	var map = {}
	for (var i = 0; i < A.length; ++i) {
		var child = A[i][0];
		var father = A[i][1];
		map[father] = map[father] || [];
		map[father].push(child);
	}
	var children = getChildren(map, name, level);
	return [children, children.length];
}

var children = module.exports([
	['Ram', 'Syam'],
	['Akil', 'Syam'],
	['Nikil', 'Ram'],
	['Subhash', 'Ram'],
	['Karthik', 'Akil']
], 'Syam', 2);

console.log(children);

- srterpe December 20, 2016 | Flag Reply
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0
of 0 vote

import java.util.ArrayList;
import java.util.List;


public class PrintChildCound {

  int countChildLevel(Node root, int level, int currentLevel, int total) {
    if(root == null) {
      return 0;
    }
    if(level == currentLevel) {
      System.out.println(root.val);
      total++;
    }
    for(Node node : root.list) {
      total = countChildLevel(node, level, currentLevel+1, total);
    }
    return total;
  }
  public static void main(String... args) {
    Node shyam = new Node("Shyam");
    Node akhil = new Node("Akhil");
    shyam.hasChild(akhil);
    Node ram = new Node("Ram");
    shyam.hasChild(ram);

    ram.hasChild(new Node("Nikil"));
    ram.hasChild(new Node("Subhash"));

    akhil.hasChild(new Node("Karthik"));

    System.out.println(new PrintChildCound().countChildLevel(shyam, 2, 0, 0));
  }
}
class Node  {
  Node(String val) {
    this.val = val;
  }
  void hasChild(Node node) {
    list.add(node);
  }
  String val;
  List<Node> list = new ArrayList();
}

- Weshall December 20, 2016 | Flag Reply
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0
of 0 vote

import java.util.*;

public class NumOfChildrenImpl{
  Map<String,String> relationships;
  
  public NumOfChildrenImpl(Map<String,String> relationships){
    this.relationships = relationships;
  }
  
  public int getNumOfChildren(String fatherName,int level){
    Set<String> directChildren = new HashSet<>();
    for(Map.Entry<String,String> entry:relationships.entrySet()){
      if(entry.getValue().equals(fatherName)){
        directChildren.add(entry.getKey());
      }
    }
    
    if(level<=1){
      //for(String childName:directChildren){
      //  System.out.print(" " + childName + " ");
      //}      
      
      return directChildren.size();
    } 
    
    int numOfIndirectChildren=0;
    
    for(String childName:directChildren){
      numOfIndirectChildren+=getNumOfChildren(childName,level-1);
    }
    
    return numOfIndirectChildren;
  }
}

- dvizelman December 21, 2016 | Flag Reply
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0
of 0 vote

One way could be iterate over the childs starting from father and increament a count. Once you reach n-1 count then push all those childs to a stack or somewhere. Now, you can simply list their childrens from hashmap or the input given.

- amiedeep December 21, 2016 | Flag Reply
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0
of 0 vote

{{
function go(name, level) {

var input = [
['aa', 'bj'],
['bb', 'bj'],
['cc', 'wk'],
['bj', 'kong'],
['wk', 'kong'],
];

var tree = {};

for(var i in input) {
var child = input[i][0];
var parent = input[i][1];

if(!tree[parent]) {
tree[parent] = {};
}

if(!tree[parent][child]) {
tree[parent][child] = {};
}
}

var count = 0;
var curKeys = [name];

while(count < level) {
var keys = [];
for(var i in curKeys) {
keys = keys.concat( Object.keys(tree[curKeys[i]]) );
}
curKeys = keys;
count++;
}
return curKeys.length;
}
}}

- kongfamily0804 January 05, 2017 | Flag Reply
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0
of 0 vote

#include<iostream>
#include<string>
using namespace std;

struct s
{
string child;
string parent;
};


int find(s cp[],string name,int level,int l,int n)
{
int i;
int count=0;
if(l==level)
return 1;
for(i=0;i<n;i++)
{
if(cp[i].parent==name)
{
count=count+find(cp,cp[i].child,level,l+1,n);
}
}
return count;
}

int main()
{
int n,i,level,l=0,count=0;
cin>>n;
s cp[n];
string name;
for(i=0;i<n;i++)
{
cin>>cp[i].child;
cin>>cp[i].parent;
}
cin>>name>>level;
cout<<find(cp,name,level,l,n);
}

- Anonymous February 27, 2018 | Flag Reply


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