Program to Find Number of Grandchildren - Zoho Programming Test Question

                                                 Grand Children

Here we will see a c program to find number of grandchildren, great 

grandchildren great-great grandchildren and so-on of a given person. In 

this program, the input is a table with two columns, 'Child' and 'Father'.

Then we enter a name whose number grandchildren, great grandchildren

etc is to be calculated. We do not have to count children. An example is:

Given a two dimensional array of strings like

<”luke”, “shaw”>

<”wayne”, “rooney”>

<”rooney”, “ronaldo”>

<”shaw”, “rooney”> 

Where in each row, the first string is “child”, second string is 

“Father”. And given “ronaldo” we have to find his no of grandchildren. 

Here “ronaldo” has total 3 grandchildren (2 grandchildren: wayne and shaw ; 

a great grandchild luke). So our output should be 3.

write a program to find the number of grandchilden for a given name.                                       

                                                   C - Solution

#include <string.h>

#include <stdio.h>

int main()

char a[10][100],b[10][100],str[100];

{

int num,k;

scanf("%d",&num);

scanf("%s",&str);

for(int i=0;i<num;i++)

{

scanf("%s%s",&a[i],&b[i]);

if(strcmp(str,b[i]) == 0)

k = i;

}

int count = 0;

for(int i=0;i<num;i++)

{

if(strcmp(a[k],b[i]) == 0)

count++;

}

printf("%d",count);

return 0;

}

(or)

#include<stdio.h>

#include<string.h>

int n;

char name[20];

struct reln

{

char child[20];

char father[20];

}r[10];

int count=0;

void countChildren(char name[])

{

    int j;

   for(j=0;j<n;j++)

        {

        if(strcmp(name,r[j].father)==0)

            {

            count++;

            countChildren(r[j].child);

            }

        }

    }

void main()

{

 int i;

printf("\nEnter the number of inputs: ");

scanf("%d",&n);

    for(i=0;i<n;i++)

    {

    scanf("%s",r[i].child);

    scanf("%s",r[i].father);

    }

printf("\nEnter name of the one whose no. of grandchildren is needed: ");

scanf("%s",name);

for(i=0;i<n;i++)

    {

    if(strcmp(r[i].father,name)==0)

        countChildren(r[i].child);

    }

printf("\nNo .of grandchildren of %s=%d",name,count);

}

scanf("%s",name);

for(i=0;i<n;i++)

    {

    if(strcmp(r[i].father,name)==0)

        countChildren(r[i].child);

    }

printf("\nNo .of grandchildren of %s=%d",name,count);

}

                                       

                                                  JAVA - Solution

import java.util.Scanner;

import java.util.ArrayList;

class Main  {

 public static void main(String args[]) { 

  Scanner scan=new Scanner(System.in);

  int n=scan.nextInt();scan.nextLine();

  ArrayList< String > arr=new  ArrayList<>();

  int i,j,flag;

  String[] input=new String[n];

  String[] father=new String[n];

  String[] child=new String[n];

   for(i=0;i<n;i++)

  {

    input[i]=scan.nextLine();

    father[i]="";

    child[i]="";

    flag=0;

    for(j=0;j<input[i].length();j++)

    {

    if(input[i].charAt(j)==',')

       flag=1;

    if(((input[i].charAt(j)>='a'&&input[i].charAt(j)<='z')||(input[i].charAt(j)>='A' &&

                input[i].charAt(j)<='Z'))&&flag==0)

        child[i]=child[i]+input[i].charAt(j);  

    else  if(((input[i].charAt(j)>='a'&&input[i].charAt(j)<='z')||

                 (input[i].charAt(j)>='A'&&input[i].charAt(j)<='Z'))&&flag==1)

        father[i]=father[i]+input[i].charAt(j);

    }

  }

  String search=scan.nextLine();

  for(i=0;i<n;i++)

  {

     if(father[i].equals(search))

     {

         arr.add(child[i]);

     }

  }

  flag=0;

  for(i=0;i<arr.size();i++)

  {

     for(j=0;j<n;j++)

     {

       if(father[j].equals(arr.get(i)))

        flag++;  

     }

   }

  System.out.println(flag);

 }

}

Sample Input:

 4

 <”luke”, “shaw”>

 <”wayne”, “rooney”>

 <”rooney”, “ronaldo”>

 <”shaw”, “rooney”>

 ronaldo

Sample Output:

 2