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Marizza181 [45]
3 years ago
13

2 Consider the sequence of keys (5,16,22,45,2,10,18,30,50,12,1). Draw the result of inserting entries with these keys (in the gi

ven order) into a. An initially empty (2,4) tree. b. An initially empty red-black tree.

Computers and Technology
1 answer:
Juliette [100K]3 years ago
6 0

Answer:

A) (2,4) tree

  • Insertion of  key 45 makes key unbalanced and this is because it violates the 2,4 tree so we split the node
  • insertion of key 10 makes key unbalanced and this is because it violates the 2,4 tree so we split the node

B) Red-black tree

Explanation:

The diagrams for the solutions are attached showing the results of inserting entries

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// code in C++

#include <bits/stdc++.h>

using namespace std;

// main function

int main()

{

   // variables

   int sum_even=0,sum_odd=0,eve_count=0,odd_count=0;

   int largest=INT_MIN;

   int smallest=INT_MAX;

   int n;

   cout<<"Enter 10 Integers:";

   // read 10 Integers

   for(int a=0;a<10;a++)

   {

       cin>>n;

       // find largest

       if(n>largest)

       largest=n;

       // find smallest

       if(n<smallest)

       smallest=n;

       // if input is even

       if(n%2==0)

       {  

           // sum of even

           sum_even+=n;

           // even count

           eve_count++;

       }

       else

       {

           // sum of odd    

          sum_odd+=n;

          // odd count

          odd_count++;

       }

   }

   

   // print sum of even

   cout<<"Sum of all even numbers is: "<<sum_even<<endl;

   // print sum of odd

   cout<<"Sum of all odd numbers is: "<<sum_odd<<endl;

   // print largest

   cout<<"largest Integer is: "<<largest<<endl;

   // print smallest

   cout<<"smallest Integer is: "<<smallest<<endl;

   // print even count

   cout<<"count of even number is: "<<eve_count<<endl;

   // print odd cout

   cout<<"count of odd number is: "<<odd_count<<endl;

return 0;

}

Explanation:

Read an integer from user.If the input is greater that largest then update the  largest.If the input is smaller than smallest then update the smallest.Then check  if input is even then add it to sum_even and increment the eve_count.If the input is odd then add it to sum_odd and increment the odd_count.Repeat this for 10 inputs. Then print sum of all even inputs, sum of all odd inputs, largest among all, smallest among all, count of even inputs and count of odd inputs.

Output:

Enter 10 Integers:1 3 4  2 10 11 12 44 5 20                                                                                

Sum of all even numbers is: 92                                                                                            

Sum of all odd numbers is: 20                                                                                              

largest Integer is: 44                                                                                                    

smallest Integer is: 1                                                                                                    

count of even number is: 6                                                                                                

count of odd number is: 4

3 0
3 years ago
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