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Hash Table of Employee Records

The document describes a C program that implements hashing and linear probing to map employee records identified by keys to memory addresses. The program takes employee records as input, hashes the keys to addresses using a modulo function, and handles collisions by linearly probing through open addresses until an empty slot is found.

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Khyathi Kiran
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38 views4 pages

Hash Table of Employee Records

The document describes a C program that implements hashing and linear probing to map employee records identified by keys to memory addresses. The program takes employee records as input, hashes the keys to addresses using a modulo function, and handles collisions by linearly probing through open addresses until an empty slot is found.

Uploaded by

Khyathi Kiran
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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12.

Given a File of N employee records with a set K of Keys (4-digit) which uniquely
determine the records in file F. Assume that file F is maintained in memory by a Hash
Table (HT) of m memory locations with L as the set of memory addresses (2-digit) of
locations in HT. Let the keys in K and addresses in L are Integers.
Develop a Program in C that uses Hash function H: K →L as H(K)=K mod m (remainder
method), and implement hashing technique to map a given key K to the address space L.
Resolve the collision (if any) using linear probing.

#include <stdio.h>
#include <stdlib.h>
#define MAX 100
int create(int);
void linear_prob(int[], int, int);
void display (int[]);
void main()
{
int a[MAX],num,key,i;
int ans=1;
printf(" collision handling by linear probing : \n");
for (i=0;i<MAX;i++)
{
a[i] = -1;
}
do
{
printf("\n Enter the data:");
scanf("%4d", &num);
key=create(num);
linear_prob(a,key,num);
printf("\n Do you wish to continue ? (1/0): ");
scanf("%d",&ans);

}while(ans);
display(a);
}

int create(int num)


{
int key;
key=num%10;
return key;

}
void linear_prob(int a[MAX], int key, int num)
{
int flag, i, count=0;
flag=0;
if(a[key]== -1)
{
a[key] = num;
}
else
{
printf("\nCollision Detected...!!!\n");
i=0;
while(i<MAX)
{
if (a[i]!=-1)
count++;
i++;
}
printf("Collision avoided successfully using LINEAR PROBING\n");
if(count == MAX)
{
printf("\n Hash table is full");
display(a);
exit(1);
}
for(i=key+1; i<MAX; i++)
{
if(a[i] == -1)
{
a[i] = num;
flag =1;
break;
}
}
i=0;
while((i<key) && (flag==0))
{
if(a[i] == -1)
{
a[i] = num;
flag=1;
break;
}
i++;
}
}
}
void display(int a[MAX])
{
int i,choice;
printf("1.Display ALL\n2.Filtered Display\n");
scanf("%d",&choice);
if(choice==1)
{
printf("\n The hash table is\n");
for(i=0; i<MAX; i++)
printf("\n %d %d ", i, a[i]);
}
else
{
printf("\n The hash table is\n");
for(i=0; i<MAX; i++)
{
if(a[i]!=-1)
{
printf("\n %d %d ", i, a[i]);
continue;
}
}
}
}

OUTPUT:
collision handling by linear probing :

Enter the data:31

Do you wish to continue ? (1/0): 1

Enter the data:43

Do you wish to continue ? (1/0): 1

Enter the data:36

Do you wish to continue ? (1/0): 1

Enter the data:44


Do you wish to continue ? (1/0): 1

Enter the data:78

Do you wish to continue ? (1/0): 1

Enter the data:57

Do you wish to continue ? (1/0): 1

Enter the data:19

Do you wish to continue ? (1/0): 1

Enter the data:77

Collision Detected...!!!
Collision avoided successfully using LINEAR PROBING

Do you wish to continue ? (1/0): 0


1.Display ALL
2.Filtered Display
2

The hash table is

1 31
3 43
4 44
6 36
7 57
8 78
9 19
10 77

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