Index Mapping (also known as Trivial Hashing) is a simple form of hashing where the data is directly mapped to an index in a hash table. The hash function used in this method is typically the identity function, which maps the input data to itself. In this case, the key of the data is used as the index in the hash table, and the value is stored at that index.
For example, if we have a hash table of size 10 and we want to store the value “apple” with the key “a”, the trivial hashing function would simply map the key “a” to the index “a” in the hash table, and store the value “apple” at that index.
One of the main advantages of Index Mapping is its simplicity. The hash function is easy to understand and implement, and the data can be easily retrieved using the key. However, it also has some limitations. The main disadvantage is that it can only be used for small data sets, as the size of the hash table has to be the same as the number of keys. Additionally, it doesn’t handle collisions, so if two keys map to the same index, one of the data will be overwritten.
Given a limited range array contains both positive and non-positive numbers, i.e., elements are in the range from -MAX to +MAX. Our task is to search if some number is present in the array or not in O(1) time.
Since the range is limited, we can use index mapping (or trivial hashing). We use values as the index in a big array. Therefore we can search and insert elements in O(1) time.
How to handle negative numbers?
The idea is to use a 2D array of size hash[MAX+1][2]
Algorithm:
Assign all the values of the hash matrix as 0.
Traverse the given array:
- If the element ele is non negative assign
- hash[ele][0] as 1.
- Else take the absolute value of ele and
- assign hash[ele][1] as 1.
To search any element x in the array.
- If X is non-negative check if hash[X][0] is 1 or not. If hash[X][0] is one then the number is present else not present.
- If X is negative take the absolute value of X and then check if hash[X][1] is 1 or not. If hash[X][1] is one then the number is present
Below is the implementation of the above idea.
// CPP program to implement direct index mapping// with negative values allowed.#include <bits/stdc++.h>using namespace std;
#define MAX 1000// Since array is global, it is initialized as 0.bool has[MAX + 1][2];
// searching if X is Present in the given array // or not.bool search(int X)
{ if (X >= 0) {
if (has[X][0] == 1)
return true;
else
return false;
}
// if X is negative take the absolute
// value of X.
X = abs(X);
if (has[X][1] == 1)
return true;
return false;
}void insert(int a[], int n)
{ for (int i = 0; i < n; i++) {
if (a[i] >= 0)
has[a[i]][0] = 1;
else
has[abs(a[i])][1] = 1;
}
}// Driver codeint main()
{ int a[] = { -1, 9, -5, -8, -5, -2 };
int n = sizeof(a)/sizeof(a[0]);
insert(a, n);
int X = -5;
if (search(X) == true)
cout << "Present";
else
cout << "Not Present";
return 0;
} |
// Java program to implement direct index // mapping with negative values allowed. class GFG
{final static int MAX = 1000;
// Since array is global, it // is initialized as 0. static boolean[][] has = new boolean[MAX + 1][2];
// searching if X is Present in // the given array or not. static boolean search(int X)
{ if (X >= 0)
{
if (has[X][0] == true)
{
return true;
}
else
{
return false;
}
}
// if X is negative take the
// absolute value of X.
X = Math.abs(X);
if (has[X][1] == true)
{
return true;
}
return false;
}static void insert(int a[], int n)
{ for (int i = 0; i < n; i++)
{
if (a[i] >= 0)
{
has[a[i]][0] = true;
}
else
{
int abs_i = Math.Abs(a[i]);
has[abs_i][1] = true;
}
}
}// Driver code public static void main(String args[])
{ int a[] = {-1, 9, -5, -8, -5, -2};
int n = a.length;
insert(a, n);
int X = -5;
if (search(X) == true)
{
System.out.println("Present");
}
else
{
System.out.println("Not Present");
}
}}// This code is contributed// by 29AjayKumar |
# Python3 program to implement direct index # mapping with negative values allowed.# Searching if X is Present in the # given array or not.def search(X):
if X >= 0:
return has[X][0] == 1
# if X is negative take the absolute
# value of X.
X = abs(X)
return has[X][1] == 1
def insert(a, n):
for i in range(0, n):
if a[i] >= 0:
has[a[i]][0] = 1
else:
has[abs(a[i])][1] = 1
# Driver codeif __name__ == "__main__":
a = [-1, 9, -5, -8, -5, -2]
n = len(a)
MAX = 1000
# Since array is global, it is
# initialized as 0.
has = [[0 for i in range(2)]
for j in range(MAX + 1)]
insert(a, n)
X = -5
if search(X) == True:
print("Present")
else:
print("Not Present")
# This code is contributed by Rituraj Jain |
<script>// JavaScript program to implement direct index // mapping with negative values allowed. let MAX = 1000;// Since array is global, it // is initialized as 0. let has = new Array(MAX+1);
for(let i=0;i<MAX+1;i++)
{ has[i]=new Array(2);
for(let j=0;j<2;j++)
has[i][j]=0;
}// searching if X is Present in // the given array or not.function search(X)
{ if (X >= 0)
{
if (has[X][0] == true)
{
return true;
}
else
{
return false;
}
}
// if X is negative take the
// absolute value of X.
X = Math.abs(X);
if (has[X][1] == true)
{
return true;
}
return false;
}function insert(a,n)
{ for (let i = 0; i < n; i++)
{
if (a[i] >= 0)
{
has[a[i]][0] = true;
}
else
{
let abs_i = Math.abs(a[i]);
has[abs_i][1] = true;
}
}
}// Driver code let a=[-1, 9, -5, -8, -5, -2];let n = a.length; insert(a, n);
let X = -5;
if (search(X) == true)
{
document.write("Present");
}
else
{
document.write("Not Present");
}
// This code is contributed by rag2127// corrected by akashish__</script> |
Present
Time Complexity: The time complexity of the above algorithm is O(N), where N is the size of the given array.
Space Complexity: The space complexity of the above algorithm is O(N), because we are using an array of max size.