Program to search a number using Linear Search

Linear search checks each element in the array one by one until the desired element is found or the array ends.

Steps:

1. Start from the first element.

2. Compare the current element with the key (number to be searched).

3. If it matches, return its position (index).

4. If not, move to the next element.

5. If the end of the array is reached without a match, the element is not present.

Example:
Array: [4, 8, 15, 16, 23, 42]
Key: 23

• Compare 4 → No match

• Compare 8 → No match

• Compare 15 → No match

• Compare 16 → No match

• Compare 23 → Match found at index 4 (or position 5 if counting from 1).

#include<iostream.h>
 main() 
{ 
int n, a[20],ser,i,flag=0; 
count<<"Enter the size of an array:"; 
cin>>n; 
count<<"Enter the elements to array:\n"; 
for(i=0; i<n; i++) cin>>a[i]; 
cout<<"\n Enter the elents to search:";
cin>>ser; 
for(i=0; i<n; i++) 
{ 
if(ser == a[i])
 { 
flag = 1;
break; 
} 
}
if(flag == 1) 
cout<<"\n Element "<<ser<<"is found in the array\n";
else cout<<"\n Element "<<ser<<"is not found in the array\n"; 
return (0); 
}

OUTPUT
Enter the size of an array: 5
Enter the elements to array:
1
4
6
9
7
Enter the element to search: 9
Element 9 is found in the array
-----
Enter the size of an array: 4
Enter the elements to array:
9
5
7
3
Enter the element to search: 4
Element 4 is not found in the array