Linear search is a very simple search algorithm. In this type of search, a sequential search is made over all items one by one. Every item is checked and if a match is found then that particular item is returned, otherwise the search continues till the end of the data collection.
Time Complexity: O(n) where n is the number of elements in the array, because the worst case scenario is that the element is not present in the array and we have to traverse the whole array.
Step 1: Set i to 1
Step 2: if i > n then go to step 7
Step 3: if A[i] = x then go to step 6
Step 4: Set i to i + 1
Step 5: Go to Step 2
Step 6: Print Element x Found at index i and go to step 8
Step 7: Print element not found
Step 8: Exit
A = [2, 4, 0, 1, 9]
x = 1
n = 5
Step 1: i = 1
Step 2: 1 <= 5
Step 3: A[1] = 4 != 1
Step 4: i = 2
Step 5: 2 <= 5
Step 3: A[2] = 0 != 1
Step 4: i = 3
Step 5: 3 <= 5
Step 3: A[3] = 1 = 1
Step 6: Print Element 1 Found at index 3
procedure linear_search (list, value)
for each item in the list
if match item == value
return the item's location
end if
end for
end procedure
#include <stdio.h>
// Search an element in an array using linear search
void search(int arr[], int n, int x){
int i;
for(i=0; i<n; i++){
if(arr[i] == x){
printf("Element found at index %d", i+1);
return;
}
}
printf("Element not found");
}
// Main function
int main(){
int arr[] = {10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60};
int n = sizeof(arr)/sizeof(arr[0]);
int x = 35;
search(arr, n, x);
return 0;
}
public class LinearSearchExample
{
public static int linearSearch(int[] arr, int key)
{
for(int i=0;i<arr.length;i++)
{
if(arr[i] == key)
{
return i;
}
}
return -1;
}
public static void main(String a[])
{
int[] a1= {10,20,30,50,70,90};
int key = 50;
System.out.println(key+" is found at index: "+linearSearch(a1, key));
}
}
def linear_search(arr, x):
for i in range(len(arr)):
if arr[i] == x:
return i
return -1
arr = [2, 0, 1, 9]
x = 1
print("Element found at index " + str(linear_search(arr, x)))
#include <iostream>
using namespace std;
int linearSearch(int arr[], int n, int x)
{
for (int i = 0; i < n; i++)
if (arr[i] == x)
return i;
return -1;
}
int main(void)
{
int arr[] = { 2, 0, 1, 9 };
int x = 1;
int n = sizeof(arr) / sizeof(arr[0]);
int result = linearSearch(arr, n, x);
(result == -1)
? cout << "Element is not present in array"
: cout << "Element is present at index " << result;
return 0;
}