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  • Difficulty Level : Basic
  • Last Updated : 02 Aug, 2022
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Given an array arr[] of N elements, the task is to write a function to search a given element x in arr[].

Examples :  

Input: arr[] = {10, 20, 80, 30, 60, 50,110, 100, 130, 170}, x = 110;
Output: 6
Explanation: Element x is present at index 6

Input: arr[] = {10, 20, 80, 30, 60, 50,110, 100, 130, 170}, x = 175;
Output: -1
Explanation: Element x is not present in arr[].

Approach:

Iterate from 0 to N-1 and compare the value of every index with key if they match return index, if the loop terminates then return -1.

Follow the below steps to Implement the above approach:  

  • Start from the leftmost element of arr[] and one by one compare x with each element of arr[]
  • If x matches with an element, return the index.
  • If x doesn’t match with any of the elements, return -1.

Below is the implementation of the above approach:

C++




// C++ code to linearly search x in arr[]. If x
// is present then return its location, otherwise
// return -1
 
#include <iostream>
using namespace std;
 
int search(int arr[], int n, int x)
{
    int i;
    for (i = 0; i < n; i++)
        if (arr[i] == x)
            return i;
    return -1;
}
 
// Driver code
int main(void)
{
    int arr[] = { 2, 3, 4, 10, 40 };
    int x = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
   
    // Function call
    int result = search(arr, n, x);
    (result == -1)
        ? cout << "Element is not present in array"
        : cout << "Element is present at index " << result;
    return 0;
}


C




// C code to linearly search x in arr[]. If x
// is present then return its location, otherwise
// return -1
 
#include <stdio.h>
 
int search(int arr[], int n, int x)
{
    int i;
    for (i = 0; i < n; i++)
        if (arr[i] == x)
            return i;
    return -1;
}
 
// Driver code
int main(void)
{
    int arr[] = { 2, 3, 4, 10, 40 };
    int x = 10;
    int n = sizeof(arr) / sizeof(arr[0]);
   
    // Function call
    int result = search(arr, n, x);
    (result == -1)
        ? printf("Element is not present in array")
        : printf("Element is present at index %d", result);
    return 0;
}


Java




// Java code for linearly searching x in arr[]. If x
// is present then return its location, otherwise
// return -1
 
class GFG
{
    public static int search(int arr[], int x)
    {
        int n = arr.length;
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == x)
                return i;
        }
        return -1;
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 2, 3, 4, 10, 40 };
        int x = 10;
 
        // Function call
        int result = search(arr, x);
        if (result == -1)
            System.out.print(
                "Element is not present in array");
        else
            System.out.print("Element is present at index "
                             + result);
    }
}


Python3




# Python3 code to linearly search x in arr[].
# If x is present then return its location,
# otherwise return -1
 
 
def search(arr, n, x):
 
    for i in range(0, n):
        if (arr[i] == x):
            return i
    return -1
 
 
# Driver Code
arr = [2, 3, 4, 10, 40]
x = 10
n = len(arr)
 
# Function call
result = search(arr, n, x)
if(result == -1):
    print("Element is not present in array")
else:
    print("Element is present at index", result)


C#




// C# code to linearly search x in arr[]. If x
// is present then return its location, otherwise
// return -1
using System;
 
class GFG {
    public static int search(int[] arr, int x)
    {
        int n = arr.Length;
        for (int i = 0; i < n; i++)
        {
            if (arr[i] == x)
                return i;
        }
        return -1;
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 2, 3, 4, 10, 40 };
        int x = 10;
 
        // Function call
        int result = search(arr, x);
        if (result == -1)
            Console.WriteLine(
                "Element is not present in array");
        else
            Console.WriteLine("Element is present at index "
                              + result);
    }
}
 
// This code is contributed by DrRoot_


PHP




<?php
// PHP code for linearly search x in arr[].
// If x is present then return its location,
// otherwise return -1
 
function search($arr, $x)
{
    $n = sizeof($arr);
    for($i = 0; $i < $n; $i++)
    {
        if($arr[$i] == $x)
            return $i;
    }
    return -1;
}
 
// Driver Code
$arr = array(2, 3, 4, 10, 40);
$x = 10;
 
// Function call
$result = search($arr, $x);
if($result == -1)
    echo "Element is not present in array";
else
    echo "Element is present at index " ,
                                 $result;
 
// This code is contributed
// by jit_t
?>


Javascript




<script>
 
// Javascript code to linearly search x in arr[]. If x
// is present then return its location, otherwise
// return -1
 
function search(arr, n, x)
{
    let i;
    for (i = 0; i < n; i++)
        if (arr[i] == x)
            return i;
    return -1;
}
 
// Driver code
 
    let arr = [ 2, 3, 4, 10, 40 ];
    let x = 10;
    let n = arr.length;
 
    // Function call
    let result = search(arr, n, x);
    (result == -1)
        ? document.write("Element is not present in array")
        : document.write("Element is present at index " + result);
 
// This code is contributed by Manoj
 
</script>


Output

Element is present at index 3

Time complexity: O(N)
Auxiliary Space: O(1)

Linear search is rarely used practically because other search algorithms such as the binary search algorithm and hash tables allow significantly faster searching compared to Linear search.

Improve Linear Search Worst-Case Complexity:

  1. If element Found at last O(N) to O(1).
  2. It is the same as the previous method because here we are performing two steps ‘if’ operations in one iteration of the loop and in the last method we performed only 1 ‘if’ operation. This makes both the time complexities the same.

Below is the implementation of the above approach:

C++14




// C++ program for linear search
#include<bits/stdc++.h>
using namespace std;
 
void search(vector<int> arr, int search_Element)
{
    int left = 0;
    int length = arr.size();
    int position = -1;
      int right = length - 1;
       
    // Run loop from 0 to right
    for(left = 0; left <= right;)
    {
         
        // If search_element is found with
        // left variable
        if (arr[left] == search_Element)
        {
             
            position = left;
            cout << "Element found in Array at "
                 << position + 1 << " Position with "
                 << left + 1 << " Attempt";
                
            break;
        }
       
        // If search_element is found with
        // right variable
        if (arr[right] == search_Element)
        {
            position = right;
            cout << "Element found in Array at "
                 << position + 1 << " Position with "
                 << length - right << " Attempt";
                
            break;
        }
        left++;
        right--;
    }
 
    // If element not found
    if (position == -1)
        cout << "Not found in Array with "
             << left << " Attempt";
}
 
// Driver code
int main()
{
    vector<int> arr{ 1, 2, 3, 4, 5 };
    int search_element = 5;
     
    // Function call
    search(arr, search_element);
}
     
// This code is contributed by mayanktyagi1709


Java




// Java program for linear search
 
import java.io.*;
 
class GFG
{
 
    public static void search(int arr[], int search_Element)
    {
        int left = 0;
        int length = arr.length;
        int right = length - 1;
        int position = -1;
 
        // run loop from 0 to right
        for (left = 0; left <= right;)
        {
             
            // if search_element is found with left variable
            if (arr[left] == search_Element)
            {
                position = left;
                System.out.println(
                    "Element found in Array at "
                    + (position + 1) + " Position with "
                    + (left + 1) + " Attempt");
                break;
            }
           
            // if search_element is found with right variable
            if (arr[right] == search_Element)
            {
                position = right;
                System.out.println(
                    "Element found in Array at "
                    + (position + 1) + " Position with "
                    + (length - right) + " Attempt");
                break;
            }
             
            left++;
            right--;
        }
 
        // if element not found
        if (position == -1)
            System.out.println("Not found in Array with "
                               + left + " Attempt");
    }
   
    
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 2, 3, 4, 5 };
        int search_element = 5;
       
        // Function call
        search(arr,search_element);
    }
}


Python3




# Python3 program for linear search
def search(arr, search_Element):
    left = 0
    length = len(arr)
    position = -1
    right = length - 1
 
    # Run loop from 0 to right
    for left in range(0, right, 1):
 
        # If search_element is found with
        # left variable
        if (arr[left] == search_Element):
            position = left
            print("Element found in Array at ", position +
                  1, " Position with ", left + 1, " Attempt")
            break
 
        # If search_element is found with
        # right variable
        if (arr[right] == search_Element):
            position = right
            print("Element found in Array at ", position + 1,
                  " Position with ", length - right, " Attempt")
            break
        left += 1
        right -= 1
 
    # If element not found
    if (position == -1):
        print("Not found in Array with ", left, " Attempt")
 
# Driver code
arr = [1, 2, 3, 4, 5]
search_element = 5
 
# Function call
search(arr, search_element)
 
# This code is contributed by Dharanendra L V.


C#




// C# program for linear search
using System;
class GFG
{
 
    public static void search(int []arr,
                              int search_Element)
    {
        int left = 0;
        int length = arr.Length;
        int right = length - 1;
        int position = -1;
 
        // run loop from 0 to right
        for (left = 0; left <= right;)
        {
             
            // if search_element is found with left variable
            if (arr[left] == search_Element)
            {
                position = left;
                Console.WriteLine(
                    "Element found in Array at "
                    + (position + 1) + " Position with "
                    + (left + 1) + " Attempt");
                break;
            }
           
            // if search_element is found with right variable
            if (arr[right] == search_Element)
            {
                position = right;
                Console.WriteLine(
                    "Element found in Array at "
                    + (position + 1) + " Position with "
                    + (length - right) + " Attempt");
                break;
            }
             
            left++;
            right--;
        }
 
        // if element not found
        if (position == -1)
            Console.WriteLine("Not found in Array with "
                               + left + " Attempt");
    }
   
    
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = { 1, 2, 3, 4, 5 };
        int search_element = 5;
       
        // Function call
        search(arr,search_element);
    }
}
 
// This code is contributed by 29AjayKumar


Javascript




<script>
     
// Javascript program for linear search
 
function search(arr, search_Element)
{
    let left = 0;
    let length = arr.length;
    let right = length - 1;
    let position = -1;
 
    // Run loop from 0 to right
    for(left = 0; left <= right;)
    {
         
        // If search_element is found
        // with left variable
        if (arr[left] == search_Element)
        {
            position = left;
            document.write(
                "Element found in Array at " +
                (position + 1) + " Position with " +
                    (left + 1) + " Attempt");
            break;
        }
        
        // If search_element is found
        // with right variable
        if (arr[right] == search_Element)
        {
            position = right;
            document.write(
                "Element found in Array at " +
                (position + 1) + " Position with " +
                (length - right) + " Attempt");
            break;
        }
        left++;
        right--;
    }
 
    // If element not found
    if (position == -1)
        document.write("Not found in Array with " +
                       left + " Attempt");
}
      
// Driver code
let arr = [ 1, 2, 3, 4, 5 ];
let search_element = 5;
 
// Function call
search(arr, search_element);
 
// This code is contributed by code_hunt
 
</script>


Output

Element found in Array at 5 Position with 1 Attempt

Time Complexity: O(n)
Auxiliary Space: O(1)

Another method: Recursive Approach For Linear Search

Approach: 

  • If the size of the array is zero then, return -1 representation that the element is not found. This can also be treated as the base condition of a recursion call).
  • Otherwise, check if the element at the current index in the array is equals to the key or not i.e, arr[size – 1] == key.
    • =If equal, then return the index of the found key.

Below is the implementation of the above approach:

C++14




// Recursive Code For Linear Search
#include <iostream>
using namespace std;
int linearsearch(int arr[], int size, int key)
{
    if (size == 0) {
        return -1;
    }
    if (arr[size - 1] == key) {
        // Return the index of found key.
        return size - 1;
    }
    else {
        int ans = linearsearch(arr, size - 1, key);
        return ans;
    }
}
 
// Driver Code
int main()
{
    int arr[5] = { 5, 15, 6, 9, 4 };
    int key = 4;
    int ans = linearsearch(arr, 5, key);
    if (ans == -1) {
        cout << "The element " << key << " is not found."
             << endl;
    }
    else {
        cout << "The element " << key << " is found at "
             << ans << " index of the given array." << endl;
    }
    return 0;
}
// Code contributed by pragatikohli


Python3




"""Python Program to Implement Linear Search Recursively"""
 
 
def linear_search(arr, key, size):
        # If the array is empty we will return -1
 
    if size == 0:
        return -1
 
    # Otherwise if the array consists of only one element and that element is not the one
    # we are searching for then it will also return  -1
 
    elif size == 1 and arr[0] != key:
        return -1
 
    # ELse , if the element at the size index is same as the element we are searching for
    # Then return the size. This will return the index position is 0 index manner.
    # i.e if the element is present at 6th position it will return 5.
    # To get the exact position in human readble format (counting starts from 1 not 0)
    # Then just return size + 1
 
    elif arr[size] == key:
        return size
 
    # If none of the conditions are True then in else condition we will call the
    # function recursively by decreasing the size by 1 each time.
 
    else:
        return linear_search(arr, key, size-1)
 
 
arr = [5, 15, 6, 9, 4]
key = 4
size = len(arr)-1
 
# Calling the Function
print("The element ", key, " is found at index: ",
      linear_search(arr, key, size), " of given array")
 
 
# Code Contributed By - DwaipayanBandyopadhyay


Javascript




// JavaScript Recursive Code For Linear Search
 
let linearsearch = (arr, size, key) => {
  if (size == 0) {
    return -1;
  }
  if (arr[size - 1] == key)
  {
   
    // Return the index of found key.
    return size - 1;
  }
  else
  {
    let ans = linearsearch(arr, size - 1, key);
    return ans;
  }
};
 
// Driver Code
let main = () => {
  let arr = [5, 15, 6, 9, 4];
  let key = 4;
  let ans = linearsearch(arr, 5, key);
  if (ans == -1) {
    console.log(`The element ${key} is not found.`);
  } else {
    console.log(
      `The element ${key} is found at ${ans} index of the given array.`
    );
  }
  return 0;
};
 
main();
 
// This code is contributed by Aman Singla...


Output

The element 4 is found at 4 index of the given array.

Time Complexity: O(n)
Auxiliary Space: O(1)

Also See – Binary Search

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
 


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