Linear Search

Problem: Given an array arr[] of n elements, 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
Element x is present at index 6

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

A simple approach is to do a linear search, i.e  

  • 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 elements, return -1.
     

Example: 



C++

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// 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;
}

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C

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// 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;
}

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Java

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// 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);
    }
}

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Python3

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# 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)

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C#

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// 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_

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PHP

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<?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
?>

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Output

Element is present at index 3

The time complexity of the above algorithm is O(n). 

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

Improve Linear Search Worst-Case Complexity

  1. if element Found at last  O(n) to O(1)
  2. if element Not found O(n) to O(n/2)

Below is the implementation:

Java

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// 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 varaible
            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 varaible
            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);
    }
}

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Output

Element found in Array at 5 Position with 1 Attempt


Also See – Binary Search

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