Linear Search - GeeksforGeeks

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[].

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A simple approach is to do 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++

// 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; 
} 
  
int main(void) 
{ 
    int arr[] = { 2, 3, 4, 10, 40 }; 
    int x = 10; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    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; 
} 
  
int main(void) 
{ 
    int arr[] = { 2, 3, 4, 10, 40 }; 
    int x = 10; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    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; 
} 
  
public static void main(String args[]) 
{ 
    int arr[] = { 2, 3, 4, 10, 40 };  
    int x = 10; 
      
    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); 
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; 
    } 
      
    public static void Main() 
    { 
        int[] arr = { 2, 3, 4, 10, 40 };  
        int x = 10; 
          
        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; 
      
$result = search($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 
?> 

Output:

Element is present at index 3

The time complexity of 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.

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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My Personal Notes

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

C

Java

Python3

C#

PHP

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