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++
// 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
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
- if element Found at last O(n) to O(1)
- It is the same as previous method because here we are performing 2 ‘if’ operations in one iteration of the loop and in previous method we performed only 1 ‘if’ operation. This makes both the time complexities same.
Below is the implementation:
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 mayanktyagi1709Java
// 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
Also See – Binary Search
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