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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 codeint 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 codeint 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 -1class 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 -1def search(arr, n, x): for i in range(0, n): if (arr[i] == x): return i return -1# Driver Codearr = [2, 3, 4, 10, 40]x = 10n = len(arr)# Function callresult = 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 -1using 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 -1function 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 -1function 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 codeint 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 searchimport 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 searchdef 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 codearr = [1, 2, 3, 4, 5]search_element = 5# Function callsearch(arr, search_element)# This code is contributed by Dharanendra L V. |
C#
// C# program for linear searchusing 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 searchfunction 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 codelet arr = [ 1, 2, 3, 4, 5 ];let search_element = 5;// Function callsearch(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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