Linear Search
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 6Input: 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> |
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:
- If element Found at last O(N) to O(1).
- 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> |
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... |
The element 4 is found at 4 index of the given array.
Time Complexity: O(n)
Auxiliary Space: O(1)
Also See – Binary Search
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