Linear Search Algorithm (with Code)

# Linear Search Algorithm (with Code)

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Updated on Dec 13, 2023 12:26 IST

Have you ever wondered how to find a specific item in a list without sorting it? The linear search algorithm does just that, sequentially checking each element of the list until it finds the target item or reaches the end of the list. This simple yet effective method is widely used when data is unsorted or constantly changing. Let us understand more!

The linear search algorithm is a simple and straightforward method used to find a particular element in a list. It works by sequentially checking each element of the list until the desired element is found or the end of the list is reached.

## What is Linear Search?

Linear search is the simplest type of searching. It is a sequential searching algorithm where we start from one end and check every element of the list until the searched element is found.

## Implementation code in C, C++, Python, and Java

Here, we are going to implement Linear Search in C, C++, Python, and Java using two approaches.

## Simple Approach to Implement Linear Search

• Time Complexity: O(n)
• Space Complexity: O(1)

### C Program

```#include <stdio.h> // Function to perform linear searchint search(int array[], int n, int x) { for (int i = 0; i < n; i++) { if (array[i] == x) { return i; // Return the index where the element is found } } return -1; // Element not found} // Main functionint main() { int array[] = {2, 4, 0, 8, 6, 10, 23, 1, 9}; int x = 23; int n = sizeof(array) / sizeof(array[0]); // Calculate the number of elements in the array int result = search(array, n, x); // Perform the search if (result == -1) { printf("Element not found"); } else { printf("Element found at index: %d", result); } return 0;}Copy code```

Output

Element found at index: 6

### C++ Program

```#include <iostream> using namespace std; // Function to perform linear searchint search(int array[], int n, int x) { // Going through array sequentially for (int i = 0; i < n; i++) { if (array[i] == x) { return i; // Return the index where the element is found } } return -1; // Element not found} // Main functionint main() { int array[] = {2, 4, 7, 8, 0, 1, 9}; int x = 8; int n = sizeof(array) / sizeof(array[0]); // Calculate the number of elements in the array int result = search(array, n, x); // Perform the search if (result == -1) { cout << "Element not found"; } else { cout << "Element found at index: " << result; } return 0;}Copy code```

Output

Element found at index: 3

### Python Program

```# Linear Search in Pythondef linearSearch(array1, n, x): # Going through array sequentially for i in range(0, n): if array1[i] == x: return i return -1 # Test array and search elementarray = [2, 4, 5, 6, 8, 0, 1, 9]x = 1n = len(array) # Perform the linear searchres = linearSearch(array, n, x) # Output the resultif res == -1: print("Element not found")else: print("Element found at index: ", res)Copy code```

Output

Element found at index:  6

### Java Program

```class LinearSearch { public static int linearSearch(int[] array, int x) { int n = array.length; for (int i = 0; i < n; i++) { if (array[i] == x) { return i; // Return the index where the element is found } } return -1; // Element not found } public static void main(String args[]) { int[] array = {2, 4, 5, 8, 0, 1, 9}; int x = 8; int result = linearSearch(array, x); if (result == -1) { System.out.print("Element not found"); } else { System.out.print("Element found at index: " + result); } }}Copy code```

Output

Element found at index:  6

## Improved Approach to Implementing Linear Search

Improve Linear Search Worst-Case Complexity – where the search_element is at the end of the array.

### C Program

```#include <stdio.h> void search(int arr[], int length, int Element) { int left = 0; 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] == Element) { position = left; printf("Element found in Array at %d Position with %d attempts\n", position + 1, left + 1); break; } // If search_element is found with right variable if (arr[right] == Element) { position = right; printf("Element found in Array at %d Position with %d attempts\n", position + 1, length - right); break; } left++; right--; } if (position == -1) printf("Not found element in Array\n");} int main() { int arr[] = {2, 4, 7, 8, 0, 1, 9}; int element = 1; int length = sizeof(arr) / sizeof(arr[0]); // Correct way to get the number of elements in an array search(arr, length, element);}Copy code```

Output

Element found in Array at 6 Position with 2 attempts

### C++ Program

```#include <iostream> using namespace std; void search(int arr[], int length, int Element) { int left = 0; 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] == Element) { position = left; cout << "Element found in Array at " << position + 1 << " Position with " << left + 1 << " Attempt" << endl; break; } // If search_element is found with right variable if (arr[right] == Element) { position = right; cout << "Element found in Array at " << position + 1 << " Position with " << length - right << " Attempt" << endl; break; } left++; right--; } if (position == -1) cout << "Not found element in Array" << endl;} int main() { int arr[] = {2, 4, 7, 8, 0, 1, 9}; int element = 0; int length = sizeof(arr) / sizeof(arr[0]); // Correctly calculate the length search(arr, length, element);}Copy code```

Output

Element found in Array at 5 Position with 3 Attempt

### Python Program

```def search(arr, Element): left = 0 length = len(arr) position = -1 right = length - 1 # Run loop from 0 to right while left <= right: # If element is found with left variable if arr[left] == Element: position = left print("Element found in Array at", position + 1, "Position with", left + 1, "Attempt") break # If element is found with right variable if arr[right] == Element: position = right print("Element found in Array at", position + 1, "Position with", length - right, "Attempt") break left += 1 right -= 1 # If element is not found if position == -1: print("Not found in Array with", left, "Attempt") # Driver codearr = [1, 2, 3, 4, 5, 7, 8]element = 5 # Function callsearch(arr, element)Copy code```

Output

Element found in Array at 5 Position with 3 Attempt

### Java Program

```import java.io.*; class Improved_LS { public static void search(int arr[], int 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 Element is found with left variable if (arr[left] == Element) { position = left; System.out.println("Element found in Array at " + (position + 1) + " Position with " + (left + 1) + " Attempt"); break; } // If Element is found with right variable if (arr[right] == Element) { position = right; System.out.println("Element found in Array at " + (position + 1) + " Position with " + (length - right) + " Attempt"); break; } left++; right--; } if (position == -1) System.out.println("Not found in Array with " + left + " Attempt"); } public static void main(String[] args) { int arr[] = {1, 2, 3, 4, 5, 8, 9}; int element = 5; search(arr, element); }}Copy code```

Output

Element found in Array at 5 Position with 3 Attempt

## Which type of search is better, Linear Search or Binary Search?

Linear search can be used on single and multidimensional arrays, whereas binary search can only be implemented on the one-dimensional array. Linear search is less efficient when we consider large data sets. Binary search is more efficient than linear search in the case of large data sets.

However, actually, both types of searches have their own merits and demerits. The answer to this question depends on the dataset entirely. If we’re searching for an element present at the extreme ends of the list, then a Linear search will run better than a binary search.

## Conclusion

Hope the above article helped you understand linear search better. If you have any queries feel free to reach us at the link below. For more such articles, stay tuned to Shiksha Online.