We are given an array in which we need to find a given element and return its index if present and -1, if not present.
So, we just start from the first index and traverse the whole array up to the (n-1)th element and compare each element with the given number.
If at some index they both match, just return the index.
If we reach the end of the array and no element from the array matches with the given number, return -1.

Code

#include <iostream>
using namespace std;

int LinearSearch(int arr[], int n, int element){
    
    for(int i=0;i<n;i++){
        
        if(arr[i] == element){
            
            // Return index, if the given element 
            // matches with any element of array.
            return i;
        }
    }
    
    // If the given number not found.
    return -1;
    
}
int main() {
	
	// Let size of array be 5 and element 
	// to be searched for be 7.
	int n = 5, element = 7;
	int arr[n] = {1,3,5,7,8};
	cout<<LinearSearch(arr,n,element);
	
	return 0;
}
public class Main {
    public static int linearSearch(int[] arr, int n, int element) {
        for (int i = 0; i < n; i++) {
            // Check if current element matches the target
            if (arr[i] == element) {
                return i; // Return the index
            }
        }
        return -1; // If not found
    }

    public static void main(String[] args) {
        int n = 5;
        int element = 7;
        int[] arr = {1, 3, 5, 7, 8};

        // Call the fu