Searching in One dimention array

This algorithm is used to search an element in a one dimention array.

CC++Python

    #include <stdio.h>
    int main()
    {
        int arr[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
        int i, n, flag = 0;
        printf("Enter the number to be searched: ");
        scanf("%d", &n);
        for (i = 0; i < 10; i++)
        {
            if (arr[i] == n)
            {
                flag = 1;
                break;
            }
        }
        if (flag == 1)
        {
            printf("The number is found at position %d", i + 1);
        }
        else
        {
            printf("The number is not found");
        }
        return 0;
    }

    #include <stdio.h>
    int main()
    {
        int arr[10] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
        int i, n, flag = 0;
        cout<<"Enter the number to be searched: ";
        cin>>n;
        for (i = 0; i < 10; i++)
        {
            if (arr[i] == n)
            {
                flag = 1;
                break;
            }
        }
        if (flag == 1)
        {
            cout<<"The number is found at position <<   i + 1";
        }
        else
        {
            cout<<"The number is not found";
        }
        return 0;
    }

    arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
    n = int(input("Enter the number to be searched: "))
    flag = 0
    for i in range(10):
        if arr[i] == n:
            flag = 1
            break
    if flag == 1:
        print("The number is found at position ", i + 1)
    else:
        print("The number is not found")

Output

Enter the number to be searched: 5
The number is found at position 5

Time Complexity

The time complexity of this algorithm is O(n).