Answer:
def corresponding_of_lists(lst1, lst2):
c = ""
corresponding_list = []
if (lst1 is None) or (lst2 is None):
return None
else:
for i in lst1:
for j in lst2:
c = "(" + str(i) + "," + str(j) + ")"
corresponding_list.append(c)
return corresponding_list
Explanation:
- Create a function called corresponding_of_lists that takes two lists as parameter
- Initialize an empty list to hold the corresponding values
- Check if any of the lists are null, if they are return None
- Otherwise, get the corresponding elements and put them into the corresponding_list
Return the corresponding_list
Answer:
"Web designer" is the appropriate answer.
Explanation:
- If we follow the market pattern, the number of purchases made by online shopping will continue to rise from 2002 to 2018 as well as the times to obtain no indication of slowing or stopping.
- Because of the same, web designers are expected to have the greatest increase in jobs, as it is very important to customize the website so that online customers can appear more interactive.
So that the above is the correct answer.
Answer:
See explaination
Explanation:
MinMax.java
import java.util.*;
public class MinMax
{
static void MinMax(int[] arr)
{
int Min=arr[0]; // initializinf min and max with 1st value of array
int Max=arr[0];
for(int i=0;i<arr.length;i++) // iterating loop only once
{
if(arr[i]>Max) // checking max value
{
Max=arr[i];
}
if(arr[i]<Min) // checking min value
{
Min=arr[i];
}
}
System.out.println("Min Number is "+Min); //printing min value
System.out.println("Max Number is "+Max); //printing max value
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
System.out.print("Enter N value: "); // taking n value
int n=sc.nextInt();
int[] arr=new int[n];
System.out.println("Enter N elements:"); // taking n elements into array
for(int i=0;i<n;i++)
{
arr[i]=sc.nextInt(); // each element into the array
}
MinMax(arr); // calling MinMax() method.
}
}
