<span>Validation of electronic signatures was designed to encourage a paperless society.</span>
Answer:
- public class FindDuplicate{
-
- public static void main(String[] args) {
- Scanner input = new Scanner(System.in);
-
- int n = 5;
- int arr[] = new int[n];
-
- for(int i=0; i < arr.length; i++){
- int inputNum = input.nextInt();
- if(inputNum >=1 && inputNum <=n) {
- arr[i] = inputNum;
- }
- }
-
- for(int j =0; j < arr.length; j++){
- for(int k = 0; k < arr.length; k++){
- if(j == k){
- continue;
- }else{
- if(arr[j] == arr[k]){
- System.out.println("True");
- return;
- }
- }
- }
- }
- System.out.println("False");
- }
- }
Explanation:
Firstly, create a Scanner object to get user input (Line 4).
Next, create an array with n-size (Line 7) and then create a for-loop to get user repeatedly enter an integer and assign the input value to the array (Line 9 - 14).
Next, create a double layer for-loop to check the each element in the array against the other elements to see if there is any duplication detected and display "True" (Line 21 - 22). If duplication is found the program will display True and terminate the whole program using return (Line 23). The condition set in Line 18 is to ensure the comparison is not between the same element.
If all the elements in the array are unique the if block (Line 21 - 23) won't run and it will proceed to Line 28 to display message "False".
All I will say is good luck. because it can be very difficult to find someone that is willing to do this.
It's E) both A and B
A) Control logic is invoked
B) Data flow paths are established
