# Array Sort (Ascending and Descending Order) In Java – 2022

Usually we think that, Ascending and Descending order of array will be very easy and normally we neglect to refresh this program before we going interview. Even though concepts are same Interviewer may ask same program in different way. so we need to be more careful.

One of the most frequently asked question in java interview on java program is arrange the given numbers in sorting order either ascending or descending order. Below programs clearly explains about you need.

## 1. Arrange Numbers In Ascending Order

### Program logic,

• Get the total number of element count.
• Get input data and store it in array.
• Compare each element in array one by one. if 1st element is grater than second element, store 1st element in temp variable and push 2nd element to 1st element position. Then move temp value to 2nd element position.
• Repeat the same for all elements.
```import java.util.Scanner;

public class SortingNumber {

public static void main(String[] args) {
//Create Object for Scanner class to get input from keyword
Scanner scn = new Scanner(System.in);

System.out.print("Enter number of element count: ");
//Set array size
int inputValue = scn.nextInt(), temp;

//Get input from keyword
System.out.print("Enter array values\n");
int[] arr = new int[inputValue];

//Get all input from keyword
for(int i =0; i<inputValue;i++){
arr[i]=scn.nextInt();
}

//Validate the element for ascending sort order
for(int m=0;m<arr.length;m++){
for(int n=m+1;n<arr.length;n++){
if(arr[m]>arr[n]){
temp = arr[m];
arr[m]=arr[n];
arr[n]=temp;
}
}
}

//Display the sorted order
System.out.print("Sorted order are: \n");
for(int t: arr){
System.out.println(t);
}
}
}
```
```Output:
Enter number of element count: 10
Enter array values
22
55
66
99
44
66
11
33
77
55
Sorted order are:
11
22
33
44
55
55
66
66
77
99
```

## 2. Arrange Numbers In Descending Order

### Program logic,

• Get the total number of element count.
• Get input data and store it in array.
• Compare each element in array one by one. if 1st element is less than second element, store 1st element in temp variable and push 2nd element to 1st element position. Then move temp value to 2nd element position. (for ascending order)
• Repeat the same for all elements.
```import java.util.Scanner;

public class SortingNumber {

public static void main(String[] args) {
//Create Object for Scanner class to get input from keyword
Scanner scn = new Scanner(System.in);

System.out.print("Enter number of element count: ");
//Set array size
int inputValue = scn.nextInt(), temp;

//Get input from keyword
System.out.print("Enter array values\n");
int[] arr = new int[inputValue];

//Get all input from keyword
for(int i =0; i<inputValue;i++){
arr[i]=scn.nextInt();
}

//Validate the element for ascending sort order
for(int m=0;m<arr.length;m++){
for(int n=m+1;n<arr.length;n++){
if(arr[m]<arr[n]){
temp = arr[m];
arr[m]=arr[n];
arr[n]=temp;
}
}
}

//Display the sorted order
System.out.print("Sorted order are: \n");
for(int t: arr){
System.out.println(t);
}
}
}

```
```Output:
Enter number of element count: 10
Enter array values
11 55 99 44 22 33 77 88 66 44
Sorted order are:
99
88
77
66
55
44
44
33
22
11
```

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