Page 339 - ComputerScience_Class_11

Iteration 2:
Index 0 1 2 3 4

Ar 16 12 10 21 2
Step 1: There is no interchange in the first three numbers as they are already in descending order. Thus, the third and
fourth elements are checked.

Index 0 1 2 3 4
Ar 16 12 10 21 2

Since, ar[2] < ar[3], i.e., 10 < 21 is true; the numbers are swapped.
Swapping the numbers in the position, the array becomes:

Index 0 1 2 3 4
Ar 16 12 21 10 2

Step 2: The next two elements are compared.

Index 0 1 2 3 4
Ar 16 12 21 10 2
Since, ar[3] < ar[4], i.e., 10 < 2 is false; there will be no interchange.

With this, the second smallest number is placed in the second last position.

Iteration 3:
Index 0 1 2 3 4
Ar 16 12 21 10 2

Step 1: There is no interchange in the first two numbers as they are already in descending order. Thus, the second and
third elements are compared.

Index 0 1 2 3 4
Ar 16 12 21 10 2

Since, ar[1] < ar[2], i.e., 12 < 21 is true; the numbers are swapped.
Swapping the numbers in the position, the array becomes:

Index 0 1 2 3 4
Ar 16 21 12 10 2

Step 2:
Ar 16 21 12 10 2

There is no interchange in the last three numbers as they are already in descending order.

Iteration 4:
Step 1: The first two elements are compared.

Index 0 1 2 3 4
Ar 16 21 12 10 2
Since, ar[0] < ar[1] , i.e., 16 < 21 is true; the numbers are swapped.

Arrays 337