Page 300 - Cs_withBlue_J_C11_Flipbook
P. 300
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: 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: 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.
Swapping the numbers in the position, the array becomes:
index 0 1 2 3 4
Ar 21 16 12 10 2
298298 Touchpad Computer Science-XI

