4. Only (ii)
<u>Explanation:</u>
The declaration of the array can be of two types:
1. int a[100];
2. int[] a = new int[100];
The general thing about an array is that whenever we want to undergo traversal in an array, we always have to start from the 0th position as the size of the array may be a whole number (let us say 10). So, to undergo traversal in an array, we start from 0 to n-1 (in this case 9) such that it covers the size of the array.
The size of an array can be finite or infinite. The general rule is it starts from 0 to n-1 where n is the size of the array. In the above example, the range of the index of the array will be 0 through 100 and not 1 through 100.
Explanation:
Internal gear wear must be significant (visually obvious) to be the cause of off-center sharpening. Cutter carrier is rotating but the pencil is not sharpening (doesn't feel like the cutter is engaging the pencil) This is usually caused by a foreign object (e.g., an eraser or broken pencil lead) inside the pencil bore.
Answer:
If all the character pairs match after processing both strings, one string in stack and the other in queue, then this means one string is the reverse of the other.
Explanation:
Lets take an example of two strings abc and cba which are reverse of each other.
string1 = abc
string2 = cba
Now push the characters of string1 in stack. Stack is a LIFO (last in first out) data structure which means the character pushed in the last in stack is popped first.
Push abc each character on a stack in the following order.
c
b
a
Now add each character of string2 in queue. Queue is a FIFO (first in first out) data structure which means the character inserted first is removed first.
Insert cba each character on a stack in the following order.
a b c
First c is added to queue then b and then a.
Now lets pop one character from the stack and remove one character from queue and compare each pair of characters of both the strings to each other.
First from stack c is popped as per LIFO and c is removed from queue as per FIFO. Then these two characters are compared. They both match
c=c. Next b is popped from stack and b is removed from queue and these characters match too. At the end a is popped from the stack and a is removed from queue and they both are compared. They too match which shows that string1 and string2 which are reverse of each other are matched.
