Answer Posted / abalonesoft

Preventing unauthorized access to some piece of information
or functionality.

The key money-saving insight is to separate the volatile
part of some chunk of software from the stable part.
Encapsulation puts a firewall around the chunk, which
prevents other chunks from accessing the volatile parts;
other chunks can only access the stable parts. This prevents
the other chunks from breaking if (when!) the volatile parts
are changed. In context of OO software, a "chunk" is
normally a class or a tight group of classes.

The "volatile parts" are the implementation details. If the
chunk is a single class, the volatile part is normally
encapsulated using the private and/or protected keywords. If
the chunk is a tight group of classes, encapsulation can be
used to deny access to entire classes in that group.
Inheritance can also be used as a form of encapsulation.

The "stable parts" are the interfaces. A good interface
provides a simplified view in the vocabulary of a user, and
is designed from the outside-in (here a "user" means another
developer, not the end-user who buys the completed
application). If the chunk is a single class, the interface
is simply the class's public member functions and friend
functions. If the chunk is a tight group of classes, the
interface can include several of the classes in the chunk.

Designing a clean interface and separating that interface
from its implementation merely allows users to use the
interface. But encapsulating (putting "in a capsule") the
implementation forces users to use the interface.

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You have given 2 array. You need to find whether they will create the same BST or not. For example: Array1:10 5 20 15 30 Array2:10 20 15 30 5 Result: True Array1:10 5 20 15 30 Array2:10 15 20 30 5 Result: False One Approach is Pretty Clear by creating BST O(nlogn) then checking two tree for identical O(N) overall O(nlogn) ..we need there exist O(N) Time & O(1) Space also without extra space .Algorithm ?? DevoCoder guest Posted 3 months ago # #define true 1 #define false 0 int check(int a1[],int a2[],int n1,int n2) { int i; //n1 size of array a1[] and n2 size of a2[] if(n1!=n2) return false; //n1 and n2 must be same for(i=0;ia1[i+1]) && (a2[i]>a2[i+1]) ) ) return false; } return true;//assumed that each array doesn't contain duplicate elements in themshelves }

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