Conversion of Basic Data Type to string in C++ \ VC++

Current post is on how to convert basic data type into string in C++ \ VC++

Have you ever tried of converting basic data types to string in most simple and elegant way!!!

Till now, I used to call 'itoa()' or similar function after which I used to append the char* buffer to string variable.

Lots of temp variable to initalize, worry about charbuff limited size declaration and ugly looking code. I came across this simple and clean function calls which really pleased my eyes :)

Let us take int to string conversion example:

BEFORE:

int myint=0;
char mycharbuf[64]="";
string mystr;

... // Get the int value

itoa(mycharbuff,myint,10);

mystr.append(mycharbuff);

... // Use the string

NOW:

int myint;
string mystr;
stringstream mystream;

... //Get the int value

mystream << myint;

mystr = mystream.str();

... // Use the string

Here you go, with clean code & faster code review of your work.


Appreciate your feedback via comments. Thanks.

Order the array of 0 & 1

Question:

Order the list with ‘0’ first and ‘1’ second, which has ‘0’ and ‘1’ stored in irregular fashion.

Constraints:
-> Time & Space complexity to be minimal

Example:

Array: 0-1-0-1-1-0-1

Answer: 0-0-0-1-1-1-1

Answer:

-> Start with one pointer from start and another pointer from back.
-> Stop if you find ‘1’ for first pointer
-> Stop if you find ‘0’ for second pointer
-> Interchange till both the pointers are not crossed over.

Complexity:

Time: O(2N) -> O(N) , where N represents longest link list length

Example:

Array: 0-1-0-1

Answer: 0-0-1-1

Tracing:

Step 1:

-> Let Pointer 1 points to head of array
-> Let Pointer 2 points to tail of array

Step 2:

-> Pointer 1 will stop at node 2 as it has '1'

Step 3:

-> Pointer 2 will stop at node 3 as it has '0'

Step 4:

-> Interchange the values of node 2 & 3 with 0 & 1 respectively

Step 5:

-> Pointer 1 crosses over Pointer 2. Stop

Answer:

0-0-1-1

Reach i-th position from tail of link list

Question:

To reach i-th position from tail of link-list by traversing only once.

Example:

1-2-3-4-5-6-7-8-9

To reach 3rd node from tail. i.e Node 7

Answer:

-> Take two pointers.

-> Take first pointer to i-th position from head

-> Start incrementing both the pointers

-> When the first pointer reaches the tail, at that point second pointer will be pointing to i-th node from tail

Example:

1-2-3-4-5-6-7-8-9, to reach 3rd node from tail i.e. Node 7.

Tracing:

Step 1:

-> Pointer 1 & Pointer 2 points to node 1

Step 2:

-> Increment Pointer 1 '3' times, as we want 3rd node from tail

Step 3:

-> Increment both the pointers

Movement of Pointer 1: 3-4-5-6-7-8-9
Movement of Pointer 2: 1-2-3-4-5-6-7

Step 4:

Stop when Pointer 1 points to 9, at that time the Pointer 2 points to Node 7

Answer:

Node 7

Test whether two link list have common nodes or not?

Question:

To test whether two link list have common nodes or not?

Hints:
-> There may be more then 1 common nodes
-> The length of both the link list might not be same

Constraints:
-> Time & Space complexity to be minimal


Example 1:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Answer: Common nodes from "Node 6"

Example 2:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-d-e

Answer: No common nodes

Answer:

-> Take two pointers, each pointing to head of each list
-> Take two counter variables for each list to store the length of list

-> Traverse both the list and increment the count on each node traversal
-> Get the length difference by subtracting the counter with higher value to lower one and put to some difference variable

-> Increment the pointer "difference variable" times whose length is greater

-> Increment both the pointers and compare at each iteration, till any node reaches the tail

If they reach tail and both pointers didn't got common value -> No common nodes
Else -> Common nodes from that node

Complexity:

Time: O(2N) -> O(N) , where N represents longest link list length
Space: O(3) -> O(1)

Example 1:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Answer: Common nodes from "Node 6"

Tracing:

See the previous post.

Example 2:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-d-e

Answer: No common nodes

Tracing:

Step 1:

-> Let Pointer 1 points to head of List 1 and counter is Length 1
-> Let Pointer 2 points to head of List 2 and counter is Length 2

Step 2:

-> The Length 1 value will be: 7
-> The Length 2 value will be: 5
-> Difference variable value will be: 2

Step 3:

-> List 1 length is bigger, hence Pointer 1 will be incremented two time to point to Node 2

Step 4:

Movement of Pointer 1: 3-4-5-6-7
Movement of Pointer 2: a-b-c-d-e

We reached the tail, hence no common node

Answer:

No common nodes

Find common meet point node of two link lists?

Question:

There are two link list, which meet at some node X,find the meet point?

Hints:
-> There may be more then 1 common nodes
-> The length of both the link list might not be same

Constraints:
-> Time & Space complexity to be minimal

Example:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Here "Node 6" is the meet point.

Answer:

-> Take two pointers, each pointing to head of each list
-> Take two counter variables for each list to store the length of list

-> Traverse both the list and increment the count on each node traversal
-> Get the length difference by subtracting the counter with higher value to lower one and put to some difference variable

-> Increment the pointer "difference variable" times whose length is greater

-> Increment both the pointers and compare at each iteration, till any node reaches the tail
-> The point at which both the pointers get same value is the common node meet point

Complexity:

Time: O(2N) -> O(N) , where N represents longest link list length
Space: O(3) -> O(1)

Example:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Here "Node 6" is the meet point.

Tracing:

Step 1:

-> Let Pointer 1 points to head of List 1 and counter is Length 1
-> Let Pointer 2 points to head of List 2 and counter is Length 2

Step 2:

-> The Length 1 value will be: 7
-> The Length 2 value will be: 5
-> Difference variable value will be: 2

Step 3:

-> List 1 length is bigger, hence Pointer 1 will be incremented two time to point to Node 2

Step 4:

Movement of Pointer 1: 3-4-5-6
Movement of Pointer 2: a-b-c-6

Node 6 is the common meet point.

Answer:

Node 6

Difference between "Struct" of "C" & "C++" ?

Difference:

C Struct :
-> Cannot have member functions.
-> Cannot exhibit polymorphic behaviour.
-> "Struct MyStruct a". "Struct" Keyword required while declaration.

C++ Struct:
-> Can have member functions.
-> Can exhibit polymorphic behaviour.
-> "MyStruct a". "Struct" Keyword not required while declaration.




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Difference between Class and Struct in C++ ?

Current post is on difference between Class and Structure in C++ language.

Difference:

Struct:
- By default access is public

Class :
- By default access is private

It means when:
- member variable is declared
- member function is declared
- It is derived

and one does not specific the access specifier, it will get defaulted to mentioned access specifier

Structure Key Points:

- Struct can have member functions.
- Struct can be derived.
- Struct can have polymorphic behavior.



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Check whether computer is Little Endian / Big Endian ?

Current post is on how to check whether computer is Big Endian or Little Endian.

It is a common technical interview question to test your knowledge on computer architecture.

Endianness refers - the order in which computer stores the bytes.

Little Endian - Stores LSB (Least Significant Byte) in lowest address.

Big Endian - Stores MSB (Most Significant Byte) in highest address.

Answer:

/* Function returns whether the computer system is Little or Big Endian*/

bool IsLittleEndian()
{
       int TestNumber = 1;
       char * ptr;

       ptr = (char *) &TestNumber;

       return (*ptr);    //True if little Endian.
}

Appreciate your feedback via comments. Thanks.