Friday, December 23, 2011

Part 1: How inheritance, encapsulation and polymorphism work in C++

How inheritance, encapsulation and polymorphism work in C++

Posted on September 18, 2008, 6:15 pm, by Alexander Sandler, under Programming Articles.

Introduction
Encapsulation
How methods work
How overloading works
How mangling solves the problem
Structure and size of the object
How inheritance and polymorphism work
How basic polymorphism works
How multiple inheritance works
Difference between different casting types
Polymorphism and multiple inheritance
What if we try something even more complicated
Few words about C++ constructors
Conclusion

IntroductionBACK TO TOC

Inheritance, encapsulation and polymorphism are undoubtedly the cornerstones of OOP/OOD in general and C++ in particular.
When programming C, it is very easy to remember how things work. You know that when you add an int variable to a structure it mostly grows by four bytes. You know that long is either four or eight bytes long depending on the architecture you’re working with.
Things are less obvious when moving to C++. OOP brings more abstractions to the program. As a result you are no longer sure if a+b sums two numbers or calls some overloaded operator method that concatenates contents of two files together.
In this article, I would like to give you a short insight into what’s going on behind the scenes. In particular we’ll see how the three whales of OOP work in C++.
Things that I am going to show in this article may differ from compiler to compiler. I will talk mostly about g++ (version 4.2.3). Note however, that same ideas apply everywhere.

EncapsulationBACK TO TOC

As you know, encapsulation is a principle by which same entity, the object, encapsulates data and methods that manipulate the data. You may be surprised to find out that underneath, class methods are just plain functions.

How methods workBACK TO TOC

In C++ there’s one fundamental difference between plain functions and class methods. Class methods receive one additional argument and that is the pointer to the object whose data the method is expected to manipulate. I.e. first argument to a method is pointer to this.
To speed things up, C++ developers used single CPU register (ECX/RCX on x86/x86_64) to pass pointer to this, instead of passing it via stack as if it was a regular function argument (no longer true in x86_64).
Otherwise, objects know nothing about methods that operate on them.

How overloading worksBACK TO TOC

Another thing that we have to take care of in C++ is how to distinguish between some_function() and some_class::some_function(). Or between some_class::some_function( int ) and some_class::some_function() I.e. what’s the difference between two methods with the same name that receive different number and type of arguments? What is the difference between method and function that has same name?
Obviously, out of linker, compiler and preprocessor, linker is the one that should be aware of the above difference. This is because we may have some_function() in some distant object file. Linker is the component that should find this distant function and interconnect the call to the function and the actual function. Linker uses function name as a unique identifier of the function.
To make things work, g++ and any other modern compiler, mangles the name of the method/function and makes sure that:

Mangled method name includes name of the class it belongs to (if it belongs to any class).
Mangled method name includes number and type of arguments method receives.
Mangled method name includes namespace it belongs to.

With these three, some_class::some_function() and some_function() will have totally different mangled name. See the following code sample.

01namespace some_namespace
02{
03    class some_class
04    {
05    public:
06        some_class() { }
07        void some_method() { }
08    };
09};
10 
11class some_class
12{
13public:
14    some_class() { }
15    void some_method() { }
16};
17 
18void some_method()
19{
20    int a;
21}

g++ will turn:

void some_class::some_method() into _ZN10some_class11some_methodEv
void some_namespace::some_class::some_method() into _ZN14some_namespace10some_class11some_methodEv
void some_method() into _Z11some_methodv

Adding integer argument to void some_method() will turn it from _Z11some_methodv to _Z11some_methodi.

How mangling solves the problemBACK TO TOC

So when you create two methods with same name, but with different arguments, compiler turns them into two functions with different names. Later, when linker links the code together it doesn’t know that these are two methods of the same class. From linkers standpoint, these are two different functions.

Structure and size of the objectBACK TO TOC

You probably already know that C++ class and good old C structures are nearly the same thing. Perhaps the only difference is that all class members are private unless specified otherwise. On the contrary, all structure members are public.
When looking at the memory layout of the object, it is very similar to C structure.

Differences begin when you add virtual methods. Once you add virtual methods to the class, compiler will create virtual methods table for the class. Then it will place pointer to the table in the beginning of each instance of this class.

Object vs. structure and virtual methods table

So, bear in mind that once your class has virtual methods, each object of this class will be four or eight bytes (depends on whether you have 64-bit support or not) bigger.
Actually, pointer to the virtual methods table does not have to be at the beginning of the object. It is just handy to keep it at the beginning, so g++ and most of the modern compilers do it this way.
Adding virtual methods to the class will also increase amount of RAM your program consumes and its size on your hard drive.

How inheritance and polymorphism workBACK TO TOC

Lets say we have two classes. A and B. Class B inherits from class A.

01#include <iostream>
02 
03using namespace std;
04 
05class A
06{
07public:
08    A() { a_member = 0; }
09    int a_member;
10};
11 
12class B : public A
13{
14public:
15    B() : A() { b_member = 0; };
16    int b_member;
17};
18 
19int main()
20{
21    A *a = new B;
22    a->a_member = 10;
23 
24    return 0;
25}

The interesting thing to notice here is that a actually points to instance of class B. When dereferencing a_member, we’re actually dereferencing a_member that defined in class A, but belongs to class B (via inheritance). To make this happen, compiler has to make sure that common part of both classes (a_member in our case) located at the same offset in the object.

Now what if we have some virtual methods.

How basic polymorphism worksBACK TO TOC

Let’s change our example a bit and add some virtual methods.

01#include <iostream>
02 
03using namespace std;
04 
05class A
06{
07public:
08    A() { a_member = 0; }
09    virtual int reset() { a_member = 0; }
10    void set_a_member( int a ) { a_member = a; }
11    int get_a_member() { return a_member; }
12protected:
13    int a_member;
14};
15 
16class B : public A
17{
18public:
19    B() : A() { b_member = 0; };
20    virtual int reset() { a_member = b_member = 0; }
21    virtual void some_virtual_method() { }
22    void set_b_member(int b ) { b_member = b; }
23    int get_b_member() { return b_member; }
24protected:
25    int b_member;
26};
27 
28int main()
29{
30    B *b = new B;
31    A *a = b;
32 
33    b->set_b_member( 20 );
34    b->set_a_member( 10 );
35 
36    a->reset();
37 
38    cout << b->get_a_member() << " " << b->get_b_member() <<
39        endl;
40 
41    return 0;
42}

If you compile and run this program it will obviously print “0 0″. But how, you may ask. After all we did a->reset(). Without our understanding of polymorphism we could think that we’re calling method that belongs to class A.
The reason it works is because when compiler sees code that dereferences pointer to A it expects certain internal object structure and when it dereferences pointer to B it expects different object structure. Let us take a look at both of them.

Classes A and B with virtual methods tables

However even more important here is the structure of the virtual methods tables of both classes.

It is because of the virtual methods table structure compilers knows what virtual method to call. When it generates the code that dereferences pointer to A, it expects that first method in the virtual methods table of the object will be pointer to right reset() routine. It doesn’t really care if the pointer actually points to B object. It will call first method of the virtual methods table anyway.

How multiple inheritance worksBACK TO TOC

Multiple inheritance makes things much more complicated. The problem is that when class C inherits from both A and B, we should have both members of class A and class B in the instance of class C.

01#include <iostream>
02 
03using namespace std;
04 
05class A
06{
07public:
08    A() { a_member = 0; }
09protected:
10    int a_member;
11};
12 
13class B
14{
15public:
16    B() { b_member = 0; }
17protected:
18    int b_member;
19};
20 
21class C : public A, public B
22{
23public:
24    C() : A(), B() { c_member = 0; }
25protected:
26    int c_member;
27};
28 
29int main()
30{
31    C c;
32 
33    A *a1 = &c;
34    B *b1 = &c;
35 
36    A *a2 = reinterpret_cast<A *>( &c );
37    B *b2 = reinterpret_cast<B *>( &c );
38 
39    printf( "%p %p %p\n", &c, a1, b1 );
40    printf( "%p %p %p\n", &c, a2, b2 );
41 
42    return 0;
43}

Once we cast pointer to class C into class B, we cannot keep the value of the pointer as is because first fields in the object occupied by fields defined in class A (a_member). Therefore, when we do casting we have to do a very special kind of casting – the one that changes the actual value of the pointer.
If you compile and run above code snippet, you will see that all the values are the same except for b1, which should be 4 bytes bigger than other values.

This is what (C style casting in our case) casting does – it increments the value of the pointer to make sure that it points to the beginning of the, inherited from B, part of the object.
In case you wonder what other types of casting will do, here is a short description.

Difference between different casting typesBACK TO TOC

There are five types of casting in C++.

reinterpret_cast<>()
static_cast<>()
dynamic_cast<>()
const_cast<>()
C style cast.

I guess you know already what const_cast<>() does. Also, it is only a compile time casting. C style cast is same as static_cast<>(). This leaves us with three types of casting.

reinterpret_cast<>()
static_cast<>()
dynamic_cast<>()

From the above example we learn that reinterpret_cast<>() does nothing to the pointer value and leaves it as is.
static_cast<>() and dynamic_cast<>() both modify value of the pointer. The difference between two is that the later relies on RTTI to see if the casting is legal – it looks inside the object to see if it truly belongs to the type we’re trying to cast from. static_cast<>() on the other hand, simply increments the value of the pointer.

Polymorphism and multiple inheritanceBACK TO TOC

Things getting even more complicated when we have virtual methods in each one of the classes A, B and C that we already met. Let’s add following virtual methods to the classes.

virtual void set_a( int new_a ) { a_member = new_a; }

To class A.

virtual void set_b( int new_b ) { b_member = new_b; }

To class B and

virtual void set_c( int new_c ) { c_member = new_c; }

To class C.

Dual inheritance with virtual methods table

You could have assumed that even in this case class C objects will have only one virtual tables methods, but this is not true. When you static_cast class C object into class B object, class B object must have its own virtual tables method. If we want to use same casting method as with regular objects (that is adding few bytes to the pointer to reach right portion of the object), then we have no choice but to place another virtual tables method in the middle of the object.

As a result, you can have many different virtual methods tables for the same class. The above diagram shows very simple case of inheritance and the truth is that it does not get more complicated than this. Take a look at the following, more complex, class hierarchy.

It may surprise you, but structure of the class X object will be quiet simple. In our previous example inheritance hierarchy had two branches. This one has three:

A-C-F-X
D-G-X
B-E-H-X

All end up with X of course. They are a little longer than in our previous example, but there is nothing special about them. The structure of the object will be the following:

Structure of derivative of the hierarchy

As a rule of thumb, g++ (and friends) calculates the branches that lead to the target class, class X in our case. Next it creates a virtual methods table for each branch and places all virtual methods from all classes in the branch into virtual methods table. This includes pointer to virtual methods of the class itself.
If we project this rule onto our last example. A-C-F-X branch virtual methods table will include pointers to virtual methods from classes A, C, F and X. Same with other two branches.

When should static_cast, dynamic_cast and reinterpret_cast be used?

Source

static_cast is the first cast you should attempt to use. It does things like implicit conversions between types (such as int to float, or pointer to void*), and it can also call explicit conversion functions (or implicit ones). In many cases, explicitly stating static_cast isn't necessary, but it's important to note that the T(something) syntax is equivalent to (T)something and should be avoided (more on that later). A T(something, something_else) is safe, however, and guaranteed to call the constructor.
static_cast can also cast through inheritance hierarchies. It is unecessary when casting upwards (towards a base class), but when casting downwards it can be used as long as it doesn't cast through virtual inheritance. It does not do checking, however, and it is undefined behavior to static_cast down a hierarchy to a type that isn't actually the type of the object.

const_cast can be used to remove or add const to a variable; no other C++ cast is capable of removing it (not even reinterpret_cast). It is important to note that using it is only undefined if the orginial variable is const; if you use it to take the const of a reference to something that wasn't declared with const, it is safe. This can be useful when overloading member functions based on const, for instance. It can also be used to add const to an object, such as to call a member function overload.
const_cast also works similarly on volatile, though that's less common .

dynamic_cast is almost exclusively used for handling polymorphism. You can cast a pointer or reference to any polymorphic type to any other class type (a polymorphic type has at least one virtual function, declared or inherited). You don't have to use it to cast downwards, you can cast sideways or even up another chain. The dynamic_cast will seek out the desired object and return it if possible. If it can't, it will return NULL in the case of a pointer, or throw std::bad_cast in the case of a reference.
dynamic_cast has some limitations, though. It doesn't work if there are multiple objects of the same type in the inheritance hierarchy (the so-called 'dreaded diamond') and you aren't using virtual inheritance. It also can only go through public inheritance - it will always fail to travel through protected or private inheritance. This is rarely an issue, however, as such forms of inheritance are rare.

reinterpret_cast is the most dangerous cast, and should be used very sparingly. It turns one type directly into another - such as casting the value from one pointer to another, or storing a pointer in an int, or all sorts of other nasty things. Largely, the only guarantee you get with reinterpret_cast is that if you cast the result back to the original type, you will get the same value. Other than that, you're on your own. reinterpret_cast cannot do all sorts of conversions; in fact it is relatively limited. It should almost never be used (even interfacing with C code using void* can be done with static_cast).

C casts are casts using (type)object or type(object). A C-style cast is defined as the first of the following which succeeds:

const_cast
static_cast
static_cast, then const_cast
reinterpret_cast
reinterpret_cast, then const_cast

It can therefore be used as a replacement for other casts in some instances, but can be extremely dangerous because of the ability to devolve into a reinterpret_cast, and the latter should be preferred when explicit casting is needed, unless you are sure static_cast will succeed or reinterpret_cast will fail. Even then, consider the longer, more explicit option.
C-style casts also ignore access control when performing a static_cast, which means that they have the ability to perform an operation that no other cast can. This is mostly a kludge, though, and in my mind is just another reason to avoid C-style casts.
I hope this helps!

Finding all unique triplets that sums to zero

Source

Given a set S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the set which gives the sum of zero.
For example, given set S = {-1 0 1 2 -1 -4},
One possible solution set is:
(-1, 0, 1)
(-1, 2, -1)
Note that (0, 1, -1) is not part of the solution above, because (0, 1, -1) is the duplicate of (-1, 0, 1). Same with (-1, -1, 2).
For a set S, there is probably no “the” solution, another solution could be:
(0, 1, -1)
(2, -1, -1)

This is also known as the 3sum problem. The 3sum problem is the extension of the problem below:

Given a set S of n integers, find all pairs of integers of a and b in S such that a + b = k?

The above problem can be solved in O(n) time, assuming that the set S is already sorted. By using two index first and last, each pointing to the first and last element, we look at the element pointed by first, which we call A. We know that we need to find B = k – A, the complement of A. If the element pointed by last is less than B, we know that the choice is to increment pointer first by one step. Similarly, if the element pointed by last is greater than B, we decrement pointer last by one step. We are progressively refining the sum step by step. Since each step we move a pointer one step, there are at most n steps, which gives the complexity of O(n).
By incorporating the solution above, we can solve the 3sum problem in O(n^2) time, which is a straight forward extension.

set<vector<int> > find_triplets(vector<int> arr) {
  sort(arr.begin(), arr.end());
  set<vector<int> > triplets;
  vector<int> triplet(3);
  int n = arr.size();
  for (int i = 0;i < n; i++) {
    int j = i + 1;
    int k = n - 1;
    while (j < k) {
      int sum_two = arr[i] + arr[j];
      if (sum_two + arr[k] < 0) {
        j++;
      } else if (sum_two + arr[k] > 0) {
        k--;
      } else {
        triplet[0] = arr[i];
        triplet[1] = arr[j];
        triplet[2] = arr[k];
        triplets.insert(triplet);
        j++;
        k--;
      }
    }
  }
  return triplets;
}

Note that a set is chosen to store the triplets, because we are only interested in unique triplets. Since the set S is already sorted, and we don’t look back as it progresses forward, we can guarantee there will be no duplicate triplets (Even though the set might have duplicate elements.)

Friday, December 16, 2011

Maximum Length Bitonic Subarray

Source

Given an array A[0 ... n-1] containing n positive integers, a subarray A[i ... j] is bitonic if there is a k with i <= k <= j such that A[i] <= A[i + 1] ... <= A[k] >= A[k + 1] >= .. A[j - 1] > = A[j]. Write a function that takes an array as argument and returns the length of the maximum length bitonic subarray.
Expected time complexity of the solution is O(n)
Simple Examples
1) A[] = {12, 4, 78, 90, 45, 23}, the maximum length bitonic subarray is {4, 78, 90, 45, 23} which is of length 5.
2) A[] = {20, 4, 1, 2, 3, 4, 2, 10}, the maximum length bitonic subarray is {1, 2, 3, 4, 2} which is of length 5.
Extreme Examples
1) A[] = {10}, the single element is bitnoic, so output is 1.
2) A[] = {10, 20, 30, 40}, the complete array itself is bitonic, so output is 4.
3) A[] = {40, 30, 20, 10}, the complete array itself is bitonic, so output is 4.
Solution
Let us consider the array {12, 4, 78, 90, 45, 23} to understand the soultion.
1) Construct an auxiliary array inc[] from left to right such that inc[i] contains length of the nondecreaing subarray ending at arr[i].
For for A[] = {12, 4, 78, 90, 45, 23}, inc[] is {1, 1, 2, 3, 1, 1}
2) Construct another array dec[] from right to left such that dec[i] contains length of nonincreasing subarray starting at arr[i].
For A[] = {12, 4, 78, 90, 45, 23}, dec[] is {2, 1, 1, 3, 2, 1}.
3) Once we have the inc[] and dec[] arrays, all we need to do is find the maximum value of (inc[i] + dec[i] – 1).
For {12, 4, 78, 90, 45, 23}, the max value of (inc[i] + dec[i] – 1) is 5 for i = 3.

#include<stdio.h>
#include<stdlib.h>
 
int bitonic(int arr[], int n)
{
    int i;
    int *inc = new int[n];
    int *dec = new int[n];
    int max;
    inc[0] = 1; // The length of increasing sequence ending at first index is 1
    dec[n-1] = 1; // The length of increasing sequence starting at first index is 1
 
    // Step 1) Construct increasing sequence array
    for(i = 1; i < n; i++)
    {
        if (arr[i] > arr[i-1])
            inc[i] = inc[i-1] + 1;
        else
            inc[i] = 1;
    }
 
    // Step 2) Construct decreasing sequence array
    for (i = n-2; i >= 0; i--)
    {
        if (arr[i] > arr[i+1])
            dec[i] = dec[i+1] + 1;
        else
            dec[i] = 1;
    }
 
    // Step 3) Find the length of maximum length bitonic sequence
    max = inc[0] + dec[0] - 1;
    for (i = 1; i < n; i++)
    {
        if (inc[i] + dec[i] - 1 > max)
        {
            max = inc[i] + dec[i] - 1;
        }
    }
 
    // free dynamically allocated memory
    delete [] inc;
    delete [] dec;
 
    return max;
}
 
/* Driver program to test above function */
int main()
{
    int arr[] = {12, 4, 78, 90, 45, 23};
    int n = sizeof(arr)/sizeof(arr[0]);
    printf("\n Length of max length Bitnoic Subarray is %d", bitonic(arr, n));
    getchar();
    return 0;
}

Time Complexity: O(n)
Auxiliary Space: O(n)

Simulating final class in C++

Source

Ever wondered how can you design a class in C++ which can’t be inherited. Java and C# programming languages have this feature built-in. You can use final keyword in java, sealed in C# to make a class non-extendable.
Below is a mechanism using which we can achieve the same behavior in C++. It makes use of private constructor, virtual inheritance and friend class.
In the following code, we make the Final class non-inheritable. When a class Derived tries to inherit from it, we get compilation error.
An extra class MakeFinal (whose default constructor is private) is used for our purpose. Constructor of Final can call private constructor of MakeFinal as Final is a friend of MakeFinal .
Note that MakeFinal is also a virtual base class. The reason for this is to call the constructor of MakeFinal through the constructor of Derived, not Final (The constructor of a virtual base class is not called by the class that inherits from it, instead the constructor is called by the constructor of the concrete class).

/* A program with compilation error to demonstrate that Final class cannot
   be inherited */
#include<iostream>
using namespace std;
 
class Final;  // The class to be made final
 
class MakeFinal // used to make the Final class final
{
private:
    MakeFinal() { cout << "MakFinal constructor" << endl; }
friend class Final;
};
 
class Final : virtual MakeFinal
{
public:
    Final() { cout << "Final constructor" << endl; }
};
 
class Derived : public Final // Compiler error
{
public:
    Derived() { cout << "Derived constructor" << endl; }
};
 
int main(int argc, char *argv[])
{
    Derived d;
    return 0;
}

Output: Compiler Error

In constructor 'Derived::Derived()':
  error: 'MakeFinal::MakeFinal()' is private

In the above example, Derived‘s constructor directly invokes MakeFinal’s constructor, and the constructor of MakeFinal is private, therefore we get the compilation error.
You can create the object of Final class as it is friend class of MakeFinal and has access to its constructor. For example, the following program works fine.

/* A program without any compilation error to demonstrate that instances of
   the Final class can be created */
#include<iostream>
using namespace std;
 
class Final;
 
class MakeFinal
{
private:
    MakeFinal() { cout << "MakeFinal constructor" << endl; }
    friend class Final;
};
 
class Final : virtual MakeFinal
{
public:
    Final() { cout << "Final constructor" << endl; }
};
 
int main(int argc, char *argv[])
{
    Final f;
    return 0;
}

Output: Compiles and runs fine

MakeFinal constructor
Final constructor

Length of the longest substring without repeating characters

Source

Given a string, find the length of the longest substring without repeating characters. For example, the longest substrings without repeating characters for “ABDEFGABEF” are “BDEFGA” and “DEFGAB”, with length 6. For “BBBB” the longest substring is “B”, with length 1. For “GEEKSFORGEEKS”, there are two longest substrings shown in the below diagrams, with length 7.

The desired time complexity is O(n) where n is the length of the string.
Method 1 (Brute Force)
Let us cover the Brute-Force solution as the first method, no one would like this method though. The Brute-Force solution is to generate all possible substrings, find substrings with all unique characters and return the length of longest such substring. Time complexity of such a solution will be exponential which is not practical.
Method 2 (Linear Time)
Let us talk about the linear time solution now. This solution uses extra space to store the last indexes of already visited characters. The idea is to scan the string from left to right, keep track of the maximum length Non-Repeating Character Substring (NRCS) seen so far. Let the maximum length be max_len. When we traverse the string, we also keep track of length of the current NRCS using cur_len variable. For every new character, we look for it in already processed part of the string (A temp array called visited[] is used for this purpose). If it is not present, then we increase the cur_len by 1. If present, then there are two cases:
a) The previous instance of character is not part of current NRCS (The NRCS which is under process). In this case, we need to simply increase cur_len by 1.
b) If the previous instance is part of the current NRCS, then our current NRCS changes. It becomes the substring staring from the next character of previous instance to currently scanned character. We also need to compare cur_len and max_len, before changing current NRCS (or changing cur_len).
Implementation

#include<stdlib.h>
#include<stdio.h>
#define NO_OF_CHARS 256
 
int min(int a, int b);
 
int longestUniqueSubsttr(char *str)
{
    int n = strlen(str);
    int cur_len = 1;  // To store the lenght of current substring
    int max_len = 1;  // To store the result
    int prev_index;  // To store the previous index
    int i;
    int *visited = (int *)malloc(sizeof(int)*NO_OF_CHARS);
 
    /* Initialize the visited array as -1, -1 is used to indicate that
       character has not been visited yet. */
    for (i = 0; i < NO_OF_CHARS;  i++)
        visited[i] = -1;
 
    /* Mark first character as visited by storing the index of first
       character in visited array. */
    visited[str[0]] = 0;
 
    /* Start from the second character. First character is already processed
       (cur_len and max_len are initialized as 1, and visited[str[0]] is set */
    for (i = 1; i < n; i++)
    {
        prev_index =  visited[str[i]];
 
        /* If the currentt character is not present in the already processed
           substring or it is not part of the current NRCS, then do cur_len++ */
        if (prev_index == -1 || i - cur_len > prev_index)
            cur_len++;
 
        /* If the current character is present in currently considered NRCS,
           then update NRCS to start from the next character of previous instance. */
        else
        {
            /* Also, when we are changing the NRCS, we should also check whether
              length of the previous NRCS was greater than max_len or not.*/
            if (cur_len > max_len)
                max_len = cur_len;
 
            cur_len = i - prev_index;
        }
 
        visited[str[i]] = i; // update the index of current character
    }
 
    // Compare the length of last NRCS with max_len and update max_len if needed
    if (cur_len > max_len)
        max_len = cur_len;
 
    free(visited); // free memory allocated for visited
 
    return max_len;
}
 
/* A utility function to get the minimum of two integers */
int min(int a, int b)
{
    return (a>b)?b:a;
}
 
/* Driver program to test above function */
int main()
{
    char str[] = "ABDEFGABEF";
    printf("The input string is %s \n", str);
    int len =  longestUniqueSubsttr(str);
    printf("The length of the longest non-repeating character substring is %d", len);
 
    getchar();
    return 0;
}

Output

The input string is ABDEFGABEF
  The length of the longest non-repeating character substring is 6

Time Complexity: O(n + d) where n is length of the input string and d is number of characters in input string alphabet. For example, if string consists of lowercase English characters then value of d is 26.
Auxiliary Space: O(d)
Algorithmic Paradigm: Dynamic Programming

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