# Find Missing Number with Vector This one is simple, given an array containing all numbers from 1 to N with the exception of one print the missing number to the standard output. My solution makes use of vector. Expected complexity is O(N). As a side note, it has been said that this problem has been asked on Microsoft interviews.

```
#include <iostream>
#include <vector>

using namespace std;

void find_missing_number(const vector<int> &v) {
int total, i;
total  = (v.size()+1)*(v.size()+2)/2;
for (i = 0; i < v.size(); i++)
total -= v[i];
cout <<  total;
}
```

# Finding the Longest Palindrome

The other day I was doing a programming challenge that required me to find the longest palindrome within a string. The challenge was to do it in at least O(N)2 complexity.  I found this to be fairly interesting. The challenge read like so:

Given a string S, find the longest substring in S that is the same in reverse and print it to the standard output.

So for example, if I had the string given was:  s= “abcdxyzyxabcdaaa” . Then the longest palindrome within it is “xyzyx “. We need to write a function to find this for us.  The nature of the challenge just needed me to write a function for a web back-end, so the example code below does not have main, or a defined string, etc. Also, there are O(N) solutions for this problem if you seek them, but they are a little more complicated.

```#include <iostream>
#include <string>

using namespace std;

void longest_palind(const string &s) {
int n = s.length();
int longestBegin = 0;
int maxLen = 1;
bool table = {false};

for (int i = 0; i < n; i++)
{
table[i][i] = true;
}
for (int i = 0; i < n-1; i++)
{
if (s[i] == s[i+1]) {
table[i][i+1] = true;
longestBegin = i;
maxLen = 2;
}
}
for (int len = 3; len <= n; len++) {
for (int i = 0; i < n-len+1; i++) {
int j = i+len-1;
if (s[i] == s[j] && table[i+1][j-1]) {
table[i][j] = true;
longestBegin = i;
maxLen = len;
}
}
}
cout << s.substr(longestBegin, maxLen);
}
```

# C++ Primer Plus Chapter 7 Exercise 9 Chapter 7 ends with a short exercise on function pointers. The simplest way to complete this, which I provided in my source, is to create the two functions add() and calculate(), then call them in main() with a loop. The text mentions if you are feeling adventurous to include other functions in addition to add. For simplicities sake, I did the minimum requirement here. Finally,  we can put this chapter to rest. See my source below:

```#include <iostream>

using namespace std;

double calculate(double x, double y, double (*pf)(double a, double b));

int main()
{
double x, y;
cout << "Enter two numbers (q to exit): ";
while(cin >> x >> y)
{
if(cin.fail())
break;
cout << "The sum is: " << calculate(x, y, add) << "\n\n";
}
return 0;
}

double calculate(double x, double y, double (*pf)(double a, double b))
{
return pf(x, y);
}

{
return x + y;
}
```

# C++ Primer Plus Chapter 7 Exercise 8 Exercise 8 has us re-writing a skeleton program to fill in the missing functions. A combination of structs, pointers, and functions will complete this exercise. Comments describing where and what additions have been made are embedded in the code. See my solution below:

8. This exercise provides practice in writing functions dealing with arrays and structures. The following is a program skeleton. Complete it by providing the described functions:

```
#include <iostream>

using namespace std;
const int SLEN = 30;
struct student {
char fullname[SLEN];
char hobby[SLEN];
int ooplevel;
};

int getinfo(student pa[], int n);
void display1(student st);
void display2(const student * ps);
void display3(const student pa[], int n);

int main(){
cout << "Enter class size: ";
int class_size;
cin >> class_size;
while (cin.get() != '\n')
continue;
student * ptr_stu = new student[class_size];
int entered = getinfo(ptr_stu, class_size);

for (int i = 0; i < entered; i++) {
display1(ptr_stu[i]);
display2(&ptr_stu[i]);
}
display3(ptr_stu, entered);
delete [] ptr_stu;
cout << "Done\n";

return 0;
}

// getinfo() has two arguments: a pointer to the first element of
// an array of student structures and an int representing the
// number of elements of the array. The function solicits and
// stores data about students. It terminates input upon filling
// the array or upon encountering a blank line for the student
// name. The function returns the actual number of array elements
// filled.
int getinfo(student pa[], int n)
{
int i;
for(i = 0; i < n; i++)
{
cout << "Student Number: " << (i+1) << "\n";
cout << "Student's Name: ";
cin.getline(pa[i].fullname, SLEN);

cout << "Enter students hobby: ";
cin.getline(pa[i].hobby, SLEN);

cout << "Enter students OOP level: ";
cin >> pa[i].ooplevel;
cin.get();
}
return i;
}

// display1() takes a student structure as an argument
// and displays its contents
void display1(student st)
{
cout << "\n=============Display 1================\n";
cout << "Student name: " << st.fullname << "\n";
cout << "Hobby: " <<  st.hobby << "\n";
cout << "OOP Level: " << st.ooplevel << "\n";
cout << "=============End Display 1==============\n";
}

// display2() takes the address of student structure as an
// argument and displays the structure's contents
void display2(const student * ps)
{
cout << "\n============Display 2================\n";
cout << "Students Name: " << ps->fullname << "\n";
cout << "Hobby: " << ps->hobby << "\n";
cout << "OOP Level" << ps->ooplevel << "\n";
cout << "==========End Display 2================\n";
}

// display3() takes the address of the first element of an array
// of student structures and the number of array elements as
// arguments and displays the contents of the structures
void display3(const student pa[], int n)
{
for(int i = 0; i < n; i++)
{
cout << "\n============Display 3================\n";
cout << "Student: " << (i+1) << "\n";
cout << "Fullname: " << pa[i].fullname << "\n";
cout << "OOP Level: " << pa[i].ooplevel << "\n";
cout << "=============End Display 3============\n";
}
}

```

# Bubble Sort Algorithm

The Bubble Sort algorithm is one the simplest sorting algorithm to implement. It works by continually looping through pairs in a list and comparing each pair. At each pair, it will perform a swap if the pair happens to be in the wrong order. The order will either be ascending or descending depending on what the programmer wants. This happens until no more swaps are needed. This algorithm apparently gets the name “bubble-sort” because you can imagine the smaller elements bubble to the top of the list. Checkout my bubble sort implementation below. When ran, it will give a good visual reference for the algorithm. It sorts in ascending order, complexity O(n^2). This algorithm approaches O(n) if the elements are already almost in order. Note that is the not the most efficient algorithm for large data sets.

```#include <iostream>

using namespace std;

int main()
{
int array;
int i, j;

for(i = 0; i <= 6; i++){
cout << "Enter unsorted numbers: " << endl;
cin >> array[i];
}
for(i = 0; i <= 5; i++){ //loop through pair of integers
for(j = i+1; j <= 6; j++){
int temp;
if(array[i] > array[j])
{
temp = array[i]; //swapping mechanisms
array[i] = array[j];
array[j] = temp;
}
}
}
for(i = 0; i <= 6; i++){
cout << array[i]; //sorted output
}
return 0;
}

```

# Fizz Buzz C++ This is a “classic” fizz-buzz solution I created. For those of you who don’t already know what fizz-buzz is, it follows this basic context: “Write a program that prints the numbers from 1 to 100. But for multiples of three print “Fizz” instead of the number and for the multiples of five print “Buzz”. For numbers which are multiples of both three and five print “FizzBuzz””. This question has been asked by recruiters as a method for determining programming competency.  While the fizz-buzz question is not really difficult itself, it does force you to think about one of the most important concepts to programmers, loop structures.

```
#include <iostream>

using namespace std;

int main(){
for (int i = 0; i <= 100; i++) {
if ((i % 15) == 0) //for multiples of both 3 and 5
cout << "FizzBuzz" << endl;
else if ((i % 3) == 0) //multiples of 3
cout << "Fizz" << endl;
else if ((i % 5) == 0) //multiples of 5
cout << "Buzz" << endl;
else
cout << i << endl; //all other numbers
}
return 0;
}

```

# C++ Primer Plus Chapter 7 Exercise 7 This exercise makes use heavier use of pointers then what we have been doing. However, much of this program can be pulled from listing 7.7 and around that part of the chapter. See source below.

Redo Listing 7.7, modifying the three array-handling functions to each use two pointer parameters to represent a range. The fill_array() function, instead of returning the actual number of items read, should return a pointer to the location after the last location filled; the other functions can use this pointer as the second argument to identify the end of the data.

```#include <iostream>

using namespace std;

const int Max = 5;

// function prototypes
double * fill_array(double *first, double *last);
void show_array(const double *first, const double *last);
void revalue(double *first, double *last, double factor);

int main()
{
double properties[Max];
double * last;

last = fill_array(properties, properties+Max);
show_array(properties, last);

cout << "Enter revaluation factor: ";
double factor;
cin >> factor;

revalue(properties, last, factor);
show_array(properties, last);
cout << "Done.\n";

return 0;
}

double *fill_array(double *first, double *last)
{
double temp;
double *pt;
int i=0;
for (pt = first; pt != last; pt++, i++)
{
cout << "Enter value #" << (i + 1) << ": ";
cin >> temp;
{
cin.clear();
while (cin.get() != '\n')
continue;
cout << "Bad input; input process terminated.\n";
break;
}
else if (temp < 0) // signal to terminate
break;
*pt = temp;
}
return pt;
}

void show_array(const double *first, const double *last)
{
const double *pt;
int i=0;
for (pt = first; pt != last; pt++, i++)
{
cout << "Property #" << (i + 1) << ": \$ \n";
cout << *pt;
}
}

void revalue(double *first, double *last, double factor)
{
double *pt;
for (pt = first; pt != last; pt++)
*pt *= factor;
}

```

# C++ Primer Plus Chapter 7 Exercise 6 Exercise 6 of chapter 7 is not so bad because the first two functions are given to us in the text. However, you do need to do some thinking as to how to reverse an array. There are a few ways to do this. One would be to implement an XOR type algorithm. Another would be to use reverse() from the std library, and then you can do it the way I have done it which is work through the array by starting on the ends. The way I have done it is one of the most common ways to solve this problem and is good to know as it is sometimes used as a skill assessment on programming interviews. See source below.

Write a program that uses the following functions:
Fill_array() takes as arguments the name of an array of double values and an array
size. It prompts the user to enter double values to be entered in the array. It ceases taking
input when the array is full or when the user enters non-numeric input, and it
returns the actual number of entries.
Show_array() takes as arguments the name of an array of double values and an array
size and displays the contents of the array.
Reverse_array() takes as arguments the name of an array of double values and an
array size and reverses the order of the values stored in the array.
The program should use these functions to fill an array, show the array, reverse the array,
show the array, reverse all but the first and last elements of the array, and then show the
array.

```#include <iostream>

using namespace std;

const int Max = 5;

int fill_array(double ar[], int limit);
void show_array(const double ar[], int n);
void reverse_array(double ar[], int n);

int main()
{
double properties[Max];
int size = fill_array(properties, Max);
cout << endl;
show_array(properties, size);
cout << endl;
reverse_array(properties, size);
show_array(properties, size);
cout << endl;
reverse_array(properties + 1, size -2);
show_array(properties, size);
return 0;
}

int fill_array(double ar[], int limit)
{
double temp;
int i;
for(i = 0; i < limit; i++)
{
cout << "Enter value #" << (i + 1) << ": ";
cin >> temp;
if(!cin)
{
cin.clear();
while(cin.get() != '\n')
continue;
cout << "Bad input; input process terminated" << endl;
break;
}
else if(temp < 0)
break;
ar[i] = temp;
}
return i;
}

void show_array(const double ar[], int n)
{
for (int i = 0; i < n; i++)
{
cout << "Property #" << (i + 1) << ": ";
cout << ar[i] << endl;
}
}

void reverse_array(double ar[], int n)
{
double temp;
for(int i = 0; i < n/2; i++)
{
temp = ar[i];
ar[i] = ar[n - i - 1];
ar[n - i - 1] = temp;
}
}
```

Note: for the sake of the solution. fill_array() could be stripped of the error handling code to make this easier to read.

# C++ Primer Plus Chapter 7 Exercise 5 To understand recursion, one must understand recursion.

Define a recursive function that takes an integer argument and returns the factorial of that argument. Recall that 3 factorial, written 3!, equals 3 × 2!, and so on, with 0! defined as 1. In general, if n is greater than zero, n! = n * (n – 1)!. Test your function in a program that uses a loop to allow the user to enter various values for which the program reports the factorial.

```#include <iostream>

using namespace std;

int factorial(int number);

int main()
{
int number;
while(number != -1)
{
cout << "Enter a number to find it's factorial (q to quit): ";
cin >> number;
if(!cin.good())
break;
else
{
int factor;
factor = factorial(number);
cout << "Factorial " << number << " is " << factor << endl;
}
}
return 0;
}

int factorial(int number)
{
if(number == 0)
return 1;
return number * factorial(number - 1);
}

```

# C++ Primer Plus Chapter 7 Exercise 4 The trick to this program is understanding what is it asking and getting it to work as it wants. In my solution I created a second function that holds all the numbers plus the probability function within it. I think this is one of the quickest solution to this exercise. The output is truncated to that of a long int, as I think the book format is quite ugly. Another way I had though of doing this originally was to calculate the odds of 1 in 47, store it. Calculate the 1 in 27 odds, store it, then find the product of those all in their own functions. That idea seemed clumsy when coding it and all roads were pointing to the solution provided for me.

Many state lotteries use a variation of the simple lottery portrayed by Listing 7.4. In
these variations you choose several numbers from one set and call them the field numbers.
For example, you might select 5 numbers from the field of 1–47). You also pick a
single number (called a mega number or a power ball, etc.) from a second range, such as
1–27. To win the grand prize, you have to guess all the picks correctly. The chance of
winning is the product of the probability of picking all the field numbers times the probability
of picking the mega number. For instance, the probability of winning the example
described here is the product of the probability of picking 5 out of 47 correctly times the
probability of picking 1 out of 27 correctly. Modify Listing 7.4 to calculate the probability
of winning this kind of lottery.

```#include <iostream>

using namespace std;

// Note: some implementations require double instead of long double
long double probability(unsigned numbers, unsigned picks);
long int TotalOdds(int, int, int, int, long double(*p)(unsigned, unsigned));

int main()
{
cout << "Welcome to the Powerball!\n";
cout << "Odds of winning are one in " << TotalOdds(47, 5, 27, 1, probability);
cout << "\nThanks for playing!" << endl;

return 0;
}
// the following function calculates the probability of picking picks
// numbers correctly from numbers choices
long double probability(unsigned numbers, unsigned picks)
{
long double result = 1.0; // here come some local variables
long double n;
unsigned p;
for (n = numbers, p = picks; p > 0; n--, p--)
result = result * n / p;
return result;
}

long int TotalOdds(int FirstSet, int x, int PowerBall, int y, long double(*p)(unsigned, unsigned))
{
long double odds;
odds = p(FirstSet, x)*p(PowerBall, y);
return odds;
}
```