When a started to do C.P. exercises, I found few patterns that were common to many exercises especially in in the basic and medium level. Plenty exercises required to know how to:
- Use the XOR operator and how it works.
- Check if a number is prime.
- Check if a string is a Palindrome.
- Print the reverse of a string
- Find frequencies of an object(number, string, pair) in a list.
- Use use the Left and Right Shift Operators
- Calculate the factorial of a number
- Represent a given number in binary format.
- Infix to Postfix conversion
- String Balancing
- Use arrays, vectors, vectors of arrays, stack and queues.
Note: I am not saying that it is all you need to know to do C.P., what this will help you to don't be stuck after the solutions of 15 easiest exercises that you will find for C.P.
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144
In mathemicatical terms, the sequence of Fibonaccia is created by the following relation:
Fn = Fn-1 + Fn-2, where F0 = 0 and F1 = 1;
- Aggregates: Structs, Unions
- Linear (Sequential): Arrays, Linked Lists, Stacks, Queues, Deques. Sequential Linear data structures leads to two different forms: Contigus: each element of the sequrnce is next to a neighbourhg until you get the end of the data structure (Ex: array). Non-contiguous: each element of the sequence stores bboth it own data & metadata about the location of the next & possibly also the previous element in the sequence.
- Linear: (Associative): Dictionaries
- Non-Linear: (Binary) Trees, Graphs
For memory optimization reasons considering that the length of the input can be different in any test case, and because vector has many built-in usefule methods (for example sort()) most of time I use vectors instead traditional arrays.
It's good to know the a vector has many useful built-in methods as sort(), but you should be able to implement the same methods on your known and then you the built-in to show your knowledge about the language and save time avoiding to rewrite the same code several times.
- XOR of a numebr with itself is 0
- XOR of a number with 0 is number itself.
Let's say that we need to check which number is repeated just once in an array, knowing that that all the other elements of the array are present twice.
int find_single(vector<int> vett)
{
int result = vett[0];
for(int i =1; i< vett.size(); i++)
result = result ^ vett[i];
return result;
}
Having two numeber and needing to print out their sum without the same the sum, very often we print directly cout<<a+b;.
Well, if you will do cout << a^b; to print the result of XOR operation between a and b, you will get an error.
For this reason, I would invite you to calculate and print the xor between two numbers in the following way:
#include <iostream>
using namespace std;
int get_xor(int a, int b)
{
return a^b;
}
int main() {
int a, b;
// Read two numbers in input
cin >> as >> b;
cout <<get_xor(a,b)<<endl;
}
void bin (unsigned int number) { if(number > 1) bin(number/2); cout<<number%2; }
- Brian Kerninghan's ALgorithm
- Popular Cisco and Adobe interview questions about count set bits of an integer
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A very popular C.P. that you can meet in the first rounds of many and many coding challenge is: Given a list of numbers (array / vectors), check whether some array elements can be added to sum to N OR return the subset of the array elements that added bettwen them gives N;
Input: Vett = {1,2,3,-1,3} N=4
- Yes : We can have in this case a valid sub set;
- Result:
{3,1}// 3+1 are both elements of the array and summed between them gives 4(N).
Well, as everything in C.P., depends. Depends by the size of the input and by the max running time allowed. Character aray is faster than string but, string contains useful built-in methods that allows us to manipulate it easily:
- append(): add a char at the end of the string
- getinline() : store a stream of characters as enters by the user (including eventually so special chars and white spaces).
- push_back() : add a char at the end of the string
- pop_back() : delete the last char
- lenght() : return the leghth of the string
- copy(Array of chars, length, position)