problems-challenges-hakerrank.py

# Esta es una lista de problemas que hice en hacker rank como prueba tecnica para una empresa.

# Arrays: Left Rotation
# a es uin arreglo, d es el número de veces a rotar.
def rotLeft(a, d):
    return a[d:] + a[:d]
  
  
# New Year Chaos
# queue: una lista de elementos
# numero de elementos a  intercambiar || 'too chaotic'
def functionResult(queue):
    lastIndex = len(queue) - 1
    swaps = 0
    swaped = False
    
    # check if the queue is too chaotic
    for i, v in enumerate(queue):
        if (v - 1) - i > 2:
            return ("Too chaotic")
    # bubble sorting to find the answer
    for i in range(0, lastIndex):
        for j in range(0, lastIndex):
            
            if queue[j] > queue[j+1]:
                temp = queue[j]
                queue[j] = queue[j+1]
                queue[j+1] = temp
                swaps += 1
                swaped = True
        
        if swaped:
            swaped = False
        else:
            break
    return( swaps)

## Dictionary and hashmap

# Hash Tables: Ransom Note
# Magazine, ransom: lista de palabras
# devuelve falso si No se puede replicar, verdadero en caso contrario
def checkMagazine(magazine, ransom):
    magazine.sort()
    ransom.sort()
    for word in ransom:
        if word not in magazine:
            return False
        magazine.remove(word)
    return True

# Sherlock and Anagrams
# recibe un lista de palabras 
# devuelve un número de  cuantos pares pueden generar 
def sherlockAndAnagrams(s):
    count = 0
    for i in range(1,len(s)+1):
        a = ["".join(sorted(s[j:j+i])) for j in range(len(s)-i+1)]
        b = Counter(a)
        for j in b:
            count+=b[j]*(b[j]-1)/2
    return int(count)


### Sorting ###
## Este código fue hecho en C++ 

#Sorting: Bubble Sort
# Complete the countSwaps function below.
void countSwaps(vector<int> a) {
    int arraySorted = 0;
    int firstElement = 0;
    int lastElement = 0;
    for(int i=0; i< a.size(); i++){
        for(int j=0; j<a.size()-1; j++)
        if (a[j] > a[j + 1]) {
            int temp = a[j];
            a[j] = a[j+ 1];
            a[j+1] = temp;
           // swap(a[j], a[j + 1]);
            arraySorted++;
        }
    }
    cout<< "Array is sorted in " <<arraySorted << " swaps." << '\n';
    cout<< "First Element: "<< a[0] << '\n';
    cout<<  "Last Element: "<< a[a.size()-1] << '\n';
}


# Sorting comparator 
##// complete this method
    static int comparator(Player a, Player b)  {
        if(a.score > b.score) return 1;
        if(a.score < b.score) return -1;
        if(a.name.compare(b.name) > 0 ) return -1;
        else return 1;
        return 0;
    }



## bloque C++  termina

## STRING Manipulation 
#bloque C
## Making anagram
int funct(char *a , char *b){

    int sum = 0;
    int i;
    
    for( i = 0; a[i] !='\0' ; i++){
        A[a[i] - 'a'] ++;
         
    }
    for( i = 0; b[i] !='\0' ; i++){
        A[b[i] - 'a']--;          
    }
    
    for(int i = 0; i < 30; i++){
            sum+= abs(A[i]);             
    }
        
    return sum;
}

## fin bloque C



# Sherlock and the Valid String
# Given , we would need to remove two characters, both c and d  aabb or a and b  abcd, to make it valid.
# We are limited to removing only one character, so  is invalid.
def isValid(s):
    freq=[v for k,v in Counter(s).items()]
    pivot=freq[0]                                      
    different_freqs=[]                                     
    for i in freq[1::]:                                       
        if(pivot!=i):                        
           different_freqs.append(i)                             
    if(len(different_freqs)>1):                                         
        return 'NO'                                                   
    else:                                        
        return 'YES'