Se't donen dues cadenes de la mateixa longitud que has de trobar Distància de Hamming entre aquestes cordes.
On la distància de Hamming entre dues cordes d'igual longitud és el nombre de posicions en què el caràcter corresponent és diferent.
Exemples:
Input : str1[] = 'geeksforgeeks' str2[] = 'geeksandgeeks' Output : 3 Explanation : The corresponding character mismatch are highlighted. 'geeks for geeks' and 'geeks and geeks' Input : str1[] = '1011101' str2[] = '1001001' Output : 2 Explanation : The corresponding character mismatch are highlighted. '10 1 1 1 01' and '10 0 1 0 01'
Aquest problema es pot resoldre amb un enfocament senzill en el qual travessem les cordes i comptem el desajust a la posició corresponent. La forma estesa d'aquest problema és editar la distància.
Algorisme:
int hammingDist(char str1[] char str2[]) { int i = 0 count = 0; while(str1[i]!=' ') { if (str1[i] != str2[i]) count++; i++; } return count; } A continuació es mostra la implementació de dues cadenes.
C++// C++ program to find hamming distance b/w two string #include using namespace std; // function to calculate Hamming distance int hammingDist(string str1 string str2) { int i = 0 count = 0; while (str1[i] != ' ') { if (str1[i] != str2[i]) count++; i++; } return count; } // driver code int main() { string str1 = 'geekspractice'; string str2 = 'nerdspractise'; // function call cout << hammingDist(str1 str2); return 0; } // This code is contributed by Sania Kumari Gupta (kriSania804)
C // C program to find hamming distance b/w two string #include // function to calculate Hamming distance int hammingDist(char* str1 char* str2) { int i = 0 count = 0; while (str1[i] != ' ') { if (str1[i] != str2[i]) count++; i++; } return count; } // driver code int main() { char str1[] = 'geekspractice'; char str2[] = 'nerdspractise'; // function call printf('%d' hammingDist(str1 str2)); return 0; } // This code is contributed by Sania Kumari Gupta // (kriSania804)
Java // Java program to find hamming distance b/w two string class GFG { // function to calculate Hamming distance static int hammingDist(String str1 String str2) { int i = 0 count = 0; while (i < str1.length()) { if (str1.charAt(i) != str2.charAt(i)) count++; i++; } return count; } // Driver code public static void main(String[] args) { String str1 = 'geekspractice'; String str2 = 'nerdspractise'; // function call System.out.println(hammingDist(str1 str2)); } } // This code is contributed by Sania Kumari Gupta // (kriSania804)
Python3 # Python3 program to find # hamming distance b/w two # string # Function to calculate # Hamming distance def hammingDist(str1 str2): i = 0 count = 0 while(i < len(str1)): if(str1[i] != str2[i]): count += 1 i += 1 return count # Driver code str1 = 'geekspractice' str2 = 'nerdspractise' # function call print(hammingDist(str1 str2)) # This code is contributed by avanitrachhadiya2155
C# // C# program to find hamming // distance b/w two string using System; class GFG { // function to calculate // Hamming distance static int hammingDist(String str1 String str2) { int i = 0 count = 0; while (i < str1.Length) { if (str1[i] != str2[i]) count++; i++; } return count; } // Driver code public static void Main () { String str1 = 'geekspractice'; String str2 = 'nerdspractise'; // function call Console.Write(hammingDist(str1 str2)); } } // This code is contributed by nitin mittal
PHP // PHP program to find hamming distance b/w // two string // function to calculate // Hamming distance function hammingDist($str1 $str2) { $i = 0; $count = 0; while (isset($str1[$i]) != '') { if ($str1[$i] != $str2[$i]) $count++; $i++; } return $count; } // Driver Code $str1 = 'geekspractice'; $str2 = 'nerdspractise'; // function call echo hammingDist ($str1 $str2); // This code is contributed by nitin mittal. ?> JavaScript <script> // JavaScript program to find hamming distance b/w // two string // function to calculate Hamming distance function hammingDist(str1 str2) { let i = 0 count = 0; while (i < str1.length) { if (str1[i] != str2[i]) count++; i++; } return count; } // driver code let str1 = 'geekspractice'; let str2 = 'nerdspractise'; // function call document.write(hammingDist (str1 str2)); // This code is contributed by Manoj. </script>
Sortida
4
Complexitat temporal: O(n)
Nota: Per a la distància de Hamming de dos nombres binaris, simplement podem retornar un recompte de bits establerts en XOR de dos nombres.
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