Jums tiek dotas divas vienāda garuma virknes, kas jums jāatrod Haminga attālums starp šīm virknēm.
Kur Haminga attālums starp divām vienāda garuma virknēm ir to pozīciju skaits, kurās atbilstošā rakstzīme atšķiras.
Piemēri:
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'
Šo problēmu var atrisināt ar vienkāršu pieeju, kurā mēs šķērsojam virknes un saskaitām neatbilstību attiecīgajā pozīcijā. Šīs problēmas paplašinātā forma ir rediģēt attālumu.
Algoritms:
int hammingDist(char str1[] char str2[]) { int i = 0 count = 0; while(str1[i]!=' ') { if (str1[i] != str2[i]) count++; i++; } return count; } Zemāk ir divu virkņu ieviešana.
C++// C++ program to find hamming distance b/w two string #include
// C program to find hamming distance b/w two string #include
// 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 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# 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 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. ?> <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>
Izvade
4
Laika sarežģītība: O(n)
Piezīme: Divu bināro skaitļu Haminga attālumam mēs varam vienkārši atgriezt iestatīto bitu skaitu divu skaitļu XOR.
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