Bei einem Array kleingeschriebener Zeichenfolgen besteht die Aufgabe darin, die Anzahl der Zeichenfolgen zu ermitteln, die unterschiedlich sind. Zwei Zeichenfolgen sind unterschiedlich, wenn bei der Anwendung der folgenden Operationen auf eine Zeichenfolge die zweite Zeichenfolge nicht gebildet werden kann.
- Ein Zeichen auf dem ungeraden Index kann nur mit einem anderen Zeichen auf dem ungeraden Index ausgetauscht werden.
- Ein Zeichen mit geradem Index kann nur mit einem anderen Zeichen mit geradem Index ausgetauscht werden.
Beispiele:
Input : arr[] = {'abcd' 'cbad' 'bacd'} Output : 2 The 2nd string can be converted to the 1st by swapping the first and third characters. So there are 2 distinct strings as the third string cannot be converted to the first. Input : arr[] = {'abc' 'cba'} Output : 1 A einfache Lösung besteht darin, zwei Schleifen auszuführen. Die äußere Schleife wählt einen String aus und die innere Schleife prüft, ob es einen früheren String gibt, der durch zulässige Transformationen in einen aktuellen String umgewandelt werden kann. Diese Lösung erfordert O(n2m) Zeit, wobei n die Anzahl der Zeichenfolgen und m die maximale Anzahl von Zeichen in einer Zeichenfolge ist.
Ein effiziente Lösung generiert für jede Eingabezeichenfolge eine codierte Zeichenfolge. Die Codierung umfasst die Anzahl der geraden und ungeraden Zeichen, die durch ein Trennzeichen getrennt sind. Zwei Zeichenfolgen gelten als gleich, wenn ihre codierten Zeichenfolgen gleich sind, andernfalls nicht. Sobald wir eine Möglichkeit haben, Strings zu kodieren, reduziert sich das Problem auf das Zählen verschiedener codierter Strings. Dies ist ein typisches Problem beim Hashing. Wir erstellen ein Hash-Set und speichern nacheinander die Codierungen der Zeichenfolgen. Wenn bereits eine Kodierung vorhanden ist, ignorieren wir die Zeichenfolge. Andernfalls speichern wir die Codierung im Hash und erhöhen die Anzahl der einzelnen Zeichenfolgen.
Durchführung:
C++#include
// Java program to count distinct strings with // even odd swapping allowed. import java.util.HashSet; import java.util.Set; class GFG { static int MAX_CHAR = 26; static String encodeString(char[] str) { // hashEven stores the count of even indexed character // for each string hashOdd stores the count of odd // indexed characters for each string int hashEven[] = new int[MAX_CHAR]; int hashOdd[] = new int[MAX_CHAR]; // creating hash for each string for (int i = 0; i < str.length; i++) { char c = str[i]; if ((i & 1) != 0) // If index of current character is odd hashOdd[c-'a']++; else hashEven[c-'a']++; } // For every character from 'a' to 'z' we store its // count at even position followed by a separator // followed by count at odd position. String encoding = ''; for (int i = 0; i < MAX_CHAR; i++) { encoding += (hashEven[i]); encoding += ('-'); encoding += (hashOdd[i]); encoding += ('-'); } return encoding; } // This function basically uses a hashing based set to // store strings which are distinct according // to criteria given in question. static int countDistinct(String input[] int n) { int countDist = 0; // Initialize result // Create an empty set and store all distinct // strings in it. Set<String> s = new HashSet<>(); for (int i = 0; i < n; i++) { // If this encoding appears first time increment // count of distinct encodings. if (!s.contains(encodeString(input[i].toCharArray()))) { s.add(encodeString(input[i].toCharArray())); countDist++; } } return countDist; } public static void main(String[] args) { String input[] = {'abcd' 'acbd' 'adcb' 'cdba' 'bcda' 'badc'}; int n = input.length; System.out.println(countDistinct(input n)); } }
# Python3 program to count distinct strings with # even odd swapping allowed. MAX_CHAR = 26 # Returns encoding of string that can be used # for hashing. The idea is to return same encoding # for strings which can become same after swapping # a even positioned character with other even characters # OR swapping an odd character with other odd characters. def encodeString(string): # hashEven stores the count of even indexed character # for each string hashOdd stores the count of odd # indexed characters for each string hashEven = [0] * MAX_CHAR hashOdd = [0] * MAX_CHAR # creating hash for each string for i in range(len(string)): c = string[i] if i & 1: # If index of current character is odd hashOdd[ord(c) - ord('a')] += 1 else: hashEven[ord(c) - ord('a')] += 1 # For every character from 'a' to 'z' we store its # count at even position followed by a separator # followed by count at odd position. encoding = '' for i in range(MAX_CHAR): encoding += str(hashEven[i]) encoding += str('-') encoding += str(hashOdd[i]) encoding += str('-') return encoding # This function basically uses a hashing based set to # store strings which are distinct according # to criteria given in question. def countDistinct(input n): countDist = 0 # Initialize result # Create an empty set and store all distinct # strings in it. s = set() for i in range(n): # If this encoding appears first time increment # count of distinct encodings. if encodeString(input[i]) not in s: s.add(encodeString(input[i])) countDist += 1 return countDist # Driver Code if __name__ == '__main__': input = ['abcd' 'acbd' 'adcb' 'cdba' 'bcda' 'badc'] n = len(input) print(countDistinct(input n)) # This code is contributed by # sanjeev2552
// C# program to count distinct strings with // even odd swapping allowed. using System; using System.Collections.Generic; class GFG { static int MAX_CHAR = 26; static String encodeString(char[] str) { // hashEven stores the count of even // indexed character for each string // hashOdd stores the count of odd // indexed characters for each string int []hashEven = new int[MAX_CHAR]; int []hashOdd = new int[MAX_CHAR]; // creating hash for each string for (int i = 0; i < str.Length; i++) { char m = str[i]; // If index of current character is odd if ((i & 1) != 0) hashOdd[m - 'a']++; else hashEven[m - 'a']++; } // For every character from 'a' to 'z' // we store its count at even position // followed by a separator // followed by count at odd position. String encoding = ''; for (int i = 0; i < MAX_CHAR; i++) { encoding += (hashEven[i]); encoding += ('-'); encoding += (hashOdd[i]); encoding += ('-'); } return encoding; } // This function basically uses a hashing based set // to store strings which are distinct according // to criteria given in question. static int countDistinct(String []input int n) { int countDist = 0; // Initialize result // Create an empty set and store all distinct // strings in it. HashSet<String> s = new HashSet<String>(); for (int i = 0; i < n; i++) { // If this encoding appears first time // increment count of distinct encodings. if (!s.Contains(encodeString(input[i].ToCharArray()))) { s.Add(encodeString(input[i].ToCharArray())); countDist++; } } return countDist; } // Driver Code public static void Main(String[] args) { String []input = {'abcd' 'acbd' 'adcb' 'cdba' 'bcda' 'badc'}; int n = input.Length; Console.WriteLine(countDistinct(input n)); } } // This code is contributed by 29AjayKumar
<script> // Javascript program to count distinct strings with // even odd swapping allowed let MAX_CHAR = 26; function encodeString(str) { // hashEven stores the count of even indexed character // for each string hashOdd stores the count of odd // indexed characters for each string let hashEven = Array(MAX_CHAR).fill(0); let hashOdd = Array(MAX_CHAR).fill(0); // creating hash for each string for (let i = 0; i < str.length; i++) { let c = str[i]; if ((i & 1) != 0) // If index of current character is odd hashOdd[c.charCodeAt() - 'a'.charCodeAt()]++; else hashEven[c.charCodeAt() - 'a'.charCodeAt()]++; } // For every character from 'a' to 'z' we store its // count at even position followed by a separator // followed by count at odd position. let encoding = ''; for (let i = 0; i < MAX_CHAR; i++) { encoding += (hashEven[i]); encoding += ('-'); encoding += (hashOdd[i]); encoding += ('-'); } return encoding; } // This function basically uses a hashing based set to // store strings which are distinct according // to criteria given in question. function countDistinct(input n) { let countDist = 0; // Initialize result // Create an empty set and store all distinct // strings in it. let s = new Set(); for (let i = 0; i < n; i++) { // If this encoding appears first time increment // count of distinct encodings. if (!s.has(encodeString(input[i].split('')))) { s.add(encodeString(input[i].split(''))); countDist++; } } return countDist; } // Driver program let input = ['abcd' 'acbd' 'adcb' 'cdba' 'bcda' 'badc']; let n = input.length; document.write(countDistinct(input n)); </script>
Ausgabe
4
Zeitkomplexität : An)
Hilfsraum: O(1)
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