Gitt en rekke små bokstaver er oppgaven å finne antall strenger som er forskjellige. To strenger er forskjellige hvis den andre strengen ikke kan dannes ved å bruke følgende operasjoner på en streng.
- Et tegn på oddeindeksen kan bare byttes med et annet tegn på oddeindeksen.
- Et tegn på jevn indeks kan bare byttes med et annet tegn på jevn indeks.
Eksempler:
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 EN enkel løsning er å kjøre to løkker. Den ytre sløyfen plukker en streng og den indre sløyfen sjekker om det er en tidligere streng som kan konverteres til en gjeldende streng ved å gjøre tillatte transformasjoner. Denne løsningen krever O(n2m) tid hvor n er antall strenger og m er maksimalt antall tegn i en streng.
An effektiv løsning genererer en kodet streng for hver inndatastreng. Den kodede har tellinger av partall og oddetall plasserte tegn atskilt med en skilletegn. To strenger anses som like hvis deres kodede strenger er de samme, ellers ikke. Når vi har en måte å kode strenger på, reduseres problemet til å telle distinkte kodede strenger. Dette er et typisk problem med hashing. Vi lager et hash-sett og en etter en lagrer kodinger av strenger. Hvis en koding allerede eksisterer, ignorerer vi strengen. Ellers lagrer vi koding i hash og inkrementantall av distinkte strenger.
Implementering:
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>
Produksjon
4
Tidskompleksitet : På)
Hjelpeområde: O(1)
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