Ņemot vērā mazo burtu virkņu masīvu, uzdevums ir atrast atšķirīgo virkņu skaitu. Divas virknes ir atšķirīgas, ja, veicot šādas darbības vienai virknei, otro virkni nevar izveidot.
- Rakstzīmi nepāra rādītājā var apmainīt ar citu rakstzīmi tikai nepāra indeksā.
- Pāra indeksa rakstzīmi var apmainīt ar citu rakstzīmi tikai pāra indeksā.
Piemēri:
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 vienkāršs risinājums ir palaist divas cilpas. Ārējā cilpa izvēlas virkni, un iekšējā cilpa pārbauda, vai iepriekš ir virkne, kuru var pārveidot par pašreizējo virkni, veicot atļautās transformācijas. Šim risinājumam nepieciešams O (n2m) laiks, kur n ir virkņu skaits un m ir maksimālais rakstzīmju skaits jebkurā virknē.
An efektīvs risinājums ģenerē kodētu virkni katrai ievades virknei. Kodētajam ir pāra un nepāra pozicionētu rakstzīmju skaits, kas atdalītas ar atdalītāju. Divas virknes tiek uzskatītas par vienādām, ja to kodētās virknes ir vienādas, tad nē. Kad mums ir veids, kā kodēt virknes, problēma tiek samazināta līdz atsevišķu kodētu virkņu skaitīšanai. Tā ir tipiska jaukšanas problēma. Mēs izveidojam hash komplektu un pa vienam uzglabājam virkņu kodējumus. Ja kodējums jau pastāv, mēs ignorējam virkni. Pretējā gadījumā mēs saglabājam kodējumu jauktā veidā un palielinām atšķirīgu virkņu skaitu.
Īstenošana:
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>
Izvade
4
Laika sarežģītība : O(n)
Palīgtelpa: O(1)
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