С обзиром на низ низова малих слова, задатак је пронаћи број стрингова који се разликују. Два низа се разликују ако се при примени следећих операција на једном низу не може формирати други низ.
- Знак на непарном индексу може се заменити другим карактером само на непарном индексу.
- Знак на парном индексу може се заменити другим карактером само на парном индексу.
Примери:
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 А једноставно решење је покретање две петље. Спољна петља бира стринг, а унутрашња проверава да ли постоји претходни низ који се може конвертовати у тренутни низ обављањем дозвољених трансформација. Ово решење захтева О(н2м) време где је н број стрингова, а м максимални број знакова у било ком низу.
Ан ефикасно решење генерише кодирани стринг за сваки улазни низ. Кодирано има број парних и непарних знакова раздвојених сепаратором. Два низа се сматрају истим ако су њихови кодирани низови исти, а у супротном не. Једном када имамо начин да кодирамо низове, проблем се своди на бројање различитих кодираних стрингова. Ово је типичан проблем хеширања. Креирамо хеш скуп и један по један чувамо кодирања низова. Ако кодирање већ постоји, игноришемо стринг. У супротном, чувамо кодирање у хеш-у и броју инкремената различитих стрингова.
Имплементација:
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>
Излаз
4
Временска сложеност : О(н)
Помоћни простор : О(1)
Креирај квиз