홀수 및 짝수 변경이 허용되는 고유한 문자열

소문자 문자열 배열이 주어지면 고유한 문자열 수를 찾는 작업이 수행됩니다. 한 문자열에 다음 작업을 적용할 때 두 번째 문자열을 구성할 수 없는 경우 두 문자열은 서로 다릅니다.

홀수 인덱스의 문자는 홀수 인덱스의 다른 문자로만 교체될 수 있습니다.
짝수 인덱스의 문자는 짝수 인덱스의 다른 문자로만 교체될 수 있습니다.

예:

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

에이 간단한 해결책 두 개의 루프를 실행하는 것입니다. 외부 루프는 문자열을 선택하고 내부 루프는 허용된 변환을 수행하여 현재 문자열로 변환할 수 있는 이전 문자열이 있는지 확인합니다. 이 솔루션에는 O(n²m) 시간 여기서 n은 문자열 수이고 m은 문자열의 최대 문자 수입니다.

안 효율적인 솔루션 모든 입력 문자열에 대해 인코딩된 문자열을 생성합니다. 인코딩된 문자에는 구분 기호로 구분된 짝수 및 홀수 위치의 문자 수가 포함됩니다. 인코딩된 문자열이 동일하면 두 문자열은 동일한 것으로 간주되고 그렇지 않으면 동일하지 않습니다. 문자열을 인코딩하는 방법이 있으면 문제는 인코딩된 개별 문자열을 계산하는 것으로 줄어듭니다. 이는 해싱의 전형적인 문제입니다. 우리는 해시 세트를 생성하고 문자열의 인코딩을 하나씩 저장합니다. 인코딩이 이미 존재하는 경우 문자열을 무시합니다. 그렇지 않으면 인코딩을 해시로 저장하고 고유 문자열의 개수를 늘립니다.

구현:

C++

#include   using namespace std; int MAX_CHAR = 26; string encodeString(char str[] int m) {  // hashEven stores the count of even indexed character  // for each string hashOdd stores the count of odd  // indexed characters for each string  int hashEven[MAX_CHAR];  int hashOdd[MAX_CHAR];  memset(hashEven0sizeof(hashEven));  memset(hashOdd0sizeof(hashOdd));  // creating hash for each string  for (int i = 0; i < m; 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. 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;  for (int i = 0; i < n; i++) {  // If this encoding appears first time increment  // count of distinct encodings.  char char_array[input[i].length()];  strcpy(char_array input[i].c_str());  if (s.find(encodeString(char_array input[i].length())) == s.end()) {  s.insert(encodeString(char_arrayinput[i].length()));  countDist++;  }  }  return countDist; } int main() {  string input[] = {'abcd' 'acbd' 'adcb' 'cdba'  'bcda' 'badc'};  int n = sizeof(input)/sizeof(input[0]);  cout << countDistinct(input n) << 'n'; } // This code is contributed by Harshit Sharma.

Java

// 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

# 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

JavaScript

<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>

산출

시간 복잡도 : 에)
보조공간 : O(1)

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