Distinct strings with odd and even changes allowed - GeeksforGeeks

07 Mar, 2022
Given an array of lower case strings, the task is to find the number of strings that are distinct. Two strings are distinct if, on applying the following operations on one string, the second string cannot be formed.  
A character on the odd index can be swapped with another character on the odd index only.
A character on even index can be swapped with another character on even index only.
Examples:  
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 simple solution is to run two loops. The outer loop picks a string and the inner loop checks if there is a previously string which can be converted to a current string by doing allowed transformations. This solution requires O(n2m) time where n is the number of strings and m is the maximum number of characters in any string.
An efficient solution generates an encoded string for every input string. The encoded has counts of even and odd positioned characters separated by a separator. Two strings are considered same if their encoded strings are the same, then else not. Once we have a way to encode strings, the problem is reduced to counting distinct encoded strings. This is a typical problem of hashing. We create a hash set and, one by one, store encodings of strings. If an encoding already exists, we ignore the string. Otherwise, we store encoding in hash and increment count of distinct strings. 
C++
Java
Python3
C#
Javascript
#include<bits/stdc++.h>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];         // 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()+1];              strcpy(char_array, input[i].c_str());            if (s.find(encodeString(char_array, input[i].length()+1)) == s.end()) {                s.insert(encodeString(char_array,input[i].length()+1));                countDist++;            }        }         return countDist;    } // Driver code    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 NishaBharti.
// 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 Codeif __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>
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krutikkhandhadiya
29AjayKumar
Akanksha_Rai
sanjeev2552
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