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Count of pairs of strings whose concatenation forms a palindromic string

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Given an array A[ ] consisting of N strings, the task is to count the number of pairs of possible strings that on merging forms a Palindromic String or can be rearranged to form a Palindromic String

Example :

Input: N = 6, A[ ] = {aab, abcac, dffe, ed, aa, aade}
Output: 6
Explanation: 
All possible pairs of strings which generates a palindromic string: 
{aab, abcac} = aabccbaa
{aab, aa} = aabaa
{abcac, aa} = aacbcaa
{dffe, ed} = fdeedf
{dffe, aade} = edfaafde
{ed, aade} = edaade or aeddea
Hence, total possible pairs = 6

 Input: N = 3, A[ ] = {aa, bb, cd}
Output: 1
Explanation: 
The only possible pair of strings which generates palindromic string: 
{aa, bb} = abba
Hence, total possible pairs = 1

Naive Approach:
The simplest approach to solve this problem is to generate all possible pairs from the given array and merge these pairs to generate new strings. For each of those merged strings, generate all possible permutations of the string and for each permutation, check whether it is a palindromic string or not and increment the count accordingly. 
Time Complexity: O(N2 * L! * L), where L is the maximum length of a string.
Auxiliary Space: O(M)

Efficient Approach:
A palindromic string can always be formed when at most one character has an odd occurrence and remaining every character has even occurrence. Since rearrangement of the characters in the merged string is allowed, so we only need to count the frequency of characters in the merged string and check if it satisfies to be a palindromic string or not.

Illustration:
1. Every character has even occurrence
Strings “aab” and “abcac” can be merged and rearranged to form a palindromic string “aababcac” with each character having even frequency

2. One character has odd occurrence
Strings “aab” and “aa” can be merged to form a palindromic string “aabaa” with a character(‘b’) having odd frequency and the rest having even frequency.

Hence, it can be observed that two pairs of strings of the following types can be merged to form a palindromic string:

  • The frequency of each character in both strings are same.
  • At most one character in both the strings has a different frequency.

Follow the steps below to solve the problem:

  • Traverse the given array and for every string,  store the frequency of each character.
  • Assign parity (0 or 1) to each frequency. Assign 0 for even frequency and 1 for odd frequency.
  • Initialize a map to store the frequencies of each frequency array generated.
  • Since, rearrangement is allowed, reverse the parity of each character and include the frequency array in map.
  • Finally, return the total count of similar frequency arrays, which will give the required count of pairs.

Below is the implementation of the above approach:

C++

// C++ Program to find
// palindromic string
#include <bits/stdc++.h>
using namespace std;
 
int getCount(int N, vector<string>& s)
{
    // Stores frequency array
    // and its count
    map<vector<int>, int> mp;
 
    // Total number of pairs
    int ans = 0;
 
    for (int i = 0; i < N; i++) {
 
        // Initializing array of size 26
        // to store count of character
        vector<int> a(26, 0);
 
        // Counting occurrence of each
        // character of current string
        for (int j = 0; j < s[i].size(); j++) {
 
            a[s[i][j] - 'a']++;
        }
 
        // Convert each count to parity(0 or 1)
        // on the basis of its frequency
        for (int j = 0; j < 26; j++) {
 
            a[j] = a[j] % 2;
        }
 
        // Adding to answer
        ans += mp[a];
 
        // Frequency of single character
        // can be possibly changed,
        // so change its parity
        for (int j = 0; j < 26; j++) {
 
            vector<int> changedCount = a;
 
            if (a[j] == 0)
                changedCount[j] = 1;
            else
                changedCount[j] = 0;
 
            ans += mp[changedCount];
        }
 
        mp[a]++;
    }
 
    return ans;
}
 
int main()
{
    int N = 6;
    vector<string> A
        = { "aab", "abcac", "dffe",
            "ed", "aa", "aade" };
 
    cout << getCount(N, A);
 
    return 0;
}

                    

Java

// Java program to find
// palindromic string
import java.util.*;
 
class GFG{
 
static int getCount(int N, List<String> s)
{
     
    // Stores frequency array
    // and its count
    Map<List<Integer>, Integer> mp = new HashMap<>();
     
    // Total number of pairs
    int ans = 0;
     
    for(int i = 0; i < N; i++)
    {
         
        // Initializing array of size 26
        // to store count of character
        List<Integer> a = new ArrayList<>(26);
        for(int k = 0; k < 26; k++)
            a.add(0);
         
        // Counting occurrence of each
        // character of current string
        for(int j = 0; j < s.get(i).length(); j++)
        {
            a.set((s.get(i).charAt(j) - 'a'),
             a.get(s.get(i).charAt(j) - 'a') + 1);
        }
 
        // Convert each count to parity(0 or 1)
        // on the basis of its frequency
        for(int j = 0; j < 26; j++)
        {
            a.set(j, a.get(j) % 2);
        }
         
        // Adding to answer
        ans += mp.getOrDefault(a, 0);
         
        // Frequency of single character
        // can be possibly changed,
        // so change its parity
        for(int j = 0; j < 26; j++)
        {
            List<Integer> changedCount = new ArrayList<>();
            changedCount.addAll(a);
             
            if (a.get(j) == 0)
                changedCount.set(j, 1);
            else
                changedCount.set(j, 0);
             
            ans += mp.getOrDefault(changedCount, 0);
        }
        mp.put(a, mp.getOrDefault(a, 0) + 1);
    }
    return ans;
}
 
// Driver code
public static void main (String[] args)
{
    int N = 6;
    List<String> A = Arrays.asList("aab", "abcac",
                                   "dffe", "ed",
                                   "aa", "aade");
     
    System.out.print(getCount(N, A));
}
}
 
// This code is contributed by offbeat

                    

Python3

# Python3 program to find
# palindromic string
from collections import defaultdict
 
def getCount(N, s):
 
    # Stores frequency array
    # and its count
    mp = defaultdict(lambda: 0)
 
    # Total number of pairs
    ans = 0
 
    for i in range(N):
 
        # Initializing array of size 26
        # to store count of character
        a = [0] * 26
 
        # Counting occurrence of each
        # character of current string
        for j in range(len(s[i])):
            a[ord(s[i][j]) - ord('a')] += 1
 
        # Convert each count to parity(0 or 1)
        # on the basis of its frequency
        for j in range(26):
            a[j] = a[j] % 2
 
        # Adding to answer
        ans += mp[tuple(a)]
 
        # Frequency of single character
        # can be possibly changed,
        # so change its parity
        for j in range(26):
            changedCount = a[:]
            if (a[j] == 0):
                changedCount[j] = 1
            else:
                changedCount[j] = 0
            ans += mp[tuple(changedCount)]
 
        mp[tuple(a)] += 1
 
    return ans
 
# Driver code
if __name__ == '__main__':
     
    N = 6
    A = [ "aab", "abcac", "dffe",
        "ed", "aa", "aade" ]
 
    print(getCount(N, A))
 
# This code is contributed by Shivam Singh

                    

C#

// C# program to find
// palindromic string
using System;
using System.Collections.Generic;
 
class GFG {
 
  static int CompareLists(List<int> a, List<int> b)
  {
     
    // Check the length of both lists first
    if (a.Count != b.Count) {
      return a.Count - b.Count;
    }
 
    // Compare each element of the lists
    for (int i = 0; i < a.Count; i++) {
      if (a[i] != b[i]) {
        return a[i] - b[i];
      }
    }
 
    return 0;
  }
 
  static int getCount(int N, List<string> s)
  {
 
    // Stores frequency array
    // and its count
    Dictionary<List<int>, int> mp
      = new Dictionary<List<int>, int>(
      new ListComparer());
 
    // Total number of pairs
    int ans = 0;
 
    for (int i = 0; i < N; i++) {
 
      // Initializing array of size 26
      // to store count of character
      List<int> a = new List<int>(new int[26]);
 
      // Counting occurrence of each
      // character of current string
      for (int j = 0; j < s[i].Length; j++) {
        a[s[i][j] - 'a']++;
      }
 
      // Convert each count to parity(0 or 1)
      // on the basis of its frequency
      for (int j = 0; j < 26; j++) {
        a[j] %= 2;
      }
 
      // Adding to answer
      ans += mp.GetValueOrDefault(a, 0);
 
      // Frequency of single character
      // can be possibly changed,
      // so change its parity
      for (int j = 0; j < 26; j++) {
        List<int> changedCount = new List<int>(a);
 
        if (a[j] == 0)
          changedCount[j] = 1;
        else
          changedCount[j] = 0;
 
        ans += mp.GetValueOrDefault(changedCount,
                                    0);
      }
      mp[a] = mp.GetValueOrDefault(a, 0) + 1;
    }
    return ans;
  }
 
  // Driver code
  static void Main(string[] args)
  {
    int N = 6;
    List<string> A
      = new List<string>{ "aab", "abcac", "dffe",
                         "ed""aa",    "aade" };
 
    Console.Write(getCount(N, A));
  }
}
 
class ListComparer : IEqualityComparer<List<int> > {
  public bool Equals(List<int> a, List<int> b)
  {
     
    // Compare the elements of both lists
    if (a.Count != b.Count) {
      return false;
    }
 
    for (int i = 0; i < a.Count; i++) {
      if (a[i] != b[i]) {
        return false;
      }
    }
 
    return true;
  }
 
  public int GetHashCode(List<int> list)
  {
     
    // Generate a hash code based on the elements of the
    // list
    int hash = 17;
 
    foreach(int element in list)
    {
      hash = hash * 31 + element;
    }
    return hash;
  }
}
 
// This code is contributed by phasing17

                    

Javascript

// JS Program to find
// palindromic string
 
function getCount(N, s) {
  // Stores frequency array
  // and its count
  let mp = new Map();
 
  // Total number of pairs
  let ans = 0;
 
  for (let i = 0; i < N; i++) {
 
    // Initializing array of size 26
    // to store count of character
    let a = [];
    for (let i = 0; i < 26; i++) {
      a.push(0);
    }
 
    // Counting occurrence of each
    // character of current string
    for (let j = 0; j < s[i].length; j++) {
      // console.log("p : ",);
 
      a[s[i][j].charCodeAt(0) - 'a'.charCodeAt(0)]++;
    }
 
    // Convert each count to parity(0 or 1)
    // on the basis of its frequency
    for (let j = 0; j < 26; j++) {
 
      a[j] = a[j] % 2;
    }
 
    // Adding to answer
    let str = "";
    for (let m = 0; m < a.length; m++) {
      str += a[m];
    }
    if (!mp.has(str)) {
      mp.set(str, 0);
      ans += mp.get(str);
    }
    else
      ans += mp.get(str);
 
 
    // Frequency of single character
    // can be possibly changed,
    // so change its parity
    for (let j = 0; j < 2; j++) {
 
      let changedCount = a;
 
      if (a[j] == 0)
        changedCount[j] = 1;
      else
        changedCount[j] = 0;
 
      let str2 = "";
      for (let p = 0; p < changedCount.length; p++) {
        str2 += changedCount[p];
      }
      if (!mp.has(str2)) {
        mp.set(str2, 0);
        ans += mp.get(str2);
      }
      else
        ans += mp.get(str2);
    }
    mp.set(str, mp.get(str)+1.5);
  }
 
  return ans;
}
 
 
let N = 6;
let A = ["aab", "abcac", "dffe", "ed", "aa", "aade"];
 
console.log(getCount(N, A));
 
// This code is contributed by ksam24000.

                    

Output: 
6

 

Time Complexity: O(N*(L+log(N))), where L is the maximum length of a string
Auxiliary Space: O(N)



Last Updated : 11 Mar, 2023
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