Count of distinct Strings possible by swapping prefixes of pairs of Strings from the Array

Given an array string[] consisting of N numeric strings of length M, the task is to find the number of distinct strings that can be generated by selecting any two strings, say i and j from the array and swap all possible prefixes between them. 
Note: Since the answer can be very large, print modulo 1000000007.

Examples:

Input: N = 2 M = 3 string[] = {“112”, “211”} 
Output:
Explanation: 
Swapping “1” and “2” between the strings generates “212” and “111“. 
Swapping “11” and “21” between the strings generates “212” and “111”. 
Swapping “112” and “211” between the strings generates “211” and “112“. 
Therefore, 4 distinct strings are generated.

Input: N = 4 M = 5 string[] = {“12121”, “23545”, “11111”, “71261”} 
Output: 216

Approach: Considering a string s of the form s1s2s3s4…sm, where s1 is the first letter of any of the strings in the array, s2 is the second letter of any of the strings, and so on, the answer to the problem is the product of count(i) where count(i) is the count of different letters placed at same index in the given strings.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ program to implement
// the above approach
#include <bits/stdc++.h>   
using namespace std; 
  
int mod = 1000000007;
  
// Function to count the distinct strings
// possible after swapping the prefixes
// between two possible strings of the array
long countS(string str[],int n, int m)
{
      
    // Stores the count of unique
    // characters for each index
    unordered_map<int
    unordered_set<char>> counts;
  
    for(int i = 0; i < n; i++) 
    {
          
        // Store current string
        string s = str[i];
      
        for(int j = 0; j < m; j++)
        {
            counts[j].insert(s[j]);
        }
    }
  
    // Stores the total number of
    // distinct strings possible
    long result = 1;
      
    for(auto index : counts)
    {
        result = (result * 
                  counts[index.first].size()) % mod;
    }
      
    // Return the answer
    return result;
}
  
// Driver Code
int main()
{
    string str[] = { "112", "211" };
  
    int N = 2, M = 3;
      
    cout << countS(str, N, M);
      
    return 0;
  
// This code is contributed by rutvik_56

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to implement
// the above approach
import java.util.*;
import java.io.*;
  
public class Main {
  
    static int mod = 1000000007;
  
    // Function to count the distinct strings
    // possible after swapping the prefixes
    // between two possible strings of the array
    public static long countS(String str[], int n, int m)
    {
  
        // Stores the count of unique
        // characters for each index
        Map<Integer, Set<Character> > counts
            = new HashMap<>();
  
        for (int i = 0; i < m; i++) {
            counts.put(i, new HashSet<>());
        }
  
        for (int i = 0; i < n; i++) {
  
            // Store current string
            String s = str[i];
  
            for (int j = 0; j < m; j++) {
                counts.get(j).add(s.charAt(j));
            }
        }
  
        // Stores the total number of
        // distinct strings possible
        long result = 1;
  
        for (int index : counts.keySet())
            result = (result
                    * counts.get(index).size())
                    % mod;
  
        // Return the answer
        return result;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
  
        String str[] = { "112", "211" };
  
        int N = 2, M = 3;
        System.out.println(countS(str, N, M));
    }
}

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
  
static int mod = 1000000007;
  
// Function to count the distinct strings
// possible after swapping the prefixes
// between two possible strings of the array
public static long countS(String []str,
                            int n, int m)
{
  
    // Stores the count of unique
    // characters for each index
    Dictionary<int
    HashSet<char> > counts = new Dictionary<int
                                    HashSet<char>>();
  
    for (int i = 0; i < m; i++)
    {
    counts.Add(i, new HashSet<char>());
    }
  
    for (int i = 0; i < n; i++) 
    {
  
    // Store current string
    String s = str[i];
  
    for (int j = 0; j < m; j++)
    {
        counts[j].Add(s[j]);
    }
    }
  
    // Stores the total number of
    // distinct strings possible
    long result = 1;
  
    foreach (int index in counts.Keys)
    result = (result * counts[index].Count) % mod;
  
    // Return the answer
    return result;
}
  
// Driver Code
public static void Main(String[] args)
{
    String []str = { "112", "211" };
  
    int N = 2, M = 3;
    Console.WriteLine(countS(str, N, M));
}
}
  
// This code is contributed by Rajput-Ji

chevron_right


Output: 

4

Time Complexity: O(N * M) 
Auxiliary Space: O(N)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : Rajput-Ji, rutvik_56