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: 4
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++ 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 |
// 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));
}
} |
# Python3 program to implement # the above approach from collections import defaultdict
mod = 1000000007
# Function to count the distinct strings # possible after swapping the prefixes # between two possible strings of the array def countS(string: list , n: int , m: int ) - > int :
# Stores the count of unique
# characters for each index
counts = defaultdict( lambda : set ())
for i in range (n):
# Store current string
s = string[i]
for j in range (m):
counts[j].add(s[j])
# Stores the total number of
# distinct strings possible
result = 1
for index in counts:
result = (result *
len (counts[index])) % mod
# Return the answer
return result
# Driver Code if __name__ = = "__main__" :
string = [ "112" , "211" ]
N = 2
M = 3
print (countS(string, N, M))
# This code is contributed by sanjeev2552 |
// 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 |
// Javascript Program to implement // the above approach let mod = 1000000007; // Function to count the distinct strings // possible after swapping the prefixes // between two possible strings of the array function countS(str, n, m) {
// Stores the count of unique
// characters for each index
let counts = new Map();
for (let i = 0; i < m; i++) {
counts.set(i, new Set());
}
for (let i = 0; i < n; i++) {
// Store current string
let s = str[i];
for (let j = 0; j < m; j++) {
let temp = counts.get(j);
temp.add(s[j])
}
}
// Stores the total number of
// distinct strings possible
let result = 1;
for (let index of counts.keys())
result = (result * counts.get(index).size) % mod;
// Return the answer
return result;
} // Driver Code let str = [ "112" , "211" ];
let N = 2, M = 3; console.log(countS(str, N, M)); // This code is contributed by saurabh_jaiswal. |
4
Time Complexity: O(N * M)
Auxiliary Space: O(N)