Pairs of strings which on concatenating contains each character of “string”

Given an array of strings arr[]. The task is to find the count of unordered pairs of strings (arr[i], arr[j]), which on concatenation contains each character of the string “string” at least once.

Examples:

Input: arr[] = { “s”, “ng”, “stri”}
Output: 1
(arr[1], arr[2]) is the only pair which on concatenation
will contain every character of the string “string”
i.e. arr[1] + arr[2] = “ngstri”

Input: arr[] = { “stri”, “ing”, “string” }
Output: 3



Approach: Store the given strings as bit masks i.e. a string “srin” will be stored as 101110 as ‘t’ and ‘g’ are missing so their corresponding bit will be 0. Now, create an array of size 64 which is the maximum possible value of the bitmasks obtained (0 (000000) to 63 (111111)). Now, the problem is reduced to finding the count of pairs of these bitmasks that give 63 (111111 in binary) as their OR value.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define MAX 64
  
// Function to return the bitmask for the string
int getBitmask(string s)
{
    int temp = 0;
    for (int j = 0; j < s.length(); j++) {
        if (s[j] == 's') {
            temp = temp | (1);
        }
        else if (s[j] == 't') {
            temp = temp | (2);
        }
        else if (s[j] == 'r') {
            temp = temp | (4);
        }
        else if (s[j] == 'i') {
            temp = temp | (8);
        }
        else if (s[j] == 'n') {
            temp = temp | (16);
        }
        else if (s[j] == 'g') {
            temp = temp | (32);
        }
    }
  
    return temp;
}
  
// Function to return the count of pairs
int countPairs(string arr[], int n)
{
  
    // bitMask[i] will store the count of strings
    // from the array whose bitmask is i
    int bitMask[MAX] = { 0 };
    for (int i = 0; i < n; i++)
        bitMask[getBitmask(arr[i])]++;
  
    // To store the count of pairs
    int cnt = 0;
    for (int i = 0; i < MAX; i++) {
        for (int j = i; j < MAX; j++) {
  
            // MAX - 1 = 63 i.e. 111111 in binary
            if ((i | j) == (MAX - 1)) {
  
                // arr[i] cannot make s pair with itself
                // i.e. (arr[i], arr[i])
                if (i == j)
                    cnt += ((bitMask[i] * bitMask[i] - 1) / 2);
                else
                    cnt += (bitMask[i] * bitMask[j]);
            }
        }
    }
    return cnt;
}
  
// Driver code
int main()
{
    string arr[] = { "strrr", "strring", "gstrin" };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countPairs(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the
// above approach
class GFG
{
  
static int MAX = 64;
  
// Function to return the bitmask for the string
static int getBitmask(char[] s)
{
    int temp = 0;
    for (int j = 0; j < s.length; j++) 
    {
        switch (s[j]) 
        {
            case 's':
                temp = temp | (1);
                break;
            case 't':
                temp = temp | (2);
                break;
            case 'r':
                temp = temp | (4);
                break;
            case 'i':
                temp = temp | (8);
                break;
            case 'n':
                temp = temp | (16);
                break;
            case 'g':
                temp = temp | (32);
                break;
            default:
                break;
        }
    }
  
    return temp;
}
  
// Function to return the count of pairs
static int countPairs(String arr[], int n)
{
  
    // bitMask[i] will store the count of strings
    // from the array whose bitmask is i
    int []bitMask = new int[MAX];
    for (int i = 0; i < n; i++)
        bitMask[getBitmask(arr[i].toCharArray())]++;
  
    // To store the count of pairs
    int cnt = 0;
    for (int i = 0; i < MAX; i++) 
    {
        for (int j = i; j < MAX; j++)
        {
  
            // MAX - 1 = 63 i.e. 111111 in binary
            if ((i | j) == (MAX - 1)) 
            {
  
                // arr[i] cannot make s pair with itself
                // i.e. (arr[i], arr[i])
                if (i == j)
                    cnt += ((bitMask[i] * bitMask[i] - 1) / 2);
                else
                    cnt += (bitMask[i] * bitMask[j]);
            }
        }
    }
    return cnt;
}
  
// Driver code
public static void main(String[] args) 
{
    String arr[] = { "strrr", "strring", "gstrin" };
    int n = arr.length;
    System.out.println(countPairs(arr, n));
}
}
  
/* This code contributed by PrinciRaj1992 */

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Python3

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# Python3 implementation of the approach
MAX = 64
  
# Function to return the bitmask
# for the string
def getBitmask(s):
  
    temp = 0
    for j in range(len(s)):
        if (s[j] == 's'):
            temp = temp | 1
        elif (s[j] == 't'):
            temp = temp | 2
        elif (s[j] == 'r'):
            temp = temp | 4
        elif (s[j] == 'i'):
            temp = temp | 8
        elif (s[j] == 'n'):
            temp = temp | 16
        elif (s[j] == 'g'):
            temp = temp | 32
  
    return temp
  
# Function to return the count of pairs
def countPairs(arr, n):
  
    # bitMask[i] will store the count of strings
    # from the array whose bitmask is i
    bitMask = [0 for i in range(MAX)]
  
    for i in range(n):
        bitMask[getBitmask(arr[i])] += 1
  
    # To store the count of pairs
    cnt = 0
    for i in range(MAX):
        for j in range(i, MAX):
  
            # MAX - 1 = 63 i.e. 111111 in binary
            if ((i | j) == (MAX - 1)):
  
                # arr[i] cannot make s pair with itself
                # i.e. (arr[i], arr[i])
                if (i == j):
                    cnt += ((bitMask[i] * 
                             bitMask[i] - 1) // 2)
                else:
                    cnt += (bitMask[i] * bitMask[j])
              
    return cnt
  
# Driver code
arr = ["strrr", "strring", "gstrin"]
n = len(arr)
print(countPairs(arr, n))
  
# This code is contributed by mohit kumar

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C#

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// C# implementation of the
// above approach 
using System;
using System.Collections.Generic;
  
class GFG
{
  
static int MAX = 64;
  
// Function to return the bitmask for the string
static int getBitmask(char[] s)
{
    int temp = 0;
    for (int j = 0; j < s.Length; j++) 
    {
        switch (s[j]) 
        {
            case 's':
                temp = temp | (1);
                break;
            case 't':
                temp = temp | (2);
                break;
            case 'r':
                temp = temp | (4);
                break;
            case 'i':
                temp = temp | (8);
                break;
            case 'n':
                temp = temp | (16);
                break;
            case 'g':
                temp = temp | (32);
                break;
            default:
                break;
        }
    }
  
    return temp;
}
  
// Function to return the count of pairs
static int countPairs(String []arr, int n)
{
  
    // bitMask[i] will store the count of strings
    // from the array whose bitmask is i
    int []bitMask = new int[MAX];
    for (int i = 0; i < n; i++)
        bitMask[getBitmask(arr[i].ToCharArray())]++;
  
    // To store the count of pairs
    int cnt = 0;
    for (int i = 0; i < MAX; i++) 
    {
        for (int j = i; j < MAX; j++)
        {
  
            // MAX - 1 = 63 i.e. 111111 in binary
            if ((i | j) == (MAX - 1)) 
            {
  
                // arr[i] cannot make s pair with itself
                // i.e. (arr[i], arr[i])
                if (i == j)
                    cnt += ((bitMask[i] * bitMask[i] - 1) / 2);
                else
                    cnt += (bitMask[i] * bitMask[j]);
            }
        }
    }
    return cnt;
}
  
// Driver code
public static void Main(String[] args) 
{
    String []arr = { "strrr", "strring", "gstrin" };
    int n = arr.Length;
    Console.WriteLine(countPairs(arr, n));
}
}
  
// This code has been contributed by 29AjayKumar

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PHP

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<?php
// PHP implementation of the approach 
$MAX = 64;
  
// Function to return the bitmask for the string 
function getBitmask($s
    $temp = 0; 
    for ($j = 0; $j < strlen($s); $j++) 
    
        if ($s[$j] == 's'
        
            $temp = $temp | (1); 
        
        else if ($s[$j] == 't')
        
            $temp = $temp | (2); 
        
        else if ($s[$j] == 'r'
        
            $temp = $temp | (4); 
        
        else if ($s[$j] == 'i')
        
            $temp = $temp | (8); 
        
        else if ($s[$j] == 'n'
        
            $temp = $temp | (16); 
        
        else if ($s[$j] == 'g'
        
            $temp = $temp | (32); 
        
    
  
    return $temp
  
// Function to return the count of pairs 
function countPairs($arr, $n
  
    // bitMask[i] will store the count of strings 
    // from the array whose bitmask is i 
    $bitMask = array_fill(0, $GLOBALS['MAX'], 0);
      
    for ($i = 0; $i < $n; $i++) 
        $bitMask[getBitmask($arr[$i])]++; 
  
    // To store the count of pairs 
    $cnt = 0; 
    for ($i = 0; $i < $GLOBALS['MAX']; $i++)
    
        for ($j = $i; $j < $GLOBALS['MAX']; $j++) 
        
  
            // MAX - 1 = 63 i.e. 111111 in binary 
            if (($i | $j) == ($GLOBALS['MAX'] - 1))
            
  
                // arr[i] cannot make s pair with itself 
                // i.e. (arr[i], arr[i]) 
                if ($i == $j
                    $cnt += floor(($bitMask[$i] * 
                                   $bitMask[$i] - 1) / 2); 
                else
                    $cnt += ($bitMask[$i] * $bitMask[$j]); 
            
        
    
    return $cnt
  
// Driver code 
$arr = array( "strrr", "strring", "gstrin" ); 
$n = count($arr);
  
echo countPairs($arr, $n); 
  
// This code is contributed by Ryuga
?>

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Output:

3


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