Given an array of strings arr[]. The task is to find the count of unordered pairs of strings (arr[i], arr[j]), which in concatenation includes 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++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;#define MAX 64// Function to return the bitmask for the stringint 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 pairsint 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 codeint main(){ string arr[] = { "strrr", "string", "gstrin" }; int n = sizeof(arr) / sizeof(arr[0]); cout << countPairs(arr, n); return 0;} |
Java
// Java implementation of the// above approachclass GFG{static int MAX = 64;// Function to return the bitmask for the stringstatic 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 pairsstatic 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 codepublic static void main(String[] args) { String arr[] = { "strrr", "string", "gstrin" }; int n = arr.length; System.out.println(countPairs(arr, n));}}/* This code contributed by PrinciRaj1992 */ |
Python3
# Python3 implementation of the approachMAX = 64# Function to return the bitmask# for the stringdef 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 pairsdef 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 codearr = ["strrr", "string", "gstrin"]n = len(arr)print(countPairs(arr, n))# This code is contributed by mohit kumar |
C#
// 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 stringstatic 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 pairsstatic 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 codepublic static void Main(String[] args) { String []arr = { "strrr", "string", "gstrin" }; int n = arr.Length; Console.WriteLine(countPairs(arr, n));}}// This code has been contributed by 29AjayKumar |
PHP
<?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", "string", "gstrin" ); $n = count($arr);echo countPairs($arr, $n); // This code is contributed by Ryuga?> |
Javascript
<script> // JavaScript implementation of the // above approach const MAX = 64; // Function to return the bitmask for the string function getBitmask(s) { var temp = 0; for (var 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 function countPairs(arr, n) { // bitMask[i] will store the count of strings // from the array whose bitmask is i var bitMask = new Array(MAX).fill(0); for (var i = 0; i < n; i++) bitMask[getBitmask(arr[i].split(""))]++; // To store the count of pairs var cnt = 0; for (var i = 0; i < MAX; i++) { for (var 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 += parseInt((bitMask[i] * bitMask[i] - 1) / 2); else cnt += bitMask[i] * bitMask[j]; } } } return cnt; } // Driver code var arr = ["strrr", "string", "gstrin"]; var n = arr.length; document.write(countPairs(arr, n));</script> |
Output:
3
Time Complexity: O(n * |s| + MAX2)
Auxiliary Space: O(MAX)
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