// C++ implementation for find pairs of complete // strings. #include    using namespace std; // Returns count of complete pairs from set[0..n-1] // and set2[0..m-1] int countCompletePairs(string set1[] string set2[]  int n int m) {  int result = 0;  // Consider all pairs of both strings  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // Create a concatenation of current pair  string concat = set1[i] + set2[j];  // Compute frequencies of all characters  // in the concatenated string.  int frequency[26] = { 0 };  for (int k = 0; k < concat.length(); k++)  frequency[concat[k] - 'a']++;  // If frequency of any character is not  // greater than 0 then this pair is not  // complete.  int i;  for (i = 0; i < 26; i++)  if (frequency[i] < 1)  break;  if (i == 26)  result++;  }  }  return result; } // Driver code int main() {  string set1[] = { 'abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc' };  string set2[] = { 'ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' };  int n = sizeof(set1) / sizeof(set1[0]);  int m = sizeof(set2) / sizeof(set2[0]);  cout << countCompletePairs(set1 set2 n m);  return 0; } 

// Java implementation for find pairs of complete // strings. class GFG {  // Returns count of complete pairs from set[0..n-1]  // and set2[0..m-1]  static int countCompletePairs(String set1[] String set2[]  int n int m)  {  int result = 0;  // Consider all pairs of both strings  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // Create a concatenation of current pair  String concat = set1[i] + set2[j];  // Compute frequencies of all characters  // in the concatenated String.  int frequency[] = new int[26];  for (int k = 0; k < concat.length(); k++) {  frequency[concat.charAt(k) - 'a']++;  }  // If frequency of any character is not  // greater than 0 then this pair is not  // complete.  int k;  for (k = 0; k < 26; k++) {  if (frequency[k] < 1) {  break;  }  }  if (k == 26) {  result++;  }  }  }  return result;  }  // Driver code  static public void main(String[] args)  {  String set1[] = { 'abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc' };  String set2[] = { 'ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' };  int n = set1.length;  int m = set2.length;  System.out.println(countCompletePairs(set1 set2 n m));  } } // This code is contributed by PrinciRaj19992 

# Python3 implementation for find pairs of complete # strings.  # Returns count of complete pairs from set[0..n-1] # and set2[0..m-1] def countCompletePairs(set1set2nm): result = 0 # Consider all pairs of both strings for i in range(n): for j in range(m): # Create a concatenation of current pair concat = set1[i] + set2[j] # Compute frequencies of all characters # in the concatenated String. frequency = [0 for i in range(26)] for k in range(len(concat)): frequency[ord(concat[k]) - ord('a')] += 1 # If frequency of any character is not # greater than 0 then this pair is not # complete. k = 0 while(k<26): if (frequency[k] < 1): break k += 1 if (k == 26): result += 1 return result # Driver code  set1=['abcdefgh' 'geeksforgeeks' 'lmnopqrst' 'abc'] set2=['ijklmnopqrstuvwxyz' 'abcdefghijklmnopqrstuvwxyz' 'defghijklmnopqrstuvwxyz'] n = len(set1) m = len(set2) print(countCompletePairs(set1 set2 n m)) # This code is contributed by shinjanpatra 

// C# implementation for find pairs of complete // strings. using System; class GFG {  // Returns count of complete pairs from set[0..n-1]  // and set2[0..m-1]  static int countCompletePairs(string[] set1 string[] set2  int n int m)  {  int result = 0;  // Consider all pairs of both strings  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // Create a concatenation of current pair  string concat = set1[i] + set2[j];  // Compute frequencies of all characters  // in the concatenated String.  int[] frequency = new int[26];  for (int k = 0; k < concat.Length; k++) {  frequency[concat[k] - 'a']++;  }  // If frequency of any character is not  // greater than 0 then this pair is not  // complete.  int l;  for (l = 0; l < 26; l++) {  if (frequency[l] < 1) {  break;  }  }  if (l == 26) {  result++;  }  }  }  return result;  }  // Driver code  static public void Main()  {  string[] set1 = { 'abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc' };  string[] set2 = { 'ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' };  int n = set1.Length;  int m = set2.Length;  Console.Write(countCompletePairs(set1 set2 n m));  } } // This article is contributed by Ita_c. 

<script> // Javascript implementation for find pairs of complete // strings.   // Returns count of complete pairs from set[0..n-1]  // and set2[0..m-1]  function countCompletePairs(set1set2nm)  {  let result = 0;    // Consider all pairs of both strings  for (let i = 0; i < n; i++) {  for (let j = 0; j < m; j++) {  // Create a concatenation of current pair  let concat = set1[i] + set2[j];    // Compute frequencies of all characters  // in the concatenated String.  let frequency = new Array(26);  for(let i= 0;i<26;i++)  {  frequency[i]=0;  }    for (let k = 0; k < concat.length; k++) {  frequency[concat[k].charCodeAt(0) - 'a'.charCodeAt(0)]++;  }    // If frequency of any character is not  // greater than 0 then this pair is not  // complete.  let k;  for (k = 0; k < 26; k++) {  if (frequency[k] < 1) {  break;  }  }  if (k == 26) {  result++;  }  }  }    return result;  }    // Driver code   let set1=['abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc'];  let set2=['ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz']  let n = set1.length;  let m=set2.length;  document.write(countCompletePairs(set1 set2 n m));    // This code is contributed by avanitrachhadiya2155   </script> 

// C++ program to find count of complete pairs #include    using namespace std; // Returns count of complete pairs from set[0..n-1] // and set2[0..m-1] int countCompletePairs(string set1[] string set2[]  int n int m) {  int result = 0;  // con_s1[i] is going to store an integer whose  // set bits represent presence/absence of characters  // in string set1[i].  // Similarly con_s2[i] is going to store an integer  // whose set bits represent presence/absence of  // characters in string set2[i]  int con_s1[n] con_s2[m];  // Process all strings in set1[]  for (int i = 0; i < n; i++) {  // initializing all bits to 0  con_s1[i] = 0;  for (int j = 0; j < set1[i].length(); j++) {  // Setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s1[i] = con_s1[i] | (1 << (set1[i][j] - 'a'));  }  }  // Process all strings in set2[]  for (int i = 0; i < m; i++) {  // initializing all bits to 0  con_s2[i] = 0;  for (int j = 0; j < set2[i].length(); j++) {  // setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s2[i] = con_s2[i] | (1 << (set2[i][j] - 'a'));  }  }  // assigning a variable whose all 26 (0..25)  // bits are set to 1  long long complete = (1 << 26) - 1;  // Now consider every pair of integer in con_s1[]  // and con_s2[] and check if the pair is complete.  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // if all bits are set the strings are  // complete!  if ((con_s1[i] | con_s2[j]) == complete)  result++;  }  }  return result; } // Driver code int main() {  string set1[] = { 'abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc' };  string set2[] = { 'ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' };  int n = sizeof(set1) / sizeof(set1[0]);  int m = sizeof(set2) / sizeof(set2[0]);  cout << countCompletePairs(set1 set2 n m);  return 0; } 

// Java program to find count of complete pairs class GFG {  // Returns count of complete pairs from set[0..n-1]  // and set2[0..m-1]  static int countCompletePairs(String set1[] String set2[]  int n int m)  {  int result = 0;  // con_s1[i] is going to store an integer whose  // set bits represent presence/absence of characters  // in string set1[i].  // Similarly con_s2[i] is going to store an integer  // whose set bits represent presence/absence of  // characters in string set2[i]  int[] con_s1 = new int[n];  int[] con_s2 = new int[m];  // Process all strings in set1[]  for (int i = 0; i < n; i++) {  // initializing all bits to 0  con_s1[i] = 0;  for (int j = 0; j < set1[i].length(); j++) {  // Setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s1[i] = con_s1[i] | (1 << (set1[i].charAt(j) - 'a'));  }  }  // Process all strings in set2[]  for (int i = 0; i < m; i++) {  // initializing all bits to 0  con_s2[i] = 0;  for (int j = 0; j < set2[i].length(); j++) {  // setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s2[i] = con_s2[i] | (1 << (set2[i].charAt(j) - 'a'));  }  }  // assigning a variable whose all 26 (0..25)  // bits are set to 1  long complete = (1 << 26) - 1;  // Now consider every pair of integer in con_s1[]  // and con_s2[] and check if the pair is complete.  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // if all bits are set the strings are  // complete!  if ((con_s1[i] | con_s2[j]) == complete) {  result++;  }  }  }  return result;  }  // Driver code  public static void main(String args[])  {  String set1[] = { 'abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc' };  String set2[] = { 'ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' };  int n = set1.length;  int m = set2.length;  System.out.println(countCompletePairs(set1 set2 n m));  } } // This code contributed by Rajput-Ji 

// C# program to find count of complete pairs using System; class GFG {  // Returns count of complete pairs from set[0..n-1]  // and set2[0..m-1]  static int countCompletePairs(String[] set1 String[] set2  int n int m)  {  int result = 0;  // con_s1[i] is going to store an integer whose  // set bits represent presence/absence of characters  // in string set1[i].  // Similarly con_s2[i] is going to store an integer  // whose set bits represent presence/absence of  // characters in string set2[i]  int[] con_s1 = new int[n];  int[] con_s2 = new int[m];  // Process all strings in set1[]  for (int i = 0; i < n; i++) {  // initializing all bits to 0  con_s1[i] = 0;  for (int j = 0; j < set1[i].Length; j++) {  // Setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s1[i] = con_s1[i] | (1 << (set1[i][j] - 'a'));  }  }  // Process all strings in set2[]  for (int i = 0; i < m; i++) {  // initializing all bits to 0  con_s2[i] = 0;  for (int j = 0; j < set2[i].Length; j++) {  // setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s2[i] = con_s2[i] | (1 << (set2[i][j] - 'a'));  }  }  // assigning a variable whose all 26 (0..25)  // bits are set to 1  long complete = (1 << 26) - 1;  // Now consider every pair of integer in con_s1[]  // and con_s2[] and check if the pair is complete.  for (int i = 0; i < n; i++) {  for (int j = 0; j < m; j++) {  // if all bits are set the strings are  // complete!  if ((con_s1[i] | con_s2[j]) == complete) {  result++;  }  }  }  return result;  }  // Driver code  public static void Main(String[] args)  {  String[] set1 = { 'abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc' };  String[] set2 = { 'ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' };  int n = set1.Length;  int m = set2.Length;  Console.WriteLine(countCompletePairs(set1 set2 n m));  } } // This code has been contributed by 29AjayKumar 

# Python3 program to find count of complete pairs # Returns count of complete pairs from set[0..n-1] # and set2[0..m-1] def countCompletePairs(set1 set2 n m): result = 0 # con_s1[i] is going to store an integer whose # set bits represent presence/absence of characters # in set1[i]. # Similarly con_s2[i] is going to store an integer # whose set bits represent presence/absence of # characters in set2[i] con_s1 con_s2 = [0] * n [0] * m # Process all strings in set1[] for i in range(n): # initializing all bits to 0 con_s1[i] = 0 for j in range(len(set1[i])): # Setting the ascii code of char s[i][j] # to 1 in the compressed integer. con_s1[i] = con_s1[i] | (1 << (ord(set1[i][j]) - ord('a'))) # Process all strings in set2[] for i in range(m): # initializing all bits to 0 con_s2[i] = 0 for j in range(len(set2[i])): # setting the ascii code of char s[i][j] # to 1 in the compressed integer. con_s2[i] = con_s2[i] | (1 << (ord(set2[i][j]) - ord('a'))) # assigning a variable whose all 26 (0..25) # bits are set to 1 complete = (1 << 26) - 1 # Now consider every pair of integer in con_s1[] # and con_s2[] and check if the pair is complete. for i in range(n): for j in range(m): # if all bits are set the strings are # complete! if ((con_s1[i] | con_s2[j]) == complete): result += 1 return result # Driver code if __name__ == '__main__': set1 = ['abcdefgh' 'geeksforgeeks' 'lmnopqrst' 'abc'] set2 = ['ijklmnopqrstuvwxyz' 'abcdefghijklmnopqrstuvwxyz' 'defghijklmnopqrstuvwxyz'] n = len(set1) m = len(set2) print(countCompletePairs(set1 set2 n m)) # This code is contributed by mohit kumar 29 

<script> // Javascript program to find count of complete pairs    // Returns count of complete pairs from set[0..n-1]  // and set2[0..m-1]  function countCompletePairs(set1set2nm)  {  let result = 0;    // con_s1[i] is going to store an integer whose  // set bits represent presence/absence of characters  // in string set1[i].  // Similarly con_s2[i] is going to store an integer  // whose set bits represent presence/absence of  // characters in string set2[i]  let con_s1 = new Array(n);  let con_s2 = new Array(m);    // Process all strings in set1[]  for (let i = 0; i < n; i++) {    // initializing all bits to 0  con_s1[i] = 0;  for (let j = 0; j < set1[i].length; j++) {    // Setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s1[i] = con_s1[i] |   (1 << (set1[i][j].charCodeAt(0) - 'a'.charCodeAt(0)));  }  }    // Process all strings in set2[]  for (let i = 0; i < m; i++) {    // initializing all bits to 0  con_s2[i] = 0;  for (let j = 0; j < set2[i].length; j++) {    // setting the ascii code of char s[i][j]  // to 1 in the compressed integer.  con_s2[i] = con_s2[i] |   (1 << (set2[i][j].charCodeAt(0) - 'a'.charCodeAt(0)));  }  }    // assigning a variable whose all 26 (0..25)  // bits are set to 1  let complete = (1 << 26) - 1;    // Now consider every pair of integer in con_s1[]  // and con_s2[] and check if the pair is complete.  for (let i = 0; i < n; i++) {  for (let j = 0; j < m; j++) {    // if all bits are set the strings are  // complete!  if ((con_s1[i] | con_s2[j]) == complete) {  result++;  }  }  }    return result;  }    // Driver code  let set1=['abcdefgh' 'geeksforgeeks'  'lmnopqrst' 'abc'];  let set2=['ijklmnopqrstuvwxyz'  'abcdefghijklmnopqrstuvwxyz'  'defghijklmnopqrstuvwxyz' ]  let n = set1.length;  let m = set2.length;  document.write(countCompletePairs(set1 set2 n m));    // This code is contributed by avanitrachhadiya2155   </script>