Pairs of complete strings in two sets of strings
Last Updated :
14 Apr, 2023
Two strings are said to be complete if on concatenation, they contain all the 26 English alphabets. For example, "abcdefghi" and "jklmnopqrstuvwxyz" are complete as they together have all characters from 'a' to 'z'.
We are given two sets of sizes n and m respectively and we need to find the number of pairs that are complete on concatenating each string from set 1 to each string from set 2.
Input : set1[] = {"abcdefgh", "geeksforgeeks",
"lmnopqrst", "abc"}
set2[] = {"ijklmnopqrstuvwxyz",
"abcdefghijklmnopqrstuvwxyz",
"defghijklmnopqrstuvwxyz"}
Output : 7
The total complete pairs that are forming are:
"abcdefghijklmnopqrstuvwxyz"
"abcdefghabcdefghijklmnopqrstuvwxyz"
"abcdefghdefghijklmnopqrstuvwxyz"
"geeksforgeeksabcdefghijklmnopqrstuvwxyz"
"lmnopqrstabcdefghijklmnopqrstuvwxyz"
"abcabcdefghijklmnopqrstuvwxyz"
"abcdefghijklmnopqrstuvwxyz"
Method 1 (Naive method): A simple solution is to consider all pairs of strings, concatenate them and then check if the concatenated string has all the characters from 'a' to 'z' by using a frequency array.
Implementation:
C++
// C++ implementation for find pairs of complete
// strings.
#include <iostream>
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(set1,set2,n,m):
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(set1,set2,n,m)
{
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>
Time Complexity: O(n * m * k)
Auxiliary Space: O(1)
Method 2 (Optimized method using Bit Manipulation): In this method, we compress frequency array into an integer. We assign each bit of that integer with a character and we set it to 1 when the character is found. We perform this for all the strings in both the sets. Finally we just compare the two integers in the sets and if on combining all the bits are set, they form a complete string pair.
Implementation:
C++14
// C++ program to find count of complete pairs
#include <iostream>
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(set1,set2,n,m)
{
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>
Time Complexity: O(n*m), where n is the size of the first set and m is the size of the second set.
Auxiliary Space: O(n)
This article is contributed by Rishabh Jain.
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