Given two strings. The task is to check that is there any common character in between two strings.
Examples:
Input: s1 = "geeksforgeeks", s2 = "geeks" Output: Yes Input: s1 = "geeks", s2 = "for" Output: No
Approach: Traverse the 1st string and map the characters of the string with its frequency, in this map characters act as a key and the frequency its value. Then traverse the second string and we will check if there is any character that is present in both the string then it is confirmed that there is a common sub-sequence.
Below is the implementation of above approach:
// C++ implementation of above approach #include <bits/stdc++.h> using namespace std;
// Function to match character bool check(string s1, string s2)
{ // Create a map to map
// characters of 1st string
map< char , int > map;
// traverse the first string
// and create a hash map
for ( int i = 0; i < s1.length(); i++)
map[s1[i]]++;
// traverse the second string
// and if there is any
// common character than return 1
for ( int i = 0; i < s2.length(); i++)
if (map[s2[i]] > 0)
return true ;
// else return 0
return false ;
} // Driver code int main()
{ // Declare two strings
string s1 = "geeksforgeeks" , s2 = "geeks" ;
// Find if there is a common subsequence
bool yes_or_no = check(s1, s2);
if (yes_or_no == true )
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
} |
// Java implementation of above approach import java.util.*;
class GFG
{ // Function to match character static boolean check(String s1, String s2)
{ // Create a map to map
// characters of 1st string
Map<Character, Integer> mp = new HashMap<>();
// traverse the first string
// and create a hash map
for ( int i = 0 ; i < s1.length(); i++)
{
mp.put(s1.charAt(i), mp.get(s1.charAt(i)) == null ? 1 : mp.get(s1.charAt(i)) + 1 );
}
// traverse the second string
// and if there is any
// common character than return 1
for ( int i = 0 ; i < s2.length(); i++)
{
if (mp.get(s2.charAt(i)) > 0 )
{
return true ;
}
}
// else return 0
return false ;
} // Driver code public static void main(String[] args)
{ // Declare two strings
String s1 = "geeksforgeeks" , s2 = "geeks" ;
// Find if there is a common subsequence
boolean yes_or_no = check(s1, s2);
if (yes_or_no == true )
{
System.out.println( "Yes" );
}
else
{
System.out.println( "No" );
}
} } /* This code contributed by PrinciRaj1992 */ |
# Python3 program to check whether # two lists are overlapping or not def is_member( List , key):
for i in range ( 0 , len ( List )):
if key = = List [i]:
return True
return False
def overlap(List1 , List2):
for key in List1:
if is_member( List2, key ):
return True
return False
# Driver Code if __name__ = = '__main__' :
s1 = 'geeksforgeeks'
s2 = 'geeks'
List1 = list ( s1 )
List2 = list ( s2 )
yes_or_no = str (overlap( List1, List2 ))
if (yes_or_no):
print ( "Yes" )
else :
print ( "No" )
# This code is contributed # by Krishna_Yadav |
// C# program to check if successive // pair of numbers in the queue are // consecutive or not using System;
using System.Collections.Generic;
class GFG
{ // Function to match character static Boolean check(String s1, String s2)
{ // Create a map to map
// characters of 1st string
Dictionary< char , int > mp = new Dictionary< char , int >();
// traverse the first string
// and create a hash map
for ( int i = 0; i < s1.Length; i++)
{
if (mp.ContainsKey(s1[i]))
{
var val = mp[s1[i]];
mp.Remove(s1[i]);
mp.Add(s1[i], val + 1);
}
else
{
mp.Add(s1[i], 1);
}
}
// traverse the second string
// and if there is any
// common character than return 1
for ( int i = 0; i < s2.Length; i++)
{
if (mp[s2[i]] > 0)
{
return true ;
}
}
// else return 0
return false ;
} // Driver code public static void Main(String[] args)
{ // Declare two strings
String s1 = "geeksforgeeks" , s2 = "geeks" ;
// Find if there is a common subsequence
Boolean yes_or_no = check(s1, s2);
if (yes_or_no == true )
{
Console.WriteLine( "Yes" );
}
else
{
Console.WriteLine( "No" );
}
} } // This code contributed by Rajput-Ji |
<script> // Javascript implementation of above approach // Function to match character function check( s1, s2)
{ // Create a map to map
// characters of 1st string
var map = new Map();
// traverse the first string
// and create a hash map
for ( var i = 0; i < s1.length; i++)
{
if (map.has(s1[i].charCodeAt(0)))
{
map[s1[i].charCodeAt(0)]++;
}
else
{
map[s1[i].charCodeAt(0)]=1;
}
}
// traverse the second string
// and if there is any
// common character than return 1
for ( var i = 0; i < s2.length; i++)
if (map[s2[i].charCodeAt(0)] > 0)
return true ;
// else return 0
return false ;
} // Driver code // Declare two strings var s1 = "geeksforgeeks" , s2 = "geeks" ;
// Find if there is a common subsequence var yes_or_no = check(s1, s2);
if (yes_or_no)
document.write( "Yes" );
else document.write( "No" );
</script> |
Yes
Time Complexity: O(n) where n is the length of the string.
Auxiliary Space: O(n) where n is the length of the string.
Another Approach:
Convert str1 and str2 into sets of characters using the set() function.
Find the intersection of the two sets using the intersection() method, which returns a new set containing only the elements that are common to both sets.
Check if the resulting set common has at least one element using the len() function.
If the length of common is greater than 0, return True, indicating that there is at least one common character between str1 and str2.
If the length of common is 0, return False, indicating that there are no common characters between str1 and str2.
#include <iostream> #include <string> #include <unordered_set> using namespace std;
// Function to check if two strings have at least one common character bool commonChars(string str1, string str2)
{ // Convert the strings into sets of characters
unordered_set< char > set1(str1.begin(), str1.end());
unordered_set< char > set2(str2.begin(), str2.end());
// Find the intersection of the sets
for ( char c : set1) {
if (set2.count(c)) {
return true ;
}
}
return false ;
} // Example usage int main() {
string string1 = "hello" ;
string string2 = "world" ;
if (commonChars(string1, string2)) {
cout << "Yes" << endl;
} else {
cout << "No" << endl;
}
return 0;
} |
import java.util.*;
public class Main {
// Function to check if two strings have at least one
// common character
public static boolean commonChars(String str1,
String str2)
{
// Convert the strings into sets of characters
Set<Character> set1 = new HashSet<>();
for ( char c : str1.toCharArray()) {
set1.add(c);
}
Set<Character> set2 = new HashSet<>();
for ( char c : str2.toCharArray()) {
set2.add(c);
} // Find the intersection of the sets
for ( char c : set1) {
if (set2.contains(c)) {
return true ;
}
}
return false ;
}
// Example usage
public static void main(String[] args)
{
String string1 = "hello" ;
String string2 = "world" ;
if (commonChars(string1, string2)) {
System.out.println(
"Yes" );
}
else {
System.out.println(
"No" );
}
}
} |
def commonChars(str1, str2):
# Convert the strings into sets of characters
set1 = set (str1)
set2 = set (str2)
# Find the intersection of the sets
common = set1.intersection(set2)
if len (common) > 0 :
return True
else :
return False
# Example usage string1 = "hello"
string2 = "world"
if commonChars(string1, string2):
print ( "Yes" )
else :
print ( "No" )
|
function commonChars(str1, str2) {
// Convert the strings into sets of characters
let set1 = new Set(str1);
let set2 = new Set(str2);
// Find the intersection of the sets
let common = new Set([...set1].filter(x => set2.has(x)));
if (common.size > 0) {
return true ;
} else {
return false ;
}
} // Example usage let string1 = "hello" ;
let string2 = "world" ;
if (commonChars(string1, string2)) {
console.log( "Yes" );
} else {
console.log( "No" );
} |
using System;
using System.Collections.Generic;
class Program {
// Function to check if two strings have at least one
// common character
static bool CommonChars( string str1, string str2)
{
// Convert the strings into sets of characters
HashSet< char > set1 = new HashSet< char >(str1);
HashSet< char > set2 = new HashSet< char >(str2);
// Find the intersection of the sets
foreach ( char c in set1)
{
if (set2.Contains(c)) {
return true ;
}
}
return false ;
}
// Example usage
static void Main( string [] args)
{
string string1 = "hello" ;
string string2 = "world" ;
if (CommonChars(string1, string2)) {
Console.WriteLine( "Yes" );
}
else {
Console.WriteLine( "No" );
}
}
} |
Yes
Time complexity:
The time complexity of the set() function used to convert each input string into a set of characters is O(n), where n is the length of the input string.
The time complexity of the intersection() method used to find the common characters between the two sets is O(min(m, n)), where m and n are the sizes of the input sets.
The len() function used to check the size of the resulting set has a time complexity of O(1).
Overall, the time complexity of the function is O(n+m), where n and m are the lengths of the input strings.
Space complexity:
The space complexity of the function is O(n+m), where n and m are the lengths of the input strings. This is because the function creates two sets, each of which stores up to n and m characters, respectively.