Given a string str and an integer K, the task is to check if it is possible to make the string K palindromes using all the characters of the string exactly once.
Examples:
Input: str = “poor”, K = 3
Output: Yes
One way of getting 3 palindromes is: oo, p, r
Input: str = “fun”, K = 5
Output: No
2 palindromes can’t be constructed using 3 distinct letters. Hence not possible.
Approach:
- If the size of the string is less than k, getting k palindromes is not possible.
- If the size of the string is equal to k, getting k palindromes is always possible. We give 1 character to each of k strings and all of them are palindromes as a string of length 1 is always a palindrome.
- If the size of the string is greater than k then some strings might have more than 1 character.
- Create a hashmap to store the frequency of every character as the characters having even number of occurrences can be distributed in groups of 2.
- Check if the number of characters having an odd number of occurrences is less than or equal to k, we can say that k palindromes can always be generated else no.
Below is the implementation of the above approach:
// C++ program to find if string is K-Palindrome // or not using all characters exactly once #include <bits/stdc++.h> using namespace std;
void iskPalindromesPossible(string s, int k)
{ // when size of string is less than k
if (s.size() < k) {
cout << "Not Possible" << endl;
return ;
}
// when size of string is equal to k
if (s.size() == k) {
cout << "Possible" << endl;
return ;
}
// when size of string is greater than k
// to store the frequencies of the characters
map< char , int > freq;
for ( int i = 0; i < s.size(); i++)
freq[s[i]]++;
// to store the count of characters
// whose number of occurrences is odd.
int count = 0;
// iterating over the map
for ( auto it : freq) {
if (it.second % 2 == 1)
count++;
}
if (count > k)
cout << "No" << endl;
else
cout << "Yes" << endl;
} // Driver code int main()
{ string str = "poor" ;
int K = 3;
iskPalindromesPossible(str, K);
str = "geeksforgeeks" ;
K = 10;
iskPalindromesPossible(str, K);
return 0;
} |
// Java program to find if String // is K-Palindrome or not using // all characters exactly once import java.util.*;
class GFG{
static void iskPalindromesPossible(String s,
int k)
{ // When size of String is less than k
if (s.length() < k)
{
System.out.print( "Not Possible" + "\n" );
return ;
}
// When size of String is equal to k
if (s.length() == k)
{
System.out.print( "Possible" + "\n" );
return ;
}
// When size of String is greater than k
// to store the frequencies of the characters
HashMap<Character,
Integer> freq = new HashMap<Character,
Integer>();
for ( int i = 0 ; i < s.length(); i++)
if (freq.containsKey(s.charAt(i)))
{
freq.put(s.charAt(i),
freq.get(s.charAt(i)) + 1 );
}
else
{
freq.put(s.charAt(i), 1 );
}
// To store the count of characters
// whose number of occurrences is odd.
int count = 0 ;
// Iterating over the map
for (Map.Entry<Character,
Integer> it : freq.entrySet())
{
if (it.getValue() % 2 == 1 )
count++;
}
if (count > k)
System.out.print( "No" + "\n" );
else
System.out.print( "Yes" + "\n" );
} // Driver code public static void main(String[] args)
{ String str = "poor" ;
int K = 3 ;
iskPalindromesPossible(str, K);
str = "geeksforgeeks" ;
K = 10 ;
iskPalindromesPossible(str, K);
} } // This code is contributed by sapnasingh4991 |
# Find if string is K-Palindrome or not using all characters exactly once # Python 3 program to find if string is K-Palindrome # or not using all characters exactly once def iskPalindromesPossible(s, k):
# when size of string is less than k
if ( len (s)<k):
print ( "Not Possible" )
return
# when size of string is equal to k
if ( len (s) = = k):
print ( "Possible" )
return
# when size of string is greater than k
# to store the frequencies of the characters
freq = dict .fromkeys(s, 0 )
for i in range ( len (s)):
freq[s[i]] + = 1
#to store the count of characters
# whose number of occurrences is odd.
count = 0
# iterating over the map
for value in freq.values():
if (value % 2 = = 1 ):
count + = 1
if (count > k):
print ( "No" )
else :
print ( "Yes" )
# Driver code if __name__ = = '__main__' :
str1 = "poor"
K = 3
iskPalindromesPossible(str1, K)
str = "geeksforgeeks"
K = 10
iskPalindromesPossible( str , K)
# This code is contributed by Surendra_Gangwar |
// C# program to find if String // is K-Palindrome or not using // all characters exactly once using System;
using System.Collections.Generic;
class GFG{
static void iskPalindromesPossible(String s,
int k)
{ // When size of String is less than k
if (s.Length < k)
{
Console.Write( "Not Possible" + "\n" );
return ;
}
// When size of String is equal to k
if (s.Length == k)
{
Console.Write( "Possible" + "\n" );
return ;
}
// When size of String is greater than k
// to store the frequencies of the characters
Dictionary< int ,
int > freq = new Dictionary< int ,
int >();
for ( int i = 0; i < s.Length; i++)
if (freq.ContainsKey(s[i]))
{
freq[s[i]] = freq[s[i]] + 1;
}
else
{
freq.Add(s[i], 1);
}
// To store the count of characters
// whose number of occurrences is odd.
int count = 0;
// Iterating over the map
foreach (KeyValuePair< int , int > it in freq)
{
if (it.Value % 2 == 1)
count++;
}
if (count > k)
Console.Write( "No" + "\n" );
else
Console.Write( "Yes" + "\n" );
} // Driver code public static void Main(String[] args)
{ String str = "poor" ;
int K = 3;
iskPalindromesPossible(str, K);
str = "geeksforgeeks" ;
K = 10;
iskPalindromesPossible(str, K);
} } // This code is contributed by Princi Singh |
<Script> // Find if string is K-Palindrome or not using all characters exactly once // JavaScript program to find if string is K-Palindrome // or not using all characters exactly once function iskPalindromesPossible(s, k)
{ // when size of string is less than k
if (s.length < k)
{
document.write( "Not Possible" , "</br>" )
return
}
// when size of string is equal to k
if (s.length == k){
document.write( "Possible" , "</br>" )
return
}
// when size of string is greater than k
// to store the frequencies of the characters
let freq = new Map()
for (let i = 0; i < s.length; i++)
{
if (freq.has(s[i])){
freq.set(s[i],freq.get(s[i])+1);
}
freq.set(s[i],1);
}
// to store the count of characters
// whose number of occurrences is odd.
let count = 0
// iterating over the map
for (let [key,value] of freq){
if (value % 2 == 1)
count += 1
}
if (count > k)
document.write( "No" , "</br>" )
else
document.write( "Yes" , "</br>" )
} // Driver code let str1 = "poor"
let K = 3 iskPalindromesPossible(str1, K) let str = "geeksforgeeks"
K = 10 iskPalindromesPossible(str, K) // This code is contributed by shinjanpatra </script> |
Yes Yes
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ? O(1), no extra space is required, so it is a constant.