Check if K palindromic strings can be formed from a given string

Last Updated : 15 Jul, 2025

Given a string S of size N and an integer K, the task is to find whether the characters of the string can be arranged to make K palindromic strings simultaneously.
Examples:

Input: S = "annabelle", K = 2
Output: Yes
Explanation:
All characters of string S can be distributed into "elble" and "anna" which are both palindromic.
Input: S ="abcd", K = 4
Output: Yes
Explanation:
Partition all characters of string as single character.

Approach

If the frequency of every character is even and K lies between 1 and N then it is always possible to form K palindrome strings.
But if there are some characters (say odd_count) with odd frequency, then K must lie between odd_count and N for K palindromic strings to be possible.

Hence, follow the steps below to solve the problem:

If K exceeds the length of the string, straightaway print “No”.

Store the frequency of all characters in a Map.
Count the number of characters having an odd frequency.
If the count is less than given K, then print “No”. Otherwise, print “Yes”.

Below is the implementation of the above approach:

C++

// C++ program to check 
// whether the string is 
// K palindrome or not 
#include <iostream> 
#include <map> 
using namespace std; 

// function to check 
// whether the string is 
// K palindrome or not 
bool can_Construct(string S, int K) 
{ 
    // map to frequency of character 
    map<char, int> m; 

    int i = 0, j = 0, p = 0; 

    // Check when k is given 
    // as same as length of string 
    if (S.length() == K) { 
        return true; 
    } 

    // iterator for map 
    map<char, int>::iterator h; 

    // storing the frequency of every 
    // character in map 
    for (i = 0; i < S.length(); i++) { 
        m[S[i]] = m[S[i]] + 1; 
    } 

    // if K is greater than size 
    // of string then return false 
    if (K > S.length()) { 
        return false; 
    } 

    else { 

        // check that number of character 
        // having the odd frequency 
        for (h = m.begin(); h != m.end(); h++) { 

            if (m[h->first] % 2 != 0) { 
                p = p + 1; 
            } 
        } 
    } 

    // if k is less than number of odd 
    // frequency character then it is 
    // again false other wise true 
    if (K < p) { 
        return false; 
    } 

    return true; 
} 

// Driver code 
int main() 
{ 
    string S = "annabelle"; 
    int K = 4; 

    if (can_Construct(S, K)) { 
        cout << "Yes"; 
    } 
    else { 
        cout << "No"; 
    } 
}

Java

// Java program to check 
// whether the string is 
// K palindrome or not 
import java.util.*;

class GFG{

// Function to check whether 
// the string is K palindrome or not 
static boolean can_Construct(String S, int K)
{ 
    
    // Map to frequency of character 
    Map<Character, Integer> m = new HashMap<>(); 
    
    int p = 0;
    
    // Check when k is given 
    // as same as length of string 
    if (S.length() == K) 
        return true;
    
    // Storing the frequency of every 
    // character in map 
    for(int i = 0; i < S.length(); i++)
        m.put(S.charAt(i),
              m.getOrDefault(S.charAt(i), 0) + 1);
              
    // If K is greater than size 
    // of then return false 
    if (K > S.length()) 
        return false;
    
    else
    { 
        
        // Check that number of character 
        // having the odd frequency 
        for(Integer h : m.values())
        {
            if (h % 2 != 0) 
                p = p + 1;
        }
    }
    
    // If k is less than number of odd 
    // frequency character then it is 
    // again false otherwise true 
    if (K < p) 
        return false;
    
    return true;
}

// Driver Code
public static void main (String[] args)
{
    String S = "annabelle";
    int K = 4;
    
    if (can_Construct(S, K))
        System.out.println("Yes"); 
    else 
        System.out.println("No");
}
}

// This code is contributed by offbeat

Python3

# Python3 program to check whether 
# the is K palindrome or not 

# Function to check whether 
# the is K palindrome or not 
def can_Construct(S, K): 
    
    # map to frequency of character 
    m = dict() 
    p = 0
    
    # Check when k is given 
    # as same as length of string 
    if (len(S) == K): 
        return True

    # Storing the frequency of every 
    # character in map 
    for i in S: 
        m[i] = m.get(i, 0) + 1

    # If K is greater than size 
    # of then return false 
    if (K > len(S)): 
        return False

    else: 

        # Check that number of character 
        # having the odd frequency 
        for h in m: 
            if (m[h] % 2 != 0): 
                p = p + 1

    # If k is less than number of odd 
    # frequency character then it is 
    # again false otherwise true 
    if (K < p): 
        return False

    return True

# Driver code 
if __name__ == '__main__': 
    
    S = "annabelle"
    K = 4

    if (can_Construct(S, K)): 
        print("Yes") 
    else: 
        print("No") 

# This code is contributed by mohit kumar 29

// C# program to check 
// whether the string is 
// K palindrome or not 
using System;
using System.Collections.Generic;
class GFG{

// Function to check whether 
// the string is K palindrome or not 
static bool can_Construct(String S, 
                          int K)
{     
  // Map to frequency of character 
  Dictionary<char, 
             int> m = new Dictionary<char, 
                                     int>(); 
  int p = 0;

  // Check when k is given 
  // as same as length of string 
  if (S.Length == K) 
    return true;

  // Storing the frequency of every 
  // character in map 
  for(int i = 0; i < S.Length; i++)
    if(!m.ContainsKey(S[i]))
      m.Add(S[i], 1);
  else
    m[S[i]] = m[S[i]] + 1;

  // If K is greater than size 
  // of then return false 
  if (K > S.Length) 
    return false;

  else
  { 
    // Check that number of character 
    // having the odd frequency 
    foreach(int h in m.Values)
    {
      if (h % 2 != 0) 
        p = p + 1;
    }
  }

  // If k is less than number of odd 
  // frequency character then it is 
  // again false otherwise true 
  if (K < p) 
    return false;

  return true;
}

// Driver Code
public static void Main(String[] args)
{
  String S = "annabelle";
  int K = 4;
  if (can_Construct(S, K))
    Console.WriteLine("Yes"); 
  else 
    Console.WriteLine("No");
}
}

// This code is contributed by shikhasingrajput

JavaScript

<script>

// JavaScript program to check 
// whether the string is 
// K palindrome or not 

// function to check 
// whether the string is 
// K palindrome or not 
function can_Construct(S, K) 
{ 
    // map to frequency of character 
    var m = new Map(); 

    var i = 0, j = 0, p = 0; 

    // Check when k is given 
    // as same as length of string 
    if (S.length == K) { 
        return true; 
    } 


    // storing the frequency of every 
    // character in map 
    for (i = 0; i < S.length; i++) { 
        if(m.has(S[i]))
            m.set(S[i], m.get(S[i])+1)
        else    
            m.set(S[i], 1)

    } 

    // if K is greater than size 
    // of string then return false 
    if (K > S.length) { 
        return false; 
    } 

    else { 

        // check that number of character 
        // having the odd frequency 
        m.forEach((value, key) => {
            

            if (value%2 != 0) { 
                p = p + 1; 
            } 
        }); 
    } 

    // if k is less than number of odd 
    // frequency character then it is 
    // again false other wise true 
    if (K < p) { 
        return false; 
    } 

    return true; 
} 

// Driver code 
var S = "annabelle"; 
var K = 4; 
if (can_Construct(S, K)) { 
    document.write( "Yes"); 
} 
else { 
    document.write( "No"); 
} 


</script>

Output

Yes

Time Complexity: O (N).
Auxiliary Space: O (N).

Analysis of Algorithms

vabzcode12

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Practice Tags :

palindrome

Check if K palindromic strings can be formed from a given string

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