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

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

Below is the implementation of the above approach:

C++

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// 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<int, 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<int, int>::iterator h;
  
    // stroing 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
    // frequecny 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";
    }
}

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Python3

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# 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
  
    # Stroing 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
    # frequecny 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

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Output:

Yes

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

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Improved By : mohit kumar 29