Check if string can be rearranged so that every Odd length Substring is Palindrome

Given a string S. The task is to check whether it is possible to rearrange the string such that every substring of odd length is a palindrome.

Examples:

Input: S = “oiooi”
Output: YES
The string can be rearranged as “oioio”

Input: S = “yuyuo”
Output: NO

Approach:



  • The very first observation is if all the characters of the string are same then every substring of odd length is a palindrome and we do not need to rearrange them.
  • Second observation is if the number of distinct characters are more than 2 then it is impossible to rearrange.
  • Now if the number of distinct characters are exactly 2 then to get all odd length substring to be a palindrome, the difference of their count must be less than or equal to 1, and if this satisfy then we rearrange the string in alternate manner means s_{i-1}=s_{i+1} for i <— ( 1 to n – 1 ). Where n is the length of the string.

Below is the implementation of above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of
// the above approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to check is it
// possible to rearrange the string
// such that every odd length
// substring is palindrome
bool IsPossible(string s)
{
  
    // Length of the string
    int n = s.length();
  
    // To count number of distinct
    // character in string
    set<char> count;
  
    // To count frequency of
    // each character
    map<char, int> map;
  
    for (int i = 0; i < n; i++) {
  
        // Inserting into set
        count.insert(s[i]);
  
        // Incrementing the frequency
        map[s[i]] += 1;
    }
  
    // All characters in
    // the string are same
    if (count.size() == 1) {
        return true;
    }
  
    // There are more than 2 different
    // character in string
    if (count.size() > 2) {
        return false;
    }
  
    // Currently there is 2 different
    // character in string
    auto it = count.begin();
  
    // Get the frequencies of the
    // characters that present
    // in string
    int x = 0, y = 0;
    x = map[*it];
  
    it++;
    y = map[*it];
  
    // Difference between their
    // count is less than or
    // equal to 1
    if (abs(x - y) <= 1) {
        return true;
    }
  
    return false;
}
  
// Driver code
int main()
{
    string s = "aaaddad";
  
    if (IsPossible(s))
        cout << "YES\n";
    else
        cout << "NO\n";
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.util.*;
  
class GFG 
{
  
    // Function to check is it
    // possible to rearrange the string
    // such that every odd length
    // substring is palindrome
    static boolean IsPossible(String s)
    {
  
        // Length of the string
        int n = s.length();
  
        // To count number of distinct
        // character in string
        HashSet<Character> count = new HashSet<>();
  
        // To count frequency of
        // each character
        HashMap<Character, Integer> map = new HashMap<>();
  
        for (int i = 0; i < n; i++) 
        {
  
            // Inserting into set
            count.add(s.charAt(i));
  
            // Incrementing the frequency
            map.put(s.charAt(i), map.get(s.charAt(i)) == 
                    null ? 1 : map.get(s.charAt(i)) + 1);
        }
  
        // All characters in
        // the string are same
        if (count.size() == 1)
            return true;
  
        // There are more than 2 different
        // character in string
        if (count.size() > 2)
            return false;
  
        String newString = count.toArray().toString();
  
        // Currently there is 2 different
        // character in string
        int j = 0;
        char it = newString.charAt(j);
  
        // Get the frequencies of the
        // characters that present
        // in string
        int x = 0, y = 0;
        x = map.get(it) == null ? 0 : map.get(it);
        j++;
  
        it = newString.charAt(j);
        y = map.get(it) == null ? 0 : map.get(it);
  
        // Difference between their
        // count is less than or
        // equal to 1
        if (Math.abs(x - y) <= 1)
            return true;
        return false;
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        String s = "aaaddad";
        if (IsPossible(s))
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}
  
// This code is contributed by
// sanjeev2552

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of 
# the above approach 
  
# Function to check is it 
# possible to rearrange the string 
# such that every odd length 
# substring is palindrome 
def IsPossible(s) : 
  
    # Length of the string 
    n = len(s); 
  
    # To count number of distinct 
    # character in string 
    count = set();
      
    # To count frequency of
    # each character
    map = dict.fromkeys(s, 0); 
  
    for i in range(n) : 
  
        # Inserting into set 
        count.add(s[i]); 
  
        # Incrementing the frequency 
        map[s[i]] += 1
  
    # All characters in 
    # the string are same 
    if (len(count) == 1) :
        return True
  
    # There are more than 2 different 
    # character in string 
    if (len(count) > 2) :
        return False
          
    # Currently there is 2 different 
    # character in string 
    j = 0
    it = list(count)[j]; 
  
    # Get the frequencies of the 
    # characters that present 
    # in string 
    x = 0; y = 0
    x = map[it];
      
    j += 1
    it = list(count)[j]; 
    y = map[it]; 
  
    # Difference between their 
    # count is less than or 
    # equal to 1 
    if (abs(x - y) <= 1) :
        return True
  
    return False
  
# Driver code 
if __name__ == "__main__"
  
    s = "aaaddad"
  
    if (IsPossible(s)) :
        print("YES"); 
    else :
        print("NO"); 
          
# This code is contributed by AnkitRai01

chevron_right


Output:

YES

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.




My Personal Notes arrow_drop_up

Recommended Posts:


Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : AnkitRai01, sanjeev2552