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++

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// 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;
}

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Python3

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# 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

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Output:

YES


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Improved By : AnkitRai01