Rearrange the string to maximize the number of palindromic substrings

Given a string S consisting of lowercase characters(a-z) only, the task is to print a new string by rearranging the string in such a way that maximizes the number of palindromic substrings. In case of multiple answers, print any one.

Note: even if some substrings coincide, count them as many times as they appear in the obtained string.

Examples:

Input: s = “aabab”
Output: ababa
string “ababa” has 9 palindromic substrings: “a”, “b”, “a”, “b”, “a”, “aba”, “bab”, “aba”, “ababa”.

Input: s = “aa”
Output: aa
The given string has the maximum number of palindromic substrings possible, “a”, “a” and “aa”.



The problem might look to be a complex one but on solving for various cases and having observations will lead to an easy solution.

A simple solution is to sort the string and print it. Sorting takes O(N * log N).

An efficient solution is to count the frequency of each character using a freq[] array and then construct the string using the freq[] array.

Below is the implementation of the above approach.

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// C++ program to rearrange the
// string such to maximize the
// number of palindromic substrings
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the newString
string newString(string s)
{
    // length of string
    int l = s.length();
  
    // hashing array
    int freq[26] = { 0 };
  
    // iterate and count
    for (int i = 0; i < l; i++) {
        freq[s[i] - 'a'] += 1;
    }
  
    // resulting string
    string ans = "";
  
    // form the resulting string
    for (int i = 0; i < 26; i++) {
  
        // number of times character appears
        for (int j = 0; j < freq[i]; j++) {
  
            // append to resulting string
            ans += (char)(97 + i);
        }
    }
  
    return ans;
}
  
// Driver Code
int main()
{
  
    string s = "aabab";
  
    cout << newString(s);
  
    return 0;
}
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// Java program to rearrange the
// string such to maximize the
// number of palindromic substrings
  
  
public class GFG {
      
    // Function to return the newString
    static String newString(String s)
    {
        // length of string
        int l = s.length();
  
        // hashing array
        int freq[] = new int [26] ;
  
        // iterate and count
        for (int i = 0; i < l; i++) {
            freq[s.charAt(i) - 'a'] += 1;
        }
  
        // resulting string
        String ans = "";
  
        // form the resulting string
        for (int i = 0; i < 26; i++) {
  
            // number of times character appears
            for (int j = 0; j < freq[i]; j++) {
  
                // append to resulting string
                ans += (char)(97 + i);
            }
        }
  
        return ans;
    }
  
  
    // Driver code
    public static void main(String args[])
    {
           String s = "aabab";
            System.out.println(newString(s));
    }
    // This Code is contributed by ANKITRAI1
}
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# Python3 program to rearrange the 
# string such to maximize the 
# number of palindromic substrings 
  
# Function to return the newString 
def newString(s): 
  
    # length of string 
    l = len(s) 
  
    # hashing array 
    freq = [0] * (26
  
    # iterate and count 
    for i in range(0, l): 
        freq[ord(s[i]) - ord('a')] += 1
  
    # resulting string 
    ans = "" 
  
    # form the resulting string 
    for i in range(0, 26): 
  
        # number of times character appears 
        for j in range(0, freq[i]): 
  
            # append to resulting string 
            ans += chr(97 + i) 
      
    return ans 
  
# Driver Code 
if __name__ == "__main__":
  
    s = "aabab"
    print(newString(s))
  
# This code is contributed by Rituraj Jain
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// C# program to rearrange the
// string such to maximize the
// number of palindromic substrings
using System;
class GFG
{
  
// Function to return the newString
static String newString(string s)
{
    // length of string
    int l = s.Length;
  
    // hashing array
    int[] freq = new int [26];
  
    // iterate and count
    for (int i = 0; i < l; i++) 
    {
        freq[s[i] - 'a'] += 1;
    }
  
    // resulting string
    string ans = "";
  
    // form the resulting string
    for (int i = 0; i < 26; i++)
    {
  
        // number of times character appears
        for (int j = 0; j < freq[i]; j++)
        {
  
            // append to resulting string
            ans += (char)(97 + i);
        }
    }
  
    return ans;
}
  
  
// Driver code
public static void Main()
{
    string s = "aabab";
    Console.Write(newString(s));
}
}
  
// This code is contributed 
// by ChitraNayal
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Output:
aaabb

Time Complexity: O(N)
Auxiliary Space: O(26)





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