Form minimum number of Palindromic Strings from a given string
Last Updated :
26 Oct, 2023
Given a string S, the task is to divide the characters of S to form minimum number of palindromic strings.
Note: There can be multiple correct answers.
Examples:
Input: S = "geeksforgeeks"
Output: {eegksrskgee, o, f}
Explanation:
There should be at least 3 strings "eegksrskgee", "o", "f". All 3 formed strings are palindrome.
Input: S = "abbaa"
Output: {"ababa"}
Explanation:
The given string S = "ababa" is already a palindrome string so, only 1 string will be formed.
Input: S = "abc"
Output: {"a", "b", "c"}
Explanation:
There should be at least 3 strings "a", "b", "c". All 3 formed strings are palindrome.
Approach:
The main observation is that a palindrome string remains the same if reversed. The length of the strings formed does not matter so try to make a string as large as possible. If some character can't be a part of palindromic string, then push this character into a new string.
- The general approach would be to store the frequency of each character and if for some character the frequency is odd then we have to make a new string and increment our answer by 1.
- Note that if there are no characters with odd frequency, at least one string is still needed, so the final answer will be max(1, odd frequency characters).
- To print the string store odd characters frequency characters in a vector and even characters in another vector.
- Form one palindromic string with strings having odd characters and append a string with an even number of characters and all other strings will be of only one character.
Below is the implementation of the above approach:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return
// minimum palindromic strings formed
int minimumPalindromicStrings(string S)
{
// Store the length of string
int N = S.length();
// Iterate over the string
// and store the frequency of
// all characters A vector
// to store the frequency
vector<int> freq(26);
for (int i = 0; i < N; i++) {
freq[S[i] - 'a']++;
}
// Store the odd frequency characters
vector<int> oddFreqCharacters;
// Store the even frequency of characters
// in the same vector by dividing
// current frequency by 2 as one element
// will be used on left as well as right side
// Iterate through all possible characters
// and check the parity of frequency
for (int i = 0; i < 26; i++) {
if (freq[i] & 1) {
oddFreqCharacters.push_back(i);
freq[i]--;
}
freq[i] /= 2;
}
// store answer in a vector
vector<string> ans;
// if there are no odd Frequency characters
// then print the string with even characters
if (oddFreqCharacters.empty()) {
// store the left part
// of the palindromic string
string left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += char(i + 'a');
}
}
string right = left;
// reverse the right part of the string
reverse(right.begin(), right.end());
// store the string in ans vector
ans.push_back(left + right);
}
else {
// take any character
// from off frequency element
string middle = "";
int c = oddFreqCharacters.back();
oddFreqCharacters.pop_back();
middle += char(c + 'a');
// repeat the above step to form
// left and right strings
string left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += char(i + 'a');
}
}
string right = left;
// reverse the right part of the string
reverse(right.begin(), right.end());
// store the string in ans vector
ans.push_back(left + middle + right);
// store all other odd frequency strings
while (!oddFreqCharacters.empty()) {
int c = oddFreqCharacters.back();
oddFreqCharacters.pop_back();
string middle = "";
middle += char(c + 'a');
ans.push_back(middle);
}
}
// Print the answer
cout << "[";
for (int i = 0; i < ans.size(); i++) {
cout << ans[i];
if (i != ans.size() - 1) {
cout << ", ";
}
}
cout << "]";
return 0;
}
// Driver Code
int main()
{
string S = "geeksforgeeks";
minimumPalindromicStrings(S);
}
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
// Function to return
// minimum palindromic Strings formed
static int minimumPalindromicStrings(String S)
{
// Store the length of String
int N = S.length();
// Iterate over the String
// and store the frequency of
// all characters A vector
// to store the frequency
int[] freq = new int[26];
for (int i = 0; i < N; i++) {
freq[S.charAt(i) - 'a']++;
}
// Store the odd frequency characters
Vector<Integer> oddFreqCharacters = new Vector<Integer>();
// Store the even frequency of characters
// in the same vector by dividing
// current frequency by 2 as one element
// will be used on left as well as right side
// Iterate through all possible characters
// and check the parity of frequency
for (int i = 0; i < 26; i++) {
if (freq[i] % 2 == 1) {
oddFreqCharacters.add(i);
freq[i]--;
}
freq[i] /= 2;
}
// store answer in a vector
Vector<String> ans = new Vector<String>();
// if there are no odd Frequency characters
// then print the String with even characters
if (oddFreqCharacters.isEmpty()) {
// store the left part
// of the palindromic String
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.add(left + right);
}
else {
// take any character
// from off frequency element
String middle = "";
int c = oddFreqCharacters.lastElement();
oddFreqCharacters.remove(oddFreqCharacters.size()-1);
middle += (char)(c + 'a');
// repeat the above step to form
// left and right Strings
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.add(left + middle + right);
// store all other odd frequency Strings
while (!oddFreqCharacters.isEmpty()) {
c = oddFreqCharacters.lastElement();
oddFreqCharacters.remove(oddFreqCharacters.size()-1);
middle = "";
middle += (char)(c + 'a');
ans.add(middle);
}
}
// Print the answer
System.out.print("[");
for (int i = 0; i < ans.size(); i++) {
System.out.print(ans.get(i));
if (i != ans.size() - 1) {
System.out.print(", ");
}
}
System.out.print("]");
return 0;
}
static String reverse(String input) {
char[] a = input.toCharArray();
int l, r = a.length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.valueOf(a);
}
// Driver Code
public static void main(String[] args)
{
String S = "geeksforgeeks";
minimumPalindromicStrings(S);
}
}
// This code is contributed by 29AjayKumar
def minimum_palindromic_strings(s):
# Store the length of the string
n = len(s)
# Iterate over the string and store the frequency of all characters in a dictionary
freq = {}
for char in s:
if char in freq:
freq[char] += 1
else:
freq[char] = 1
# Store the odd frequency characters
odd_freq_characters = []
# Store the even frequency of characters in the same dictionary
# by dividing current frequency by 2
# as one element will be used on the left as well as the right side
# Iterate through all possible characters and check the parity of frequency
for char in freq:
if freq[char] % 2 == 1:
odd_freq_characters.append(char)
freq[char] -= 1
freq[char] //= 2
# Store answers in a list
ans = []
# If there are no odd frequency characters, then print the string with even characters
if not odd_freq_characters:
# Store the left part of the palindromic string
left = ""
for char in sorted(freq):
left += char * freq[char]
right = left[::-1] # Reverse the right part of the string
# Store the string in the ans list
ans.append(left + right)
else:
# Take any character from off-frequency elements
middle = odd_freq_characters.pop()
middle = middle
# Repeat the above step to form left and right strings
left = ""
for char in sorted(freq):
left += char * freq[char]
right = left[::-1] # Reverse the right part of the string
# Store the string in the ans list
ans.append(left + middle + right)
# Store all other odd frequency strings
for char in odd_freq_characters:
ans.append(char)
# Print the answer
print(ans)
# Driver Code
if __name__ == "__main__":
S = "geeksforgeeks"
minimum_palindromic_strings(S)
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
public class GFG{
// Function to return
// minimum palindromic Strings formed
static int minimumPalindromicStrings(String S)
{
// Store the length of String
int N = S.Length;
// Iterate over the String
// and store the frequency of
// all characters A vector
// to store the frequency
int[] freq = new int[26];
for (int i = 0; i < N; i++) {
freq[S[i] - 'a']++;
}
// Store the odd frequency characters
List<int> oddFreqchars = new List<int>();
// Store the even frequency of characters
// in the same vector by dividing
// current frequency by 2 as one element
// will be used on left as well as right side
// Iterate through all possible characters
// and check the parity of frequency
for (int i = 0; i < 26; i++) {
if (freq[i] % 2 == 1) {
oddFreqchars.Add(i);
freq[i]--;
}
freq[i] /= 2;
}
// store answer in a vector
List<String> ans = new List<String>();
// if there are no odd Frequency characters
// then print the String with even characters
if (oddFreqchars.Count==0) {
// store the left part
// of the palindromic String
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.Add(left + right);
}
else {
// take any character
// from off frequency element
String middle = "";
int c = oddFreqchars[oddFreqchars.Count-1];
oddFreqchars.RemoveAt(oddFreqchars.Count-1);
middle += (char)(c + 'a');
// repeat the above step to form
// left and right Strings
String left = "";
for (int i = 0; i < 26; i++) {
for (int j = 0; j < freq[i]; j++) {
left += (char)(i + 'a');
}
}
String right = left;
// reverse the right part of the String
right = reverse(right);
// store the String in ans vector
ans.Add(left + middle + right);
// store all other odd frequency Strings
while (oddFreqchars.Count!=0) {
c = oddFreqchars[oddFreqchars.Count-1];
oddFreqchars.RemoveAt(oddFreqchars.Count-1);
middle = "";
middle += (char)(c + 'a');
ans.Add(middle);
}
}
// Print the answer
Console.Write("[");
for (int i = 0; i < ans.Count; i++) {
Console.Write(ans[i]);
if (i != ans.Count - 1) {
Console.Write(", ");
}
}
Console.Write("]");
return 0;
}
static String reverse(String input) {
char[] a = input.ToCharArray();
int l, r = a.Length - 1;
for (l = 0; l < r; l++, r--) {
char temp = a[l];
a[l] = a[r];
a[r] = temp;
}
return String.Join("",a);
}
// Driver Code
public static void Main(String[] args)
{
String S = "geeksforgeeks";
minimumPalindromicStrings(S);
}
}
// This code is contributed by 29AjayKumar
function minimumPalindromicStrings(S) {
// Store the length of the string
const N = S.length;
// Iterate over the string and store the frequency of all characters in an array
const freq = new Array(26).fill(0);
for (let i = 0; i < N; i++) {
freq[S.charCodeAt(i) - 'a'.charCodeAt(0)]++;
}
// Store the odd frequency characters
const oddFreqCharacters = [];
// Iterate through all possible characters and check the parity of frequency
for (let i = 0; i < 26; i++) {
if (freq[i] & 1) {
oddFreqCharacters.push(i);
freq[i]--;
}
freq[i] /= 2;
}
// Store answers in an array
const ans = [];
// If there are no odd frequency characters, then print the string with even characters
if (oddFreqCharacters.length === 0) {
// Store the left part of the palindromic string
let left = "";
for (let i = 0; i < 26; i++) {
for (let j = 0; j < freq[i]; j++) {
left += String.fromCharCode(i + 'a'.charCodeAt(0));
}
}
const right = left.split("").reverse().join(""); // Reverse the right part of the string
// Store the string in the ans array
ans.push(left + right);
} else {
// Take any character from off-frequency elements
let middle = "";
const c = oddFreqCharacters.pop();
middle += String.fromCharCode(c + 'a'.charCodeAt(0));
// Repeat the above step to form left and right strings
let left = "";
for (let i = 0; i < 26; i++) {
for (let j = 0; j < freq[i]; j++) {
left += String.fromCharCode(i + 'a'.charCodeAt(0));
}
}
const right = left.split("").reverse().join(""); // Reverse the right part of the string
// Store the string in the ans array
ans.push(left + middle + right);
// Store all other odd frequency strings
while (oddFreqCharacters.length > 0) {
const c = oddFreqCharacters.pop();
let middle = "";
middle += String.fromCharCode(c + 'a'.charCodeAt(0));
ans.push(middle);
}
}
// Print the answer
console.log(ans);
}
// Driver Code
const S = "geeksforgeeks";
minimumPalindromicStrings(S);
Output[eegksrskgee, o, f]
Time Complexity: O(N*26)
Space Complexity: O(N)
Similar Reads
DSA Tutorial - Learn Data Structures and Algorithms DSA (Data Structures and Algorithms) is the study of organizing data efficiently using data structures like arrays, stacks, and trees, paired with step-by-step procedures (or algorithms) to solve problems effectively. Data structures manage how data is stored and accessed, while algorithms focus on
7 min read
Quick Sort QuickSort is a sorting algorithm based on the Divide and Conquer that picks an element as a pivot and partitions the given array around the picked pivot by placing the pivot in its correct position in the sorted array. It works on the principle of divide and conquer, breaking down the problem into s
12 min read
Merge Sort - Data Structure and Algorithms Tutorials Merge sort is a popular sorting algorithm known for its efficiency and stability. It follows the divide-and-conquer approach. It works by recursively dividing the input array into two halves, recursively sorting the two halves and finally merging them back together to obtain the sorted array. Merge
14 min read
Data Structures Tutorial Data structures are the fundamental building blocks of computer programming. They define how data is organized, stored, and manipulated within a program. Understanding data structures is very important for developing efficient and effective algorithms. What is Data Structure?A data structure is a st
2 min read
Bubble Sort Algorithm Bubble Sort is the simplest sorting algorithm that works by repeatedly swapping the adjacent elements if they are in the wrong order. This algorithm is not suitable for large data sets as its average and worst-case time complexity are quite high.We sort the array using multiple passes. After the fir
8 min read
Breadth First Search or BFS for a Graph Given a undirected graph represented by an adjacency list adj, where each adj[i] represents the list of vertices connected to vertex i. Perform a Breadth First Search (BFS) traversal starting from vertex 0, visiting vertices from left to right according to the adjacency list, and return a list conta
15+ min read
Binary Search Algorithm - Iterative and Recursive Implementation Binary Search Algorithm is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(log N). Binary Search AlgorithmConditions to apply Binary Searc
15 min read
Insertion Sort Algorithm Insertion sort is a simple sorting algorithm that works by iteratively inserting each element of an unsorted list into its correct position in a sorted portion of the list. It is like sorting playing cards in your hands. You split the cards into two groups: the sorted cards and the unsorted cards. T
9 min read
Array Data Structure Guide In this article, we introduce array, implementation in different popular languages, its basic operations and commonly seen problems / interview questions. An array stores items (in case of C/C++ and Java Primitive Arrays) or their references (in case of Python, JS, Java Non-Primitive) at contiguous
4 min read
Sorting Algorithms A Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
3 min read