Split a string in equal parts such that all parts are palindromes
Last Updated :
19 Oct, 2022
Given a string str, the task is to split the string into minimum parts such that each part is of same length and each part is a palindrome. Print the required number of parts.
Examples:
Input: str = "civicbob"
Output: 8
"b", "b", "c", "c", "i", "i", "v" and "o" are the required partitions. "civic" and "bob" are also palindromes but they are not of equal length
Input: str = "noonpeep"
Output: 1
Approach:
- Count the number of odd appearing characters and store it in countOdd.
- Sum the frequencies of all the even occurring characters and store it in sumEven.
- Since, no more than one odd frequency character can be a part of the same palindrome. So, if (sumEven / 2) % countOdd = 0 then the answer is countOdd as sumEven can be divided into countOdd parts.
- Else, we can still divide the odd occurring characters into the palindrome pairs. For example, if the character a appears 3 times i.e. aaa then aa can be a part of some palindrome while leaving a as odd (without affecting the original frequency).
Below is the implementation of the above approach:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the frequency array
// for the given string
int* getFrequencies(string str)
{
static int freq[26] = { 0 };
for (int i = 0; i < str.length(); i++) {
freq[str[i] - 'a']++;
}
return freq;
}
// Function to return the required count
int countMinParts(string str)
{
int n = str.length();
int *freq = getFrequencies(str);
vector<int> oddFreq, evenFreq;
int i, sumEven = 0;
for (i = 0; i < 26; i++) {
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.push_back(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.push_back(freq[i]);
}
for (i = 0; i < evenFreq.size(); i++) {
sumEven += evenFreq[i];
}
// If there are no characters with odd frequency
if (oddFreq.size() == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0) {
// Only a single character with odd frequency
if (oddFreq.size() == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.size()) {
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.size() == 0)
return oddFreq.size();
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq[i] == 1) {
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq[i] = oddFreq[i] - 2;
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
int main()
{
string s = "noonpeep";
cout<<countMinParts(s);
}
// Java implementation of the approach
import java.util.*;
public class GFG {
// Function to return the frequency array
// for the given string
static int[] getFrequencies(String str)
{
int freq[] = new int[26];
for (int i = 0; i < str.length(); i++) {
freq[str.charAt(i) - 'a']++;
}
return freq;
}
// Function to return the required count
static int countMinParts(String str)
{
int n = str.length();
int freq[] = getFrequencies(str);
List<Integer> oddFreq = new ArrayList<>();
List<Integer> evenFreq = new ArrayList<>();
int i, sumEven = 0;
for (i = 0; i < 26; i++) {
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.add(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.add(freq[i]);
}
for (i = 0; i < evenFreq.size(); i++) {
sumEven += evenFreq.get(i);
}
// If there are no characters with odd frequency
if (oddFreq.size() == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0) {
// Only a single character with odd frequency
if (oddFreq.size() == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.size()) {
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.size() == 0)
return oddFreq.size();
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq.get(i) == 1) {
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq.set(i, oddFreq.get(i) - 2);
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
public static void main(String[] args)
{
String s = "noonpeep";
System.out.println(countMinParts(s));
}
}
// This code is contributed by chitranayal
# Python3 implementation of the approach
# Function to return the frequency array
# for the given string
def getFrequencies(string) :
freq = [0] * 26
for i in range(len(string)) :
freq[ord(string[i]) -
ord('a')] += 1
return freq
# Function to return the required count
def countMinParts(string) :
n = len(string)
freq = getFrequencies(string)
oddFreq = []
evenFreq = []
sumEven = 0
for i in range(26) :
if freq[i] == 0 :
continue
# Add frequencies of the even
# appearing characters
if freq[i] % 2 == 0 :
evenFreq.append(freq[i])
# Count of the characters that
# appeared odd number of times
else :
oddFreq.append(freq[i])
for i in range(len(evenFreq)) :
sumEven += evenFreq[i]
# If there are no characters with
# odd frequency
if len(oddFreq) == 0 :
return 1
# If there are no characters with
# even frequency
if sumEven == 0 :
# Only a single character with
# odd frequency
if len(oddFreq) == 1:
return 1
# More than 1 character with odd
# frequency string isn't a palindrome
return 0
i = 0
# All odd appearing characters can also
# contribute to the even length palindrome
# if one character is removed from the
# frequency leaving it as even
while(i < len(oddFreq)) :
# If k palindromes are possible where
# k is the number of characters with
# odd frequency
if ((sumEven / 2) % len(oddFreq) == 0) :
return len(oddFreq)
# Current character can no longer be
# an element in a string other than
# the mid character
if (oddFreq[i] == 1) :
i += 1
continue
# If current character has odd frequency > 1
# take two characters which can be used in
# any of the parts
sumEven += 2
# Update the frequency
oddFreq[i] = oddFreq[i] - 2
# If not possible, then every character of the
# string will act as a separate palindrome
return n
# Driver code
if __name__ == "__main__" :
s = "noonpeep"
print(countMinParts(s))
# This code is contributed by Ryuga
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to return the frequency array
// for the given string
static int[] getFrequencies(String str)
{
int []freq = new int[26];
for (int i = 0; i < str.Length; i++)
{
freq[str[i] - 'a']++;
}
return freq;
}
// Function to return the required count
static int countMinParts(String str)
{
int n = str.Length;
int []freq = getFrequencies(str);
List<int> oddFreq = new List<int>();
List<int> evenFreq = new List<int>();
int i, sumEven = 0;
for (i = 0; i < 26; i++)
{
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.Add(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.Add(freq[i]);
}
for (i = 0; i < evenFreq.Count; i++)
{
sumEven += evenFreq[i];
}
// If there are no characters with odd frequency
if (oddFreq.Count == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0)
{
// Only a single character with odd frequency
if (oddFreq.Count == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.Count)
{
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.Count == 0)
return oddFreq.Count;
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq[i] == 1)
{
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq.Insert(i, oddFreq[i] - 2);
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
public static void Main(String[] args)
{
String s = "noonpeep";
Console.WriteLine(countMinParts(s));
}
}
// This code has been contributed by 29AjayKumar
<script>
// Javascript implementation of the approach
// Function to return the frequency array
// for the given string
function getFrequencies(str)
{
let freq = new Array(26);
for(let i=0;i<26;i++)
freq[i]=0;
for (let i = 0; i < str.length; i++) {
freq[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++;
}
return freq;
}
// Function to return the required count
function countMinParts(str)
{
let n = str.length;
let freq = getFrequencies(str);
let oddFreq = [];
let evenFreq = [];
let i, sumEven = 0;
for (i = 0; i < 26; i++) {
if (freq[i] == 0)
continue;
// Add frequencies of the even appearing
// characters
if (freq[i] % 2 == 0)
evenFreq.push(freq[i]);
// Count of the characters that appeared
// odd number of times
else
oddFreq.push(freq[i]);
}
for (i = 0; i < evenFreq.length; i++) {
sumEven += evenFreq[i];
}
// If there are no characters with odd frequency
if (oddFreq.length == 0)
return 1;
// If there are no characters with even frequency
if (sumEven == 0) {
// Only a single character with odd frequency
if (oddFreq.length == 1)
return 1;
// More than 1 character with odd frequency
// string isn't a palindrome
return 0;
}
i = 0;
// All odd appearing characters can also contribute to
// the even length palindrome if one character
// is removed from the frequency leaving it as even
while (i < oddFreq.length) {
// If k palindromes are possible where k
// is the number of characters with odd frequency
if ((sumEven / 2) % oddFreq.length == 0)
return oddFreq.length;
// Current character can no longer be an element
// in a string other than the mid character
if (oddFreq[i] == 1) {
i++;
continue;
}
// If current character has odd frequency > 1
// take two characters which can be used in
// any of the parts
sumEven += 2;
// Update the frequency
oddFreq[i]= oddFreq[i] - 2;
}
// If not possible, then every character of the
// string will act as a separate palindrome
return n;
}
// Driver code
let s = "noonpeep";
document.write(countMinParts(s));
// This code is contributed by rag2127
</script>
Time complexity: O(n) where n is the length of the given string
Auxiliary space: O(1)
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