Find if string is K-Palindrome or not using all characters exactly once
Last Updated :
02 Jan, 2023
Given a string str and an integer K, the task is to check if it is possible to make the string K palindromes using all the characters of the string exactly once.
Examples:
Input: str = "poor", K = 3
Output: Yes
One way of getting 3 palindromes is: oo, p, r
Input: str = "fun", K = 5
Output: No
2 palindromes can't be constructed using 3 distinct letters. Hence not possible.
Approach:
- If the size of the string is less than k, getting k palindromes is not possible.
- If the size of the string is equal to k, getting k palindromes is always possible. We give 1 character to each of k strings and all of them are palindromes as a string of length 1 is always a palindrome.
- If the size of the string is greater than k then some strings might have more than 1 character.
- Create a hashmap to store the frequency of every character as the characters having even number of occurrences can be distributed in groups of 2.
- Check if the number of characters having an odd number of occurrences is less than or equal to k, we can say that k palindromes can always be generated else no.
Below is the implementation of the above approach:
C++
// C++ program to find if string is K-Palindrome
// or not using all characters exactly once
#include <bits/stdc++.h>
using namespace std;
void iskPalindromesPossible(string s, int k)
{
// when size of string is less than k
if (s.size() < k) {
cout << "Not Possible" << endl;
return;
}
// when size of string is equal to k
if (s.size() == k) {
cout << "Possible" << endl;
return;
}
// when size of string is greater than k
// to store the frequencies of the characters
map<char, int> freq;
for (int i = 0; i < s.size(); i++)
freq[s[i]]++;
// to store the count of characters
// whose number of occurrences is odd.
int count = 0;
// iterating over the map
for (auto it : freq) {
if (it.second % 2 == 1)
count++;
}
if (count > k)
cout << "No" << endl;
else
cout << "Yes" << endl;
}
// Driver code
int main()
{
string str = "poor";
int K = 3;
iskPalindromesPossible(str, K);
str = "geeksforgeeks";
K = 10;
iskPalindromesPossible(str, K);
return 0;
}
// Java program to find if String
// is K-Palindrome or not using
// all characters exactly once
import java.util.*;
class GFG{
static void iskPalindromesPossible(String s,
int k)
{
// When size of String is less than k
if (s.length() < k)
{
System.out.print("Not Possible" + "\n");
return;
}
// When size of String is equal to k
if (s.length() == k)
{
System.out.print("Possible" + "\n");
return;
}
// When size of String is greater than k
// to store the frequencies of the characters
HashMap<Character,
Integer> freq = new HashMap<Character,
Integer>();
for(int i = 0; i < s.length(); i++)
if(freq.containsKey(s.charAt(i)))
{
freq.put(s.charAt(i),
freq.get(s.charAt(i)) + 1);
}
else
{
freq.put(s.charAt(i), 1);
}
// To store the count of characters
// whose number of occurrences is odd.
int count = 0;
// Iterating over the map
for(Map.Entry<Character,
Integer> it : freq.entrySet())
{
if (it.getValue() % 2 == 1)
count++;
}
if (count > k)
System.out.print("No" + "\n");
else
System.out.print("Yes" + "\n");
}
// Driver code
public static void main(String[] args)
{
String str = "poor";
int K = 3;
iskPalindromesPossible(str, K);
str = "geeksforgeeks";
K = 10;
iskPalindromesPossible(str, K);
}
}
// This code is contributed by sapnasingh4991
# Find if string is K-Palindrome or not using all characters exactly once
# Python 3 program to find if string is K-Palindrome
# or not using all characters exactly once
def iskPalindromesPossible(s, k):
# when size of string is less than k
if (len(s)<k):
print("Not Possible")
return
# when size of string is equal to k
if (len(s) == k):
print("Possible")
return
# when size of string is greater than k
# to store the frequencies of the characters
freq = dict.fromkeys(s,0)
for i in range(len(s)):
freq[s[i]] += 1
#to store the count of characters
# whose number of occurrences is odd.
count = 0
# iterating over the map
for value in freq.values():
if (value % 2 == 1):
count += 1
if (count > k):
print("No")
else:
print("Yes")
# Driver code
if __name__ == '__main__':
str1 = "poor"
K = 3
iskPalindromesPossible(str1, K)
str = "geeksforgeeks"
K = 10
iskPalindromesPossible(str, K)
# This code is contributed by Surendra_Gangwar
// C# program to find if String
// is K-Palindrome or not using
// all characters exactly once
using System;
using System.Collections.Generic;
class GFG{
static void iskPalindromesPossible(String s,
int k)
{
// When size of String is less than k
if (s.Length < k)
{
Console.Write("Not Possible" + "\n");
return;
}
// When size of String is equal to k
if (s.Length == k)
{
Console.Write("Possible" + "\n");
return;
}
// When size of String is greater than k
// to store the frequencies of the characters
Dictionary<int,
int> freq = new Dictionary<int,
int>();
for(int i = 0; i < s.Length; i++)
if(freq.ContainsKey(s[i]))
{
freq[s[i]] = freq[s[i]] + 1;
}
else
{
freq.Add(s[i], 1);
}
// To store the count of characters
// whose number of occurrences is odd.
int count = 0;
// Iterating over the map
foreach(KeyValuePair<int, int> it in freq)
{
if (it.Value % 2 == 1)
count++;
}
if (count > k)
Console.Write("No" + "\n");
else
Console.Write("Yes" + "\n");
}
// Driver code
public static void Main(String[] args)
{
String str = "poor";
int K = 3;
iskPalindromesPossible(str, K);
str = "geeksforgeeks";
K = 10;
iskPalindromesPossible(str, K);
}
}
// This code is contributed by Princi Singh
<Script>
// Find if string is K-Palindrome or not using all characters exactly once
// JavaScript program to find if string is K-Palindrome
// or not using all characters exactly once
function iskPalindromesPossible(s, k)
{
// when size of string is less than k
if (s.length < k)
{
document.write("Not Possible","</br>")
return
}
// when size of string is equal to k
if (s.length == k){
document.write("Possible","</br>")
return
}
// when size of string is greater than k
// to store the frequencies of the characters
let freq = new Map()
for(let i = 0; i < s.length; i++)
{
if(freq.has(s[i])){
freq.set(s[i],freq.get(s[i])+1);
}
freq.set(s[i],1);
}
// to store the count of characters
// whose number of occurrences is odd.
let count = 0
// iterating over the map
for(let [key,value] of freq){
if (value % 2 == 1)
count += 1
}
if (count > k)
document.write("No","</br>")
else
document.write("Yes","</br>")
}
// Driver code
let str1 = "poor"
let K = 3
iskPalindromesPossible(str1, K)
let str = "geeksforgeeks"
K = 10
iskPalindromesPossible(str, K)
// This code is contributed by shinjanpatra
</script>
Time Complexity: O(n), where n is the length of the given string.
Auxiliary Space: O(26) ? O(1), no extra space is required, so it is a constant.
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