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 codeint main(){ string str = "poor"; int K = 3; iskPalindromesPossible(str, K); str = "neveropen"; K = 10; iskPalindromesPossible(str, K); return 0;} |
Java
// Java program to find if String// is K-Palindrome or not using // all characters exactly onceimport 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 codepublic static void main(String[] args){ String str = "poor"; int K = 3; iskPalindromesPossible(str, K); str = "neveropen"; K = 10; iskPalindromesPossible(str, K);}}// This code is contributed by sapnasingh4991 |
Python3
# 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 oncedef 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 codeif __name__ == '__main__': str1 = "poor" K = 3 iskPalindromesPossible(str1, K) str = "neveropen" K = 10 iskPalindromesPossible(str, K)# This code is contributed by Surendra_Gangwar |
C#
// C# program to find if String// is K-Palindrome or not using // all characters exactly onceusing 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 codepublic static void Main(String[] args){ String str = "poor"; int K = 3; iskPalindromesPossible(str, K); str = "neveropen"; K = 10; iskPalindromesPossible(str, K);}}// This code is contributed by Princi Singh |
Javascript
<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 oncefunction 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 codelet str1 = "poor"let K = 3iskPalindromesPossible(str1, K)let str = "neveropen"K = 10iskPalindromesPossible(str, K)// This code is contributed by shinjanpatra</script> |
Output:
Yes Yes
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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