给定一个字符串str和一个整数K，任务是检查是否有可能要使用字符串的所有字符恰好一次用字符串k回文。
例子：
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.
方法：
- 如果字符串的大小小于k，则不可能获得k回文。
- 如果字符串的大小等于k，则总是有可能获得k回文。我们给k个字符串的每个字符串赋予1个字符，它们全部都是回文，因为长度为1的字符串始终是回文。
- 如果字符串的长度大于k那么一些字符串可能有超过1个字符。
- 创建哈希表以存储每个字符的频率，因为偶数出现的字符可以以2为一组分布。
- 检查出现次数为奇数的字符数是否小于或等于k，可以说总是可以生成k回文。
下面是上述方法的实现：
C++
// C++ program to find if string is K-Palindrome
// or not using all characters exactly once
#include
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 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
// 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 freq = new HashMap();
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 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 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 once
def iskPalindromesPossible(s, k):
# when size of string is less than k
if (len(s) 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#
// 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 freq = new Dictionary();
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 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 输出：
Yes
Yes