给定一个由N 个字符串组成的数组A[] ，任务是计算合并后形成回文字符串或可以重新排列形成回文字符串的可能字符串对的数量。
例子 ：
Input: N = 6, A[ ] = {aab, abcac, dffe, ed, aa, aade}
Output: 6
Explanation:
All possible pairs of strings which generates a palindromic string:
{aab, abcac} = aabccbaa
{aab, aa} = aabaa
{abcac, aa} = aacbcaa
{dffe, ed} = fdeedf
{dffe, aade} = edfaafde
{ed, aade} = edaade or aeddea
Hence, total possible pairs = 6
Input: N = 3, A[ ] = {aa, bb, cd}
Output: 1
Explanation:
The only possible pair of strings which generates palindromic string:
{aa, bb} = abba
Hence, total possible pairs = 1
天真的方法：
解决这个问题最简单的方法是从给定的数组中生成所有可能的对并合并这些对以生成新的字符串。对于每一个这种合并字符串，产生该字符串的所有可能的排列和每个置换，检查它是否是一个回文字符串或不并相应地增加计数。
时间复杂度： O(N 2 * L! * L)，其中 L 是字符串的最大长度。
辅助空间： O(M)
有效的方法：
当最多一个字符出现奇数，其余每个字符偶数出现时，总是可以形成回文字符串。由于允许对合并字符串中的字符进行重新排列，所以我们只需要统计合并字符串中字符出现的频率，并检查是否满足回文字符串。
Illustration:
1. Every character has even occurrence
Strings “aab” and “abcac” can be merged and rearranged to form a palindromic string “aababcac” with each character having even frequency
2. One character has odd occurrence
Strings “aab” and “aa” can be merged to form a palindromic string “aabaa” with a character(‘b’) having odd frequency and the rest having even frequency.
因此，可以观察到以下类型的两对字符串可以合并形成一个回文字符串：
- 两个字符串中每个字符的出现频率相同。
- 两个字符串最多一个字符具有不同的频率。
请按照以下步骤解决问题：
- 遍历给定数组并为每个字符串存储每个字符的频率。
- 为每个频率分配奇偶校验(0或1) 。为偶数频率指定0 ，为奇数频率指定1 。
- 初始化一个映射来存储生成的每个频率数组的频率。
- 由于允许重新排列，因此反转每个字符的奇偶校验并将频率数组包含在映射中。
- 最后，返回相似频率数组的总数，这将给出所需的对数。
下面是上述方法的实现：
C++
// C++ Program to find
// palindromic string
#include
using namespace std;
int getCount(int N, vector& s)
{
// Stores frequency array
// and its count
map, int> mp;
// Total number of pairs
int ans = 0;
for (int i = 0; i < N; i++) {
// Initializing array of size 26
// to store count of character
vector a(26, 0);
// Counting occurrence of each
// character of current string
for (int j = 0; j < s[i].size(); j++) {
a[s[i][j] - 'a']++;
}
// Convert each count to parity(0 or 1)
// on the basis of its frequency
for (int j = 0; j < 26; j++) {
a[j] = a[j] % 2;
}
// Adding to anser
ans += mp[a];
// Frequency of single character
// can be possibly changed,
// so change its parity
for (int j = 0; j < 26; j++) {
vector changedCount = a;
if (a[j] == 0)
changedCount[j] = 1;
else
changedCount[j] = 0;
ans += mp[changedCount];
}
mp[a]++;
}
return ans;
}
int main()
{
int N = 6;
vector A
= { "aab", "abcac", "dffe",
"ed", "aa", "aade" };
cout << getCount(N, A);
return 0;
} Java
// Java program to find
// palindromic string
import java.util.*;
class GFG{
static int getCount(int N, List s)
{
// Stores frequency array
// and its count
Map, Integer> mp = new HashMap<>();
// Total number of pairs
int ans = 0;
for(int i = 0; i < N; i++)
{
// Initializing array of size 26
// to store count of character
List a = new ArrayList<>(26);
for(int k = 0; k < 26; k++)
a.add(0);
// Counting occurrence of each
// character of current string
for(int j = 0; j < s.get(i).length(); j++)
{
a.set((s.get(i).charAt(j) - 'a'),
a.get(s.get(i).charAt(j) - 'a') + 1);
}
// Convert each count to parity(0 or 1)
// on the basis of its frequency
for(int j = 0; j < 26; j++)
{
a.set(j, a.get(j) % 2);
}
// Adding to anser
ans += mp.getOrDefault(a, 0);
// Frequency of single character
// can be possibly changed,
// so change its parity
for(int j = 0; j < 26; j++)
{
List changedCount = new ArrayList<>();
changedCount.addAll(a);
if (a.get(j) == 0)
changedCount.set(j, 1);
else
changedCount.set(j, 0);
ans += mp.getOrDefault(changedCount, 0);
}
mp.put(a, mp.getOrDefault(a, 0) + 1);
}
return ans;
}
// Driver code
public static void main (String[] args)
{
int N = 6;
List A = Arrays.asList("aab", "abcac",
"dffe", "ed",
"aa", "aade");
System.out.print(getCount(N, A));
}
}
// This code is contributed by offbeat Python3
# Python3 program to find
# palindromic string
from collections import defaultdict
def getCount(N, s):
# Stores frequency array
# and its count
mp = defaultdict(lambda: 0)
# Total number of pairs
ans = 0
for i in range(N):
# Initializing array of size 26
# to store count of character
a = [0] * 26
# Counting occurrence of each
# character of current string
for j in range(len(s[i])):
a[ord(s[i][j]) - ord('a')] += 1
# Convert each count to parity(0 or 1)
# on the basis of its frequency
for j in range(26):
a[j] = a[j] % 2
# Adding to anser
ans += mp[tuple(a)]
# Frequency of single character
# can be possibly changed,
# so change its parity
for j in range(26):
changedCount = a[:]
if (a[j] == 0):
changedCount[j] = 1
else:
changedCount[j] = 0
ans += mp[tuple(changedCount)]
mp[tuple(a)] += 1
return ans
# Driver code
if __name__ == '__main__':
N = 6
A = [ "aab", "abcac", "dffe",
"ed", "aa", "aade" ]
print(getCount(N, A))
# This code is contributed by Shivam Singh6
时间复杂度： O(N*(L+log(N)))，其中L是字符串的最大长度
辅助空间： O(N)
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