给定一个字符串S ，任务是从给定的字符串找到长度为 3 的不同索引回文子序列的最大数量。
例子：
Input: str = “geekforg”
Output: 2
Explanation:Possible palindromic subsequences of length 3 satisfying the conditions are “gkg” and “efe”. Therefore, the required output is 2.
Input: str = “geek”
Output: 1
Explanation: Possible palindromic subsequences of length 3 satisfying the conditions are “ege” .
方法：我们的想法是计数字符串S的每个字符的频率，和计数的频率对，使得对是相同的字符，并通过除以3字符串S计数长度为3的子序列的数目。最后，将频率对的最小值打印为子序列的数量。请按照以下步骤解决问题：
- 初始化一个数组，比如freq[] ，以存储字符串S的每个字符的频率。
- 初始化一个变量，比如freqPair ，以存储具有相同字符的频率对。
- 初始化一个变量，比如len ，以存储字符串S的长度为3的子序列的数量。
- 迭代范围[0, str.length() – 1] 。对于字符串S 的每个第i个索引，将字符freq[S[i] – ‘a’]的计数增加1 。
- 迭代范围[0, 26] 。对于数组freq[] 的每个第i个索引，通过将数组元素除以2 来计算频率对。
- 最后，打印freqPair和len的最小值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to count the maximum number
// oaf palindrome subsequences of length 3
// considering the same index only once
int maxNumPalindrome(string S)
{
// Index of the string S
int i = 0;
// Stores the frequency of
// every character
int freq[26] = { 0 };
// Stores the pair of frequency
// containing same characters
int freqPair = 0;
// Number of subsequences
// having length 3
int len = S.length() / 3;
// Counts the frequency
while (i < S.length()) {
freq[S[i] - 'a']++;
i++;
}
// Counts the pair of frequency
for (i = 0; i < 26; i++) {
freqPair += (freq[i] / 2);
}
// Returns the minimum value
return min(freqPair, len);
}
// Driver Code
int main()
{
string S = "geeksforg";
cout << maxNumPalindrome(S) << endl;
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Driver Code
public static void main(String[] args)
{
String S = "geeksforg";
System.out.println(maxNumPalindrome(S));
}
// Function to count the maximum number
// of palindrome subsequences of length 3
// considering the same index only once
static int maxNumPalindrome(String S)
{
// Index of the string S
int i = 0;
// Stores the frequency of
// every character
int[] freq = new int[26];
// Stores the pair of frequency
// containing same characters
int freqPair = 0;
// Number of subsequences
// having length 3
int len = S.length() / 3;
// Counts the frequency
while (i < S.length()) {
freq[S.charAt(i) - 'a']++;
i++;
}
// Counts the pair of frequency
for (i = 0; i < 26; i++) {
freqPair += (freq[i] / 2);
}
// Returns the minimum value
return Math.min(freqPair, len);
}
}Python3
# Python3 program to implement
# the above approach
# Function to count the maximum number
# of palindrome subsequences of length 3
# considering the same index only once
def maxNumPalindrome(S):
# Index of the S
i = 0
# Stores the frequency of
# every character
freq = [0] * 26
# Stores the pair of frequency
# containing same characters
freqPair = 0
# Number of subsequences
# having length 3
ln = len(S) // 3
# Counts the frequency
while (i < len(S)):
freq[ord(S[i]) - ord('a')] += 1
i += 1
# Counts the pair of frequency
for i in range(26):
freqPair += (freq[i] // 2)
# Returns the minimum value
return min(freqPair, ln)
# Driver Code
if __name__ == '__main__':
S = "geeksforg"
print(maxNumPalindrome(S))
# This code is contributed by mohit kumar 29C#
// C# program to implement
// the above approach
using System;
class GFG
{
// Driver Code
public static void Main(String[] args)
{
string S = "geeksforg";
Console.WriteLine(maxNumPalindrome(S));
}
// Function to count the maximum number
// of palindrome subsequences of length 3
// considering the same index only once
static int maxNumPalindrome(string S)
{
// Index of the string S
int i = 0;
// Stores the frequency of
// every character
int[] freq = new int[26];
// Stores the pair of frequency
// containing same characters
int freqPair = 0;
// Number of subsequences
// having length 3
int len = S.Length / 3;
// Counts the frequency
while (i < S.Length)
{
freq[S[i] - 'a']++;
i++;
}
// Counts the pair of frequency
for (i = 0; i < 26; i++)
{
freqPair += (freq[i] / 2);
}
// Returns the minimum value
return Math.Min(freqPair, len);
}
}
// This code is contributed by susmitakundugoaldanga.输出：
2时间复杂度： O(|S| + 26)
辅助空间： O(26)
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