字符串的回文子序列个数

Problem statement:

问题陈述：

Given a string you have to count the total number of palindromic subsequences in the giving string and print the value.

给定一个字符串，您必须计算给定字符串中回文子序列的总数并打印该值。

    Input:
T Test case
T no of input string will be given to you.
E.g.
3
abc
aa
abcc
Constrain 
1≤ length (string) ≤100
Output:
Print the count of the palindromic sub sequences from the given string.

Example

例

    T=3
Input:
abc
Output:
3 ("a", "b", "c")
Input:
aa
Output:
3 ("a", "a", "aa")
Input:
abcc
Output:
5 ("a", "b", "c", "c", "cc")

Explanation with example:

举例说明：

Let there is a string str.

让我们有一个字符串str 。

Now possible arrangements are:

现在可能的安排是：

1. Single middle characters like aba

   像aba这样的单个中间字符

2. Paired middle characters like bb

   配对的中间字符，如bb

Let, f(a,b) = count of number of palindromic subsequences from index a to index b.

设f(a，b)=从索引a到索引b的回文子序列数。

Considering the above two facts:

考虑以上两个事实：

1. If there is only one character then f(a,a) = 1

   如果只有一个字符，则f(a，a)= 1

2. If there is a substring starting from index a to index b then,

   如果存在从索引a到索引b的子字符串，

   1. If str[a] and str[b] both are same thenf(a,b)=f(a+1,b)+f(a,b-1)+1
   2. 如果str [a]和str [b]都相同，则f(a，b)= f(a + 1，b)+ f(a，b-1)+1
   3. If str[a] and str[b] both are not same thenf(a,b)=f(a+1,b)+f(a,b-1)-f(a+1,b-1)
   4. 如果str [a]和str [b]都不相同，则f(a，b)= f(a + 1，b)+ f(a，b-1)-f(a + 1，b-1)

For, str = abbaa

对于， str = abbaa

From the above, it is understandable that one function is called repeatedly so for the large input the repetition will be very high. Because of this problem we are using a Dynamic programming approach to avoid repetition. We are using the memorization process to solve the problem.

从以上内容可以理解，一个函数被重复调用，因此对于大的输入，重复率会很高。 由于这个问题，我们使用动态编程方法来避免重复。 我们正在使用记忆过程来解决问题。

Problem solution:

问题方案：

Recursive algorithm:

递归算法：

int Function(str,pos,l):
if str.length()== l
return
for a=pos to str.length()-l 
if str[a]==str[a+l-1]
return Function(str,a,l-1)+Function(str,a+1,l-1)+1
else
return Function(str,a,l-1)+Function(str,a+1,l-1)-Function(str,a+1,l-2)

DP conversion:

DP转换：

int Function(str,n):
for len=1 to str.length()
for a=0 to str.length()-len
b=pos+len-1
if str[a]==str[b]
arr[a][b]=arr[a+1][b]+ arr[a][b-1]+1
else
arr[a][b]=arr[a+1][b]+arr[a][b-1]-arr[a+1][b-1]
return arr[0][len-1]

C++ Implementation:

C ++实现：

#include <bits/stdc++.h>
using namespace std;
int count_palindrome(string str)
{
int len = str.length();
int arr[len][len] = { 0 };
for (int i = 0; i < len; i++) {
arr[i][i] = 1;
}
for (int l = 2; l <= len; l++) {
for (int i = 0; i <= len - l; i++) {
int j = i + l - 1;
if (str[i] == str[j]) {
arr[i][j] = arr[i + 1][j] + arr[i][j - 1] + 1;
}
else {
arr[i][j] = arr[i + 1][j] + arr[i][j - 1] - arr[i + 1][j - 1];
}
}
}
return arr[0][len - 1];
}
int main()
{
int t;
cout << "Test Case : ";
cin >> t;
while (t--) {
cout << "Enter the string : ";
string str;
cin >> str;
cout << "Number of palindromic subsequences are : " << count_palindrome(str) << endl;
}
return 0;
}

Output

输出量

Test Case : 3
Enter the string : abc
Number of palindromic subsequences are : 3
Enter the string : aaa
Number of palindromic subsequences are : 7
Enter the string : abcc
Number of palindromic subsequences are : 5

翻译自: https://www.includehelp.com/icp/count-the-number-of-palindromic-subsequences-in-a-given-string.aspx

字符串的回文子序列个数