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Count total number of Palindromic Subsequences
Count the total number of Palindromic Subsequences: Here, we are going to learn how to find the total number of Palindromic subsequences in a string? This is a standard interview problem that can be featured in any interview coding rounds.
Submitted by Radib Kar, on June 14, 2020
Problem statement:
Given a string str, find total number of possible palindromic subsequences. A subsequence does not need to be consecutive, but for any x_{i}x_{j} i<j must be valid in the parent string too. Like "icl" is a subsequence of "includehelp" while "ple" is not.
Input:
The first line of input contains an integer T, denoting the no of test cases then T test cases follow. Each test case contains a string str.
Output:
For each test case output will be an integer denoting the total count of palindromic subsequence which could be formed from the string str.
Constraints:
1 <= T <= 100
1 <= length of string str <= 300
Example:
Input:
Test case: 2
First test case:
Input string:
"aaaa"
Output:
Total count of palindromic subsequences is: 15
Second test case:
Input string:
"abaaba"
Output:
Total count of palindromic subsequences is: 31
Explanation:
Test case 1:
Input: "aaaa"
The valid palindromic subsequences are shown below,
Marked cells are character taken in subsequence:
Count=1
Count=2
Count=3
Count=4
Count=5
Count=6
Count=7
Count=8
Count=9
Count=10
Count=11
So on...
Total 15 palindromic subsequences
Actually in this case since all the character is same each and every subsequence is palindrome here.
For the second test case
Few subsequences can be
"a"
"b"
"a"
"aba"
So on
Total 31 such palindromic subsequences
Solution approach
This can be solved by using DP bottom up approach,
 Initialize dp[n][n] where n be the string length to 0

Fill up the base case,
Base case is that each single character is a palindrome itself. And for length of two, i.e, if adjacent characters are found to be equal then dp[i][i+1]=3, else if characters are different then dp[i][i+1]=2
To understand this lets think of a string like "acaa"
Here dp[0][1]=2 because there's only two palindrome possible because of "a" and "c".
Whereas for dp[2][3] value will be 3 as possible subsequences are "a", "a", "aa".
for i=0 to n
// for single length characters
dp[i][i]=1;
if(i==n1)
break;
if(s[i]==s[i+1])
dp[i][i+1]=3;
else
dp[i][i+1]=2;
end for

Compute for higher lengths,
for len=3 to n
for start=0 to nlen
int end=start+len1;
// start and end is matching
if(s[end]==s[start])
// 1+subsequence from semaining part
dp[start][end]=1+dp[start+1][end]+dp[start][end1];
else
dp[start][end]=dp[start+1][end]+dp[start][end1]dp[start+1][end1];
end if
end for
end for
 Final result is stored in dp[0][n1];
So for higher lengths if starting and ending index is the same then we recur for the remaining characters, since we have the subproblem result stored so we computed that. In case start and end index character are different then we have added dp[start+1][end] and dp[start][end1] that's similar to recur for leaving starting index and recur for leaving end index. But it would compute dp[start+1][end1] twice and that why we have deducted that.
For proper understanding you can compute the table by hand for the string "aaaa" to understand how it's working.
C++ Implementation:
#include <bits/stdc++.h>
using namespace std;
int countPS(string s)
{
int n = s.length();
int dp[n][n];
memset(dp, 0, sizeof(dp));
for (int i = 0; i < n; i++) {
dp[i][i] = 1;
if (i == n  1)
break;
if (s[i] == s[i + 1])
dp[i][i + 1] = 3;
else
dp[i][i + 1] = 2;
}
for (int len = 3; len <= n; len++) {
for (int start = 0; start <= n  len; start++) {
int end = start + len  1;
if (s[end] == s[start]) {
dp[start][end] = 1 + dp[start + 1][end] + dp[start][end  1];
}
else {
dp[start][end] = dp[start + 1][end] + dp[start][end  1]  dp[start + 1][end  1];
}
}
}
return dp[0][n  1];
}
int main()
{
int t;
cout << "Enter number of testcases\n";
cin >> t;
while (t) {
string str;
cout << "Enter the input string\n";
cin >> str;
cout << "Total Number of palindromic Subsequences are: " << countPS(str) << endl;
}
return 0;
}
Output:
Enter number of testcases
2
Enter the input string
aaaa
Total Number of palindromic Subsequences are: 15
Enter the input string
abaaba
Total Number of palindromic Subsequences are: 31
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