Does anyone have any objection to my posting it now? I don’t want to steal @anon55659401 's thunder 
I suppose any ongoing contest can be assumed to have finished by now?
Edit:
So it looked like the consensus was that people would post solutions “in a few hours”, and that was more than a few hours ago
This is a bit seat-of-the-pants - I think the approach is sound (it survives plenty of randomised testing) but the documentation might not be 100%:
CPP Solution
#define SUBMISSION
#define BRUTE_FORCE
#ifdef SUBMISSION
#undef BRUTE_FORCE
#define NDEBUG
#endif
#include <iostream>
#include <vector>
#include <map>
#include <cassert>
#include <sys/time.h> // TODO - this is only for random testcase generation. Remove it when you don't need new random testcases!
using namespace std;
template <typename T>
T read()
{
T toRead;
cin >> toRead;
assert(cin);
return toRead;
}
int64_t solveBruteForce(const string& s)
{
int64_t result = 0;
for (int i = 0; i < s.size(); i++)
{
for (int j = i; j < s.size(); j++)
{
const int numLetters = 26;
int letterHistogram[numLetters] = {};
for (int k = i; k <= j; k++)
{
letterHistogram[s[k] - 'a']++;
}
int numLettersWithOddOccurrence = 0;
for (int letterIndex = 0; letterIndex < numLetters; letterIndex++)
{
if ((letterHistogram[letterIndex] % 2) == 1)
numLettersWithOddOccurrence++;
}
if (numLettersWithOddOccurrence == 0 || numLettersWithOddOccurrence == 1)
{
cout << "Palindromic: " << s.substr(i, j - i + 1) << endl;
result++;
}
}
}
return result;
}
int64_t solveOptimised(const string& s)
{
// Let the "bit value" of a letter a, b, c ... be defined as
//
// 2 ** (distance of letter from 'a')
//
// i.e.
//
// bit_value(a) == 2 ** 0 = 1
// bit_value(b) == 2 ** 1 = 2
// bit_value(c) == 2 ** 2 = 4
//
// etc.
//
// Let the "xor sum" of a substring
//
// a_i a_(i+1) ... a_j
//
// be defined as
//
// xorSum (a_i a_(i+1) ... a_j) = bit_value(a_i) ^ bit_value(a_(i+1)) ^ ... ^ bit_value(a_j)
//
// Then a substring s will be a palindrome if and only if the binary representation
// of xorSum(s) has *at most* one bit set.
//
// Let prefixXorSum(k) = xorSum(a1 a2 ... a_k); then note that
//
// xorSum(a_i a_(i+1) ... a_k) = prefixXorSum(i - 1) ^ prefixXorSum(k)
//
// Thus, to find the number of palindromic-substrings ending at a_k, we need to calculate prefixXorSum(k)
// and then find the number of i's such that prefixXorSum(i) ^ prefixXorSum(k) has at most one bit set
// i.e. the number of i's such that prefixXorSum(i) ^ prefixXorSum(k) is either 0, or a power of 2.
// This is pretty easy.
int64_t result = 0;
// We can probably use a large (2 ** 26 * sizeof(int) ~ 256MB) lookup table instead of a std::map, here,
// giving us asymptotically better runtime performance at the expense of memory usage.
map<uint32_t, int> numPrefixesWithXorSum;
numPrefixesWithXorSum[0] = 1; // Empty prefix.
uint32_t prefixXorSum = 0;
for (int k = 0; k < s.size(); k++)
{
const int letterIndex = s[k] - 'a';
prefixXorSum = prefixXorSum ^ (1 << letterIndex);
result += numPrefixesWithXorSum[prefixXorSum]; // Num substrings ending at k with xorSum == 0.
for (int i = 0; i < 26; i++)
{
result += numPrefixesWithXorSum[prefixXorSum ^ (1 << i)]; // Num substrings ending at k with the ith bit of their xorSum set.
}
numPrefixesWithXorSum[prefixXorSum]++;
}
return result;
}
int main(int argc, char* argv[])
{
ios::sync_with_stdio(false);
if (argc == 2 && string(argv[1]) == "--test")
{
struct timeval time;
gettimeofday(&time,NULL);
srand((time.tv_sec * 1000) + (time.tv_usec / 1000));
const int N = rand() % 1'000 + 1;
const int maxLetter = rand() % 26 + 1;
cout << 1 << endl;
//cout << N << endl;
for (int i = 0; i < N; i++)
{
cout << static_cast<char>('a' + rand() % maxLetter);
}
cout << endl;
return 0;
}
const int T = read<int>();
for (int t = 0; t < T; t++)
{
const string s = read<string>();
#ifdef BRUTE_FORCE
const auto solutionBruteForce = solveBruteForce(s);
cout << "solutionBruteForce: " << solutionBruteForce << endl;
const auto solutionOptimised = solveOptimised(s);
cout << "solutionOptimised: " << solutionOptimised << endl;
assert(solutionOptimised == solutionBruteForce);
#else
const auto solutionOptimised = solveOptimised(s);
cout << solutionOptimised << endl;
#endif
}
assert(cin);
}
