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Progressive hints first, then the full explanation and implementation when you're ready to cash out.
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Hints
Open only as much as you need to keep the solve alive.
The subsequence you remove is chosen by indices, not by taking a contiguous chunk. That makes the search space way smaller than it might look.
Since , there are only possible subsequences. That is at most . Tiny. Brute force is not a crime here; it is the intended heist.
For each subset of indices, build two strings: from chosen indices, and from unchosen indices. Then test the two required properties directly.
A binary string is non-decreasing iff it never contains a 1 before a later 0. Equivalently, there is no adjacent drop while scanning .
Enumerate masks from to . The first mask where is non-decreasing and is a palindrome gives valid output. If none works, print -1.
Because , the clean solution is to stop trying to be clever and just try every possible subsequence.
A subsequence is fully determined by which indices we choose. For each index, we either put it into or leave it in , so there are exactly choices. With , that is at most masks per test case, and even with test cases this is completely fine.
Checking one candidate
Take a bitmask mask:
Now we only need to check the statement literally:
For a binary string, non-decreasing means all 0s come before all 1s. So while scanning , there must never be a transition from 1 to 0. In code, just check whether p[i] > p[i+1] ever happens.
For palindrome checking, compare with its reverse, or use two pointers. Both are fine.
Why this is correct
Every possible subsequence corresponds to exactly one subset of indices, and every subset of indices is considered by our mask loop. So if a valid answer exists, our brute force will eventually test it.
When we accept a mask, we have directly verified both required conditions: the removed subsequence is non-decreasing, and the leftover string is a palindrome. Therefore the printed indices are valid.
If the loop finishes without finding anything, then every possible subsequence failed at least one condition, so no solution exists and -1 is correct.
That is the whole trick: the constraints are screaming “brute force me,” and we should listen.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
bool nondecreasing(const string& p) {
for (int i = 1; i < (int)p.size(); i++) {
if (p[i - 1] > p[i]) return false;
}
return true;
}
bool palindrome(const string& x) {
int l = 0, r = (int)x.size() - 1;
while (l < r) {
if (x[l++] != x[r--]) return false;
}
return true;
}
int main() {
setIO();
int t;
cin >> t;
while (t--) {
int n;
string s;
cin >> n >> s;
bool found = false;
for (int mask = 0; mask < (1 << n) && !found; mask++) {
string p, x;
vector<int> indices;
for (int i = 0; i < n; i++) {
if (mask & (1 << i)) {
p += s[i];
indices.push_back(i + 1);
} else {
x += s[i];
}
}
if (nondecreasing(p) && palindrome(x)) {
cout << indices.size() << '\n';
for (int i = 0; i < (int)indices.size(); i++) {
if (i) cout << ' ';
cout << indices[i];
}
cout << '\n';
found = true;
}
}
if (!found) cout << -1 << '\n';
}
}#include <bits/stdc++.h>
using namespace std;
using ll = long long;
void setIO() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
}
bool nondecreasing(const string& p) {
for (int i = 1; i < (int)p.size(); i++) {
if (p[i - 1] > p[i]) return false;
}
return true;
}
bool palindrome(const string& x) {
int l = 0, r = (int)x.size() - 1;
while (l < r) {
if (x[l++] != x[r--]) return false;
}
return true;
}
int main() {
setIO();
int t;
cin >> t;
while (t--) {
int n;
string s;
cin >> n >> s;
bool found = false;
for (int mask = 0; mask < (1 << n) && !found; mask++) {
string p, x;
vector<int> indices;
for (int i = 0; i < n; i++) {
if (mask & (1 << i)) {
p += s[i];
indices.push_back(i + 1);
} else {
x += s[i];
}
}
if (nondecreasing(p) && palindrome(x)) {
cout << indices.size() << '\n';
for (int i = 0; i < (int)indices.size(); i++) {
if (i) cout << ' ';
cout << indices[i];
}
cout << '\n';
found = true;
}
}
if (!found) cout << -1 << '\n';
}
}