Hard
Special Binary String — C++
Full explanation · Time O(n^2) · Space O(n)
// Time: f(n) = k * f(n/k) + n/k * klogk <= O(logn * nlogk) <= O(n^2)
// n is the length of S, k is the max number of special strings in each depth
// Space: O(n)
class Solution {
public:
string makeLargestSpecial(string S) {
vector<string> result;
int anchor = 0, count = 0;
for (int i = 0; i < S.length(); ++i) {
(S[i] == '1') ? ++count : --count;
if (count == 0) {
result.emplace_back("1");
result.back() += makeLargestSpecial(S.substr(anchor + 1, i - anchor - 1));
result.back() += "0";
anchor = i + 1;
}
}
sort(result.begin(), result.end(), greater<string>());
return accumulate(result.begin(), result.end(), string());
}
};