Question

Special binary strings are binary strings with the following two properties:

• The number of 0's is equal to the number of 1's.
• Every prefix of the binary string has at least as many 1's as 0's.

Given a special string S, a move consists of choosing two consecutive, non-empty, special substrings of S, and swapping them. (Two strings are consecutive if the last character of the first string is exactly one index before the first character of the second string.)

At the end of any number of moves, what is the lexicographically largest resulting string possible?

Example 1:

Input: S = "11011000" Output: "11100100" Explanation: The strings "10" [occuring at S[1]] and "1100" [at S[3]] are swapped. This is the lexicographically largest string possible after some number of swaps.

Note:

1. S has length at most 50.
2. S is guaranteed to be a special binary string as defined above.

Difficulty:Medium

Category:

Analyze

输入,输出都是一个特色的二进制字符串, 需要满足要求为:

• 字符串中 0 和 1 的个数要相等
• 任何一个位置, 在前面的 1 的个素都要大于等于前面的 0 的个数.

这里要求我们能够通过交换字符串的方式, 生成字母顺序最大的特色字符串,.

Solution

class Solution {
 public:
  string makeLargestSpecial(string S) {
    int cnt = 0, i = 0;
    vector<string> v;
    string ans = "";
    for (int j = 0; j < S.size(); ++j) {
      cnt += (S[j] == '1') ? 1 : -1;
      if (cnt == 0) {
        v.push_back('1' + makeLargestSpecial(S.substr(i + 1, j - i - 1)) + '0');
        i = j + 1;
      }
    }
    sort(v.begin(), v.end(), greater<string>());
    for (int i = 0; i < v.size(); ++i) ans += v[i];
    return ans;
  }
};