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Recursion - Hard Level - Question 1

Recursion - Hard Level - Question 1


Leetcode 761 Special Binary String

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.

You are given a special binary 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.

Return the lexicographically largest resulting string possible after applying the mentioned operations on the string.

Constraints:

1 <= s.length <= 50

s[i] is either '0' or '1'.

s is a special binary string.


Analysis:

One way to solve this problem is to find all the 'special binary strings', and then just sort them reversely, and then concatenate them together.

The problem becomes how to break the input string down to sub-'special binary strings'. Based on the question, we can use a counter to count the number of '1' and '0'. Once them are the same, then it should be a 'special binary string'.

But there are cases that the 'special binary string' may NOT the optimal one (or NOT the one with the largest lexicographical value). For example, "11011000". This string itself is 'special binary string', but it is NOT the largest lexicographical one (which should be "11100100" for this example). 

Now the problem becomes how to make every 'special binary string' to be lexicographically largest, which essentially the same question as the original question! 

This is good, but we have to achieve the second condition before we can develop an algorithm of recursion: smaller size or shrinking the size of the question.

One of the key observation is that: the special binary string always start with '1' and end with '0'. Now the second condition can be realized: we just need to recursively adjust the the inner substring with the size smaller by 2.


See the code below:

class Solution {
public:
    string makeLargestSpecial(string s) {
        string res = "";
        int n = s.size();
        // base case
        if(!n) return res;
        vector<string> ss;
        int ct = 0, start = 0;
        for(int i=0; i<n; ++i) {
            if(s[i] == '1') ++ct;
            else --ct;
            if(!ct) {
                // recursive call
                string t = makeLargestSpecial(s.substr(start+1, i-start-1));
                ss.push_back("1" + t + "0");
                start = i + 1;
            }
        }
        sort(ss.begin(), ss.end(), greater<string>());
        for(auto &t : ss) res += t;
        return res;
    }
};

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