#%%
"""
- Find Permutation
- https://leetcode.com/problems/find-permutation/
- Medium
"""

#%%
"""
By now, you are given a secret signature consisting of character 'D' and 'I'. 'D' represents a decreasing relationship between two numbers, 'I' represents an increasing relationship between two numbers. And our secret signature was constructed by a special integer array, which contains uniquely all the different number from 1 to n (n is the length of the secret signature plus 1). For example, the secret signature "DI" can be constructed by array [2,1,3] or [3,1,2], but won't be constructed by array [3,2,4] or [2,1,3,4], which are both illegal constructing special string that can't represent the "DI" secret signature.

On the other hand, now your job is to find the lexicographically smallest permutation of [1, 2, ... n] could refer to the given secret signature in the input.

Example 1:

Input: "I"
Output: [1,2]
Explanation: [1,2] is the only legal initial spectial string can construct secret signature "I", where the number 1 and 2 construct an increasing relationship.
Example 2:

Input: "DI"
Output: [2,1,3]
Explanation: Both [2,1,3] and [3,1,2] can construct the secret signature "DI", 
but since we want to find the one with the smallest lexicographical permutation, you need to output [2,1,3]
Note:

The input string will only contain the character 'D' and 'I'.
The length of input string is a positive integer and will not exceed 10,000
"""

#%%
# assuming that all numbers from 1 to k have been used
# if I is followed by an I, just add k+1
# if IDD..DI, add the list of number [s+k+1, s+k, ... k+1]

class S1:
    def findPermutation(self, s):
        """
        :type s: str
        :rtype List[int]
        """
        if not s:
            return [1]
        ans = []
        i, n = 0, len(s)
        while i<n and s[i] == 'D':
            i += 1
        # used denotes the largest number that has been used so far
        used = i+1
        for num in range(used, 0, -1):
            ans.append(num)
        # now s[i] == 'I' if i <n , and for each beginning of loop we want make sure of this property
        while i<n:
            if i == n-1:
                ans.append(n+1)
                return ans
            elif s[i+1] == 'I':
                used += 1
                ans.append(used)
                i += 1
                continue
            else:
                j = i
                i = i+1
                while i<n and s[i] == 'D':
                    i += 1
                for num in range(used+i-j, used, -1):
                    ans.append(num)
                used = used +i-j
        return ans
