Problem Description

In this problem, you are given a string s and you are to simulate the behavior of a robot that is generating a new string t with the ultimate goal of writing down the lexicographically smallest string possible onto paper. The robot achieves this by performing operations until both s and t are empty. There are two operations possible:

1. Take the first character from the string s and append it to the string t that the robot holds.
2. Take the last character from the string t and write it down on the paper.

The constraints are that you can only write to paper from t and you can only add to t from s. The challenge is to determine the sequence of operations that lead to the smallest possible string on the paper.

Intuition

To arrive at the lexicographically smallest string, we need to write to the paper in a way that prioritizes placing smaller characters before larger ones. However, we are limited by the arrangement of characters in s and the operations we can perform.

The solution involves keeping track of the minimum(characters in remaining string s) while processing characters from s to t. We use a stack to replicate the behavior of t, appending characters from s and popping them when they are ready to be written to the paper. The idea is to always pop the stack and write to the paper whenever the top of the stack has a character that is less than or equal to the current minimum we can still encounter in the remainder of s.

Here's the approach broken down:

• We use a counter to keep track of the frequency of characters in s.
• We iterate through s, decrementing the count for each character as we go.
• We maintain a variable mi which is set to the smallest character a initially, and it is incremented whenever the count for the current mi becomes zero, i.e., we have used all occurrences of the current smallest character.
• We push characters onto a stack, and whenever the top character is less than or equal to mi (meaning we can write this character without worrying about missing the opportunity to write a smaller character later), we pop it from the stack and append it to the result string ans.
• We repeat the above steps until all characters from s have been pushed to the stack and the stack is emptied by writing its characters onto ans.

The stack (simulating t) essentially acts as a holding area where characters have to wait if a smaller character can appear later from s. Only when we're sure that we're at the smallest possible character that is on top of the stack given the future possibilities, we pop it to ans.

This strategy guarantees that we are writing the smallest possible characters as early as possible without violating the rules of operation, thus achieving a lexicographically smallest result.

Learn more about Stack and Greedy patterns.

Solution Approach

The solution makes use of a stack data structure, a counter to keep track of character frequencies, and a variable to store the smallest character that can still appear in the string s.

Here's a step-by-step explanation of the implementation:

• First, we create a frequency counter for the characters in s using Python's Counter from the collections module. This lets us know how many times each character appears in s.

  cnt = Counter(s)

• We initialize an empty list ans that will serve as the paper onto which the robot writes the characters. We also initialize a stack stk to simulate the behavior of holding the characters from s before they are written to the paper.

  ans = []
  stk = []

• We keep track of the smallest character we have not used up in s. Initially, this is 'a', the smallest possible character.

  mi = 'a'

• We then iterate over each character c in our string s:

  • For each character c, we decrement its count in the frequency counter because we are moving it from s to t (simulated by the stack).

    cnt[c] -= 1

  • We update the mi variable if the character count for mi has reached zero (meaning all of its occurrences have been used), incrementing it to the next character.

    while mi < 'z' and cnt[mi] == 0:
        mi = chr(ord(mi) + 1)

  • The character c is then pushed onto the stack as part of the simulation of moving it from s to t.

    stk.append(c)

  • We check the stack's top element and compare it with mi. If the top (last) character in the stack is less than or equal to mi, it is safe to write this character to paper (append it to ans), so we pop it from the stack. This check is done in a while loop because multiple characters might satisfy this condition.

    while stk and stk[-1] <= mi:
        ans.append(stk.pop())

• Finally, after the loop has processed all characters from s, the stack will be empty, and the list ans will hold the lexicographically smallest string that the robot could write on the paper.

• The resulting smallest string is obtained by joining the elements of ans and returning it.

  return ''.join(ans)

This algorithm's correctness relies on the fact that at each step, we are writing the smallest possible character to the paper. Using the stack allows us to temporarily hold characters that might not be optimal to write immediately and wait until we are certain no smaller character will come before them.

Example Walkthrough

Let's take a small example to illustrate the solution approach.

Suppose we are given the following string s: "cba".

Now, let's walk through the solution:

1. Initialize the frequency counter, the stack (simulating t), and the ans list (simulating the paper).

   cnt = Counter("cba")  # cnt = {'c': 1, 'b': 1, 'a': 1}
   ans = []
   stk = []
   mi = 'a'  # Smallest character

2. We iterate over each character in s.

   • For the first character c:

     cnt['c'] -= 1  # cnt = {'c': 0, 'b': 1, 'a': 1}
     stk.append('c')  # stk = ['c']

     We don't update mi because cnt['a'] is not zero. The stack's top c is not less than or equal to mi ('a'), so we don't pop from the stack.

   • For the second character b:

     cnt['b'] -= 1  # cnt = {'c': 0, 'b': 0, 'a': 1}
     stk.append('b')  # stk = ['c', 'b']

     We now update mi as 'b' has been used up, but mi remains 'a' because cnt['a'] is not zero. Similarly to the previous step, we do not write to ans because stk[-1] (b) is not less than or equal to mi (a).

   • For the third character a:

     cnt['a'] -= 1  # cnt = {'c': 0, 'b': 0, 'a': 0}
     stk.append('a')  # stk = ['c', 'b', 'a']

     mi remains 'a' since all characters have been used at least once. Now the condition stk[-1] <= mi holds true, so we write to ans:

     # We enter the while loop because stk[-1] ('a') is less than or equal to 'a' (mi).
     ans.append(stk.pop())  # ans = ['a']
     # We continue in the loop, now 'b' is at the top of the stack and mi is 'a'
     # Since 'b' is greater than 'a', we exit the loop.

   After steps 1-3, we have two letters in stk, one letter in ans, and mi remains 'a'. The loop ends since we traversed the entire s. We now have remaining characters in our stack, which need to be appended to ans in reverse order (because we pop them from the stack).

   So we pop b and then c from stk, appending each to ans:

   ans.append(stk.pop())  # ans = ['a', 'b']
   ans.append(stk.pop())  # ans = ['a', 'b', 'c']

3. After processing all the characters, we join the ans list to form the result string.

   return ''.join(ans)  # 'abc'

This example demonstrates how the given solution approach results in the lexicographically smallest string by strategically moving characters from s to t (stack stk) and then deciding the right time to pop them onto the paper (ans). The final output of this process is the string "abc", which is the smallest possible arrangement of the given string "cba".

Solution Implementation

Time and Space Complexity

Time Complexity

The given Python function robotWithString has several key operations contributing to its time complexity:

1. The construction of the Counter object - This happens once and takes O(n) time, where n is the length of string s, since the Counter has to iterate through all characters in the string.

2. The first for loop - This loop runs for each character in the string s, hence it iterates n times.

3. The inner while loop to find the current minimum character mi - Since there are 26 English letters, in the worst case, it might check every character until 'z', taking O(1) time regardless of the size of s as it's not directly dependent on n.

4. The second inner while loop where elements are popped from stk and added to ans - Each character can be pushed and popped at most once, leading to O(n) operations overall.

Given these, we see that the main complexity comes from the operations linked directly to the length of s. No nested loops are dependent on the size of s, thus, the overall time complexity is O(n).

Space Complexity

Space complexity considerations involve the additional memory used by the algorithm:

1. The Counter object cnt - Its space complexity is O(n) in the worst case when all characters in the s are unique.

2. The list ans - Worst case, this list will contain all characters from s, thus O(n) space.

3. The list stk - In the worst case, this can also grow to include all characters from s, hence O(n) space.

Adding all this up - we have the Counter, ans, and stk all contributing a linear amount of space complexity with respect to the length of string s. Therefore, the total space complexity is O(n), where n is the length of the string s.

Learn more about how to find time and space complexity quickly using problem constraints.