Check If String Is Transformable With Substring Sort Operations

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Strings

Given two strings s and t, transform string s into string t using the following operation any number of times:

Choose a non-empty substring in s and sort it in place so the characters are in ascending order.
- For example, applying the operation on the underlined substring in "14234" results in "12344".

Return true if it is possible to transform s into t. Otherwise, return false.

A substring is a contiguous sequence of characters within a string.

Example 1:

Input: s = "84532", t = "34852"
Output: true
Explanation: You can transform s into t using the following sort operations:
"84532" (from index 2 to 3) -> "84352"
"84352" (from index 0 to 2) -> "34852"

Example 2:

Input: s = "34521", t = "23415"
Output: true
Explanation: You can transform s into t using the following sort operations:
"34521" -> "23451"
"23451" -> "23415"

Example 3:

Input: s = "12345", t = "12435"
Output: false

Constraints:

s.length == t.length
1 <= s.length <= 10⁵
s and t consist of only digits.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

Can you provide more details on the constraints for the string lengths of `s` and `t`?
Are `s` and `t` guaranteed to contain only digits, or can they contain other characters?
If `s` cannot be transformed into `t`, what should the function return?
Can you clarify what 'substring sort operations' means? Specifically, is there a limit to the number of times I can sort a single substring?
If multiple sequences of substring sort operations can transform `s` into `t`, is there a specific sequence I should aim to find, or is any valid transformation acceptable?

Brute Force Solution

Approach

The brute force approach solves this by trying every possible series of substring sorts. It's like shuffling cards repeatedly to see if you can get them in the right order, but it will take a long time if the starting string is large.

Here's how the algorithm would work step-by-step:

Start with the original string.
Find every possible substring within the string.
For each substring, sort the characters within that substring.
This produces a new string.
Check if this new string is equal to the target string. If it is, you're done!
If not, repeat the process from the beginning, using this new string as the starting point.
Keep doing this, generating new strings by sorting substrings and checking for equality to the target. Stop if you find a solution or when a certain amount of tries has passed or depth is reached.

Code Implementation

def check_if_string_is_transformable_brute_force(source_string, target_string, max_depth=10):
    if source_string == target_string:
        return True

    queue = [(source_string, 0)]
    visited = {source_string}

    while queue:
        current_string, depth = queue.pop(0)

        if depth > max_depth:
            continue

        for i in range(len(current_string)): 
            for j in range(i + 1, len(current_string) + 1):
                substring_to_sort = current_string[i:j]
                sorted_substring = "".join(sorted(substring_to_sort))
                new_string = current_string[:i] + sorted_substring + current_string[j:]

                # Goal string reached, so return
                if new_string == target_string:
                    return True

                # Prevent infinite loops
                if new_string not in visited:
                    queue.append((new_string, depth + 1))
                    visited.add(new_string)

        # Limit search depth to avoid excessive computation
        if depth > max_depth:
            return False

    # If no transformation is found, return
    return False

Big(O) Analysis

Time Complexity

O(n! * n log n) – The brute force approach explores all possible sequences of substring sort operations. In the worst case, we might explore a vast number of combinations. Finding all substrings within a string of length n takes O(n^2) time. Sorting each substring of length at most n takes O(n log n) time. The crucial factor is the number of possible string transformations which could approach n! due to the different sorting possibilities. Therefore, the overall time complexity is approximately O(n! * n log n).

Space Complexity

O(N!) – The brute force approach explores all possible substring sorts. In the worst-case scenario, to avoid redundant computations and infinite loops, the algorithm would need to keep track of all the strings it has already generated and checked. The number of possible strings grows factorially with the input string length N because each substring sort operation can create a distinct string. Storing these visited strings requires space proportional to the number of possible permutations, which can reach up to N!. Therefore, the auxiliary space complexity is O(N!).

Optimal Solution

Approach

The key idea is to process each digit from the target string one by one. We maintain a waiting line for each digit, ensuring that we use digits from the source string in the correct relative order. This avoids the need to explore all possible substring sort operations.

Here's how the algorithm would work step-by-step:

Imagine having 'waiting lines' for each possible digit (0 through 9).
Go through the source string, placing each digit into its corresponding waiting line.
Now, go through the target string, one digit at a time.
For each digit in the target string, check if its waiting line has any digits available.
If the waiting line is empty, it's impossible to transform the source to the target; stop and report failure.
If the waiting line is not empty, take the next digit from that waiting line.
Repeat this process until you've gone through the entire target string.
If you reach the end of the target string without running into an empty waiting line, it means it's possible to transform the source to the target.

Code Implementation

def is_transformable(source_string, target_string):
    waiting_lines = [[] for _ in range(10)]

    for digit_char in source_string:
        digit = int(digit_char)
        waiting_lines[digit].append(digit)

    for digit_char in target_string:
        digit = int(digit_char)
        # If a digit's waiting line is empty, transformation is impossible.
        if not waiting_lines[digit]:
            return False

        # Check that all digits in waiting lines
        # before the current digit are greater or equal to current digit
        for i in range(digit):
            if waiting_lines[i] and waiting_lines[i][0] < digit:
                return False

        waiting_lines[digit].pop(0)

    return True

Big(O) Analysis

Time Complexity

O(n + m) – The algorithm iterates through the source string of length n to populate the waiting lines for each digit. Then, it iterates through the target string of length m, checking the waiting lines. The time complexity is dominated by these two linear iterations. Therefore, the total time complexity is O(n + m), where n is the length of the source string and m is the length of the target string. If we assume n and m are approximately the same size, this can be simplified to O(n).

Space Complexity

O(N) – The algorithm uses a list of waiting lines, one for each digit from 0 to 9. In the worst case, the source string could consist of only one unique digit, causing all the characters from the source string to be stored in a single waiting line. Since the waiting lines store characters from the source string, the auxiliary space used is proportional to the length of the source string, which we can define as N. Therefore, the space complexity is O(N).

Edge Cases

Case	How to Handle
s is null or t is null	Return false immediately since a null string cannot be transformed.
s or t is empty string	If both are empty, return true; if only one is empty, return false.
s.length != t.length	Return false immediately because strings of different lengths cannot be transformed.
s and t are identical strings	Return true immediately because no transformation is needed.
s and t have same characters but different frequencies (e.g., s = 'aabb', t = 'abab')	Return false because substring sort operations can't change character frequencies; verify character counts are identical.
s and t are reverse sorted	The string 's' needs to be transformable into 't' via substring sorts so we need to verify that character order is possible from 's' to 't'.
Long strings exceeding memory limits or causing performance issues	Consider optimizations like using character counts or early exit if character frequency analysis shows impossible transformation; also consider potential integer overflow risks while summing.
s can be transformed into t with several operations	The solution should be able to handle such cases because character counts and valid order matters for transformability rather than number of operations.

s is null or t is null

How to Handle:

Return false immediately since a null string cannot be transformed.

s or t is empty string

How to Handle:

If both are empty, return true; if only one is empty, return false.

s.length != t.length

How to Handle:

Return false immediately because strings of different lengths cannot be transformed.

s and t are identical strings

How to Handle:

Return true immediately because no transformation is needed.

s and t have same characters but different frequencies (e.g., s = 'aabb', t = 'abab')

How to Handle:

Return false because substring sort operations can't change character frequencies; verify character counts are identical.

s and t are reverse sorted

How to Handle:

The string 's' needs to be transformable into 't' via substring sorts so we need to verify that character order is possible from 's' to 't'.

Long strings exceeding memory limits or causing performance issues

How to Handle:

Consider optimizations like using character counts or early exit if character frequency analysis shows impossible transformation; also consider potential integer overflow risks while summing.

s can be transformed into t with several operations

How to Handle:

The solution should be able to handle such cases because character counts and valid order matters for transformability rather than number of operations.