Rearrange Characters to Make Target String

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You are given two 0-indexed strings s and target. You can take some letters from s and rearrange them to form new strings.

Return the maximum number of copies of target that can be formed by taking letters from s and rearranging them.

Example 1:

Input: s = "ilovecodingonleetcode", target = "code"
Output: 2
Explanation:
For the first copy of "code", take the letters at indices 4, 5, 6, and 7.
For the second copy of "code", take the letters at indices 17, 18, 19, and 20.
The strings that are formed are "ecod" and "code" which can both be rearranged into "code".
We can make at most two copies of "code", so we return 2.

Example 2:

Input: s = "abcba", target = "abc"
Output: 1
Explanation:
We can make one copy of "abc" by taking the letters at indices 0, 1, and 2.
We can make at most one copy of "abc", so we return 1.
Note that while there is an extra 'a' and 'b' at indices 3 and 4, we cannot reuse the letter 'c' at index 2, so we cannot make a second copy of "abc".

Example 3:

Input: s = "abbaccaddaeea", target = "aaaaa"
Output: 1
Explanation:
We can make one copy of "aaaaa" by taking the letters at indices 0, 3, 6, 9, and 12.
We can make at most one copy of "aaaaa", so we return 1.

Constraints:

1 <= s.length <= 100
1 <= target.length <= 10
s and target consist of lowercase English letters.

Note: This question is the same as 1189: Maximum Number of Balloons.

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 the input strings `s` and `target` contain characters outside the standard ASCII range, or are they limited to a specific character set like lowercase English letters?
Are there any constraints on the length of the input strings `s` and `target`? Could either string be empty or null?
If it's impossible to rearrange characters in `s` to form `target`, what should the function return (e.g., `null`, an empty string, or a specific error code)?
Does the order of characters in the `target` string matter, or are we just checking if all the required characters are present in `s` with sufficient frequency?
If there are multiple ways to rearrange characters in `s` to form `target`, does the function need to return a specific arrangement, or can it return any valid arrangement?

Brute Force Solution

Approach

The brute force method for this problem involves trying out every possible way to rearrange the source characters and then checking if that rearrangement exactly matches our target string. It's like trying to spell the target word by randomly picking letters from a bag and seeing if we get it right by accident. This strategy guarantees finding the answer if it exists, but it might take a very, very long time.

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

Consider all possible arrangements of the source characters.
For each arrangement, compare it to the target string.
If an arrangement is exactly the same as the target string, then the source characters can be rearranged to make the target string. We are done!
If we check all possible arrangements and none of them match the target string, then it's impossible to create the target string with those source characters.

Code Implementation

from itertools import permutations

def can_rearrange_to_make_target_brute_force(source_string, target_string):
    # Generate all possible permutations of the source string
    all_permutations = permutations(source_string)

    for possible_arrangement in all_permutations:
        # Convert the tuple of characters to a string for comparison
        rearranged_string = ''.join(possible_arrangement)

        # If a permutation matches the target, we're done
        if rearranged_string == target_string:
            return True

    # If no permutation matched, it's not possible
    return False

Big(O) Analysis

Time Complexity

O(n!) – The brute force approach considers all possible arrangements of the source string, where 'n' is the length of the source string. Generating all permutations of a string of length 'n' takes O(n!) time. For each of these n! permutations, we need to compare it with the target string, which takes O(n) time. Therefore, the overall time complexity is O(n! * n), but since n! dominates n, we approximate the complexity as O(n!).

Space Complexity

O(N!) – The brute force approach described explores all possible arrangements (permutations) of the source characters. While the algorithm itself might not explicitly store *all* permutations simultaneously, the recursive calls needed to generate each permutation implicitly consume space on the call stack. In the worst-case scenario, for a source string of length N, the recursion depth can reach N. For each level of recursion we might need to pass in parameters such as the partial permutation and the remaining characters which take O(N) space. This results in O(N) * N levels which is more than O(N!). Generating all permutations contributes dominantly to an auxiliary space roughly proportional to the number of permutations (N!).

Optimal Solution

Approach

The optimal approach counts how many times each character appears in both the input string and the target string. Then, it figures out the maximum number of target strings you can create by looking at which character is the most limiting factor.

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

Count how often each character shows up in the initial string and the target string.
For each character in the target string, determine how many full target strings you can make based solely on the number of this character available in the initial string.
Find the smallest of all these counts. This number represents the maximum number of complete target strings you can create.
This is because the character with the lowest count will limit how many times you can fully form the target string.

Code Implementation

def rearrange_characters_to_make_target(initial_string, target_string):
    initial_string_char_counts = {}
    target_string_char_counts = {}

    for char in initial_string:
        initial_string_char_counts[char] = initial_string_char_counts.get(char, 0) + 1

    for char in target_string:
        target_string_char_counts[char] = target_string_char_counts.get(char, 0) + 1

    # We'll track the limiting factor to determine the max number of target strings
    maximum_number_of_target_strings = float('inf')

    for char, count in target_string_char_counts.items():
        # Determine if the char exists in initial string.
        if char not in initial_string_char_counts:
            return 0

        # Determine how many target strings we can create based on char.
        number_of_target_strings_for_char =
            initial_string_char_counts[char] // count

        # Finding the minimum limits the target strings
        maximum_number_of_target_strings =
            min(maximum_number_of_target_strings, number_of_target_strings_for_char)

    return maximum_number_of_target_strings

Big(O) Analysis

Time Complexity

O(m + t) – The algorithm first counts character frequencies in the input string 's' of length 'm'. This takes O(m) time. Then, it counts character frequencies in the target string 'target' of length 't', taking O(t) time. The remaining steps involve iterating through the target string's character counts, which is bounded by the size of the target string. Therefore, the overall time complexity is dominated by O(m + t), where m is the length of the input string and t is the length of the target string. Since we process each character from both strings at most a constant number of times, the combined runtime is linear with respect to the input string lengths.

Space Complexity

O(1) – The solution creates two hash maps to store character counts for the input string and the target string. The size of these hash maps is bounded by the number of unique characters, which is at most the size of the alphabet, a constant. Therefore, the auxiliary space used is independent of the length of the input strings. Thus, the space complexity is O(1).

Edge Cases

Case	How to Handle
Null or empty source string	Return 0 or an appropriate error code to indicate no target string can be formed.
Null or empty target string	Return 1 (or a suitable value indicating the target can always be formed) as an empty string is always a substring.
Source string is shorter than target string	Return 0 or an appropriate error code because the target cannot be formed.
Target string contains characters not present in the source string	Return 0 because the target cannot be formed.
Source string and target string are identical	Check the frequency of characters, it is rearrangeable if the frequency in source is greater or equal to target's frequency.
Source string has insufficient frequency of a character required by target	Return 0 or an appropriate error if the available quantity of a needed character is insufficient.
Very long source and target strings, potential memory issues	The frequency map approach can be memory-intensive; using optimized data structures might be needed for extremely long strings.
Source string contains Unicode characters	Ensure the character counting mechanism correctly handles Unicode characters, which might require larger data types or different encoding considerations.

Null or empty source string

How to Handle:

Return 0 or an appropriate error code to indicate no target string can be formed.

Null or empty target string

How to Handle:

Return 1 (or a suitable value indicating the target can always be formed) as an empty string is always a substring.

Source string is shorter than target string

How to Handle:

Return 0 or an appropriate error code because the target cannot be formed.

Target string contains characters not present in the source string

How to Handle:

Return 0 because the target cannot be formed.

Source string and target string are identical

How to Handle:

Check the frequency of characters, it is rearrangeable if the frequency in source is greater or equal to target's frequency.

Source string has insufficient frequency of a character required by target

How to Handle:

Return 0 or an appropriate error if the available quantity of a needed character is insufficient.

Very long source and target strings, potential memory issues

How to Handle:

The frequency map approach can be memory-intensive; using optimized data structures might be needed for extremely long strings.

Source string contains Unicode characters

How to Handle:

Ensure the character counting mechanism correctly handles Unicode characters, which might require larger data types or different encoding considerations.