Construct the Longest New String

Medium

Guidewire

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Topics:

Greedy Algorithms

You are given three integers x, y, and z.

You have x strings equal to "AA", y strings equal to "BB", and z strings equal to "AB". You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring.

Return the maximum possible length of the new string.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: x = 2, y = 5, z = 1
Output: 12
Explanation: We can concatenate the strings "BB", "AA", "BB", "AA", "BB", and "AB" in that order. Then, our new string is "BBAABBAABBAB". 
That string has length 12, and we can show that it is impossible to construct a string of longer length.

Example 2:

Input: x = 3, y = 2, z = 2
Output: 14
Explanation: We can concatenate the strings "AB", "AB", "AA", "BB", "AA", "BB", and "AA" in that order. Then, our new string is "ABABAABBAABBAA". 
That string has length 14, and we can show that it is impossible to construct a string of longer length.

Constraints:

1 <= x, y, z <= 50

Try coding it in LeetCode

Construct the Longest New String

Medium

Guidewire

3 views

Topics:

Greedy Algorithms

You are given three integers x, y, and z.

You have x strings equal to "AA", y strings equal to "BB", and z strings equal to "AB". You want to choose some (possibly all or none) of these strings and concatenate them in some order to form a new string. This new string must not contain "AAA" or "BBB" as a substring.

Return the maximum possible length of the new string.

A substring is a contiguous non-empty sequence of characters within a string.

Example 1:

Input: x = 2, y = 5, z = 1
Output: 12
Explanation: We can concatenate the strings "BB", "AA", "BB", "AA", "BB", and "AB" in that order. Then, our new string is "BBAABBAABBAB". 
That string has length 12, and we can show that it is impossible to construct a string of longer length.

Example 2:

Input: x = 3, y = 2, z = 2
Output: 14
Explanation: We can concatenate the strings "AB", "AB", "AA", "BB", "AA", "BB", and "AA" in that order. Then, our new string is "ABABAABBAABBAA". 
That string has length 14, and we can show that it is impossible to construct a string of longer length.

Constraints:

1 <= x, y, z <= 50

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:

What are the constraints on the values of `a`, `b`, and `c`? Can they be zero or negative?
If it's impossible to construct a string that satisfies the given `a`, `b`, and `c`, what value should I return?
Are `a`, `b`, and `c` guaranteed to be integers?
Could you provide a small example or two to illustrate the expected behavior?
Are there any relationships or dependencies between `a`, `b`, and `c` that are guaranteed? For example, is `c` always less than or equal to `a + b`?

Brute Force Solution

Approach

The brute force method for constructing the longest new string involves trying every conceivable combination of the available characters to see which one produces the longest valid string. We systematically explore each possibility and check if it adheres to the specified rules. We keep the longest valid string found along the way.

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

Start by considering building a string of length 1 using the provided character counts.
Examine every possible character at the first position to see if we have enough of that character to use it.
If we do, move on to consider the second character of the string. Check every possible valid character to see if it extends the string according to the rules.
Continue adding characters one at a time, each time checking all possibilities while following the combination rules and ensuring we don't exceed the allowed count of any character.
As we build each string, check if the string adheres to any constraints of character adjacencies.
If we encounter a string that is longer than any string we have built so far, save it as the best string we have seen.
Once we have exhausted all possibilities, return the best/longest string we identified.

Code Implementation

def construct_longest_new_string_brute_force(count_a, count_b, count_c):
    longest_string_found = ""

    def build_string(current_string, remaining_a, remaining_b, remaining_c):
        nonlocal longest_string_found

        # Update the longest valid string found so far.
        if len(current_string) > len(longest_string_found):
            longest_string_found = current_string

        # Check if we can add 'a'.
        if remaining_a > 0 and (not current_string or current_string[-1] != 'a'):
            build_string(current_string + 'a', remaining_a - 1, remaining_b, remaining_c)

        # Check if we can add 'b'.
        if remaining_b > 0 and (not current_string or current_string[-1] != 'b'):
            build_string(current_string + 'b', remaining_a, remaining_b - 1, remaining_c)

        # Explore the option of adding the character 'c'.
        if remaining_c > 0 and (not current_string or current_string[-1] != 'c'):
            build_string(current_string + 'c', remaining_a, remaining_b, remaining_c - 1)

    build_string("", count_a, count_b, count_c)
    return longest_string_found

Big(O) Analysis

Time Complexity

O(3^(a+b+c)) – The brute force algorithm explores all possible combinations of characters 'a', 'b', and 'c' to build the longest valid string. In the worst-case scenario, where a, b, and c are roughly equal, the algorithm essentially tries all possible strings of length a+b+c. For each position in the string, it has up to 3 choices ('a', 'b', or 'c'), leading to a branching factor of 3. Thus, the total number of possible strings explored grows exponentially with the sum of a, b, and c. Therefore, the time complexity is O(3^(a+b+c)).

Space Complexity

O(N) – The brute force approach constructs strings of potentially varying lengths, up to a maximum length dictated by the counts of available characters. In the worst case, a string of length N (where N is the maximum possible length of the string constructed from the input counts) will be stored as the 'best string'. The recursion involved in exploring combinations can also lead to a call stack of depth proportional to N in the worst-case scenario. Therefore, the auxiliary space is O(N).

Optimal Solution

Approach

The key is to strategically build the longest possible string by maximizing the use of the available characters 'x', 'y', and 'z'. We figure out how to connect them in the most efficient way possible, given the rules, without trying every single possible arrangement.

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

First, compare the counts of 'x' and 'y'.
If the counts are the same, the optimal string will be formed by alternating 'xy' as many times as possible and adding two 'z' at the end.
If one character ('x' or 'y') has a count one greater than the other, again alternate 'xy' as many times as possible, but this time start with the character with the higher count. Add two 'z' at the end.
If the difference between the counts of 'x' and 'y' is two or more, alternate 'xy' or 'yx' as many times as possible, starting with the character with the larger count. Add the other character to the end, and also add two 'z' at the end.
The final string is the combination of all these parts.
If any of x, y or z has a count of 0, it can be omitted.

Code Implementation

def construct_longest_new_string(x_count, y_count, z_count):
    result = ""

    if x_count == y_count:
        for _ in range(x_count):
            result += "xy"
        if z_count > 0:
            result += "zz"

    elif abs(x_count - y_count) == 1:
        if x_count > y_count:
            for _ in range(y_count):
                result += "xy"
            result += "x"
        else:
            for _ in range(x_count):
                result += "yx"
            result += "y"
        if z_count > 0:
            result += "zz"

    else:
        # x_count and y_count differ by 2 or more
        if x_count > y_count:
            for _ in range(y_count):
                result += "xy"
            result += "x"
            result += "x"
        else:
            for _ in range(x_count):
                result += "yx"
            result += "y"
            result += "y"

        # Append z to the result string
        if z_count > 0:
            result += "zz"

    return result

Big(O) Analysis

Time Complexity

O(max(x, y)) – The dominant operations involve constructing the string by alternating 'x' and 'y' based on their counts. The number of iterations required for this alternation is directly proportional to the larger of the counts of 'x' and 'y', denoted as max(x, y). Appending the final 'z' characters and potentially a single 'x' or 'y' takes constant time and doesn't affect the overall complexity. Therefore, the time complexity is determined by the string construction loop, resulting in O(max(x, y)).

Space Complexity

O(N) – The algorithm constructs a new string. The maximum length of this string depends on the counts of 'x', 'y', and 'z', and in the worst-case scenario, the length could be proportional to the sum of these counts. Therefore, the auxiliary space used for storing the resulting string is proportional to the input size, N, where N represents the length of the constructed string. This leads to a space complexity of O(N).

Edge Cases

Case	How to Handle
a, b, and c are all 0	Return 0 since no substrings can be formed and the string must be empty.
a, b, and c are very large (close to integer limits)	Ensure the intermediate calculations don't overflow, potentially use long or alternative methods for calculation.
a = 0, b = 0, c > 0	The only possible solution is a string alternating A and B, so the length is c + 1.
a > 0, b = 0, c = 0 or a = 0, b > 0, c = 0	The string will consist entirely of either 'A's or 'B's respectively, with length equal to 1+2*count.
c = 0, but a > 0 and b > 0	No solution, since you must have AB present for a string to contain both AA and BB.
a = 0, b > 0, c > 0	String construction starts with BB and then AB, determine the number of A and B, or 'B' followed by repeating 'BA' until exhausted 'B' count, then 'AA' is not possible, the string will be 'B'+'BA'*min(b,c) and the remaining length will be accounted for
a > 0, b = 0, c > 0	String construction starts with AA and then AB, determine the number of A and B, or 'A' followed by repeating 'AB' until exhausted 'A' count, then 'BB' is not possible, the string will be 'A'+'AB'*min(a,c) and the remaining length will be accounted for
a = b = c	The length of the string would be 2*a + 1, with 'A' and 'B' almost equally distributed.