Construct String with Minimum Cost

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Dynamic ProgrammingStrings

You are given a string target, an array of strings words, and an integer array costs, both arrays of the same length.

Imagine an empty string s.

You can perform the following operation any number of times (including zero):

Choose an index i in the range [0, words.length - 1].
Append words[i] to s.
The cost of operation is costs[i].

Return the minimum cost to make s equal to target. If it's not possible, return -1.

Example 1:

Input: target = "abcdef", words = ["abdef","abc","d","def","ef"], costs = [100,1,1,10,5]

Output: 7

Explanation:

The minimum cost can be achieved by performing the following operations:

Select index 1 and append "abc" to s at a cost of 1, resulting in s = "abc".
Select index 2 and append "d" to s at a cost of 1, resulting in s = "abcd".
Select index 4 and append "ef" to s at a cost of 5, resulting in s = "abcdef".

Example 2:

Input: target = "aaaa", words = ["z","zz","zzz"], costs = [1,10,100]

Output: -1

Explanation:

It is impossible to make s equal to target, so we return -1.

Constraints:

1 <= target.length <= 5 * 10⁴
1 <= words.length == costs.length <= 5 * 10⁴
1 <= words[i].length <= target.length
The total sum of words[i].length is less than or equal to 5 * 10⁴.
target and words[i] consist only of lowercase English letters.
1 <= costs[i] <= 10⁴

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 length of the string we need to construct, and the cost to add each character?
What characters are allowed in the string? Is it only lowercase English letters, or are other characters possible?
If it's impossible to construct the string, what should I return? Should I return an empty string, null, or throw an exception?
Is the cost associated with adding a character always non-negative?
Are there any constraints on the total cost allowed? Could the optimal cost potentially overflow an integer?

Brute Force Solution

Approach

The brute force approach is about trying every single possible string we can make and checking if it is the cheapest. We generate every possible string that meets the given length requirement. Then we compare the cost of all the strings and choose the one that has the minimum cost.

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

Start by creating every possible string of the given length.
For each possible string, calculate its cost using the rules provided.
Compare the cost of each string to the current minimum cost found so far.
If the cost of the current string is less than the minimum cost so far, then update the minimum cost and remember the string.
After checking all possible strings, return the string with the minimum cost.

Code Implementation

def construct_string_brute_force(length, cost_a, cost_b):

    minimum_cost = float('inf')
    best_string = ""

    # Generate all possible strings of given length
    for i in range(2**(length)):
        current_string = bin(i)[2:].zfill(length)
        current_cost = 0

        # Calculate the cost of the current string
        for j in range(length):
            if current_string[j] == '0':
                current_cost += cost_a
            else:
                current_cost += cost_b

        # Compare current cost to the current min cost
        if current_cost < minimum_cost:
            minimum_cost = current_cost
            best_string = current_string

    return best_string

Big(O) Analysis

Time Complexity

O(26^n * n) – The brute force approach generates every possible string of length n where each character can be one of 26 lowercase letters. This generation takes O(26^n) time. For each of these strings, the cost is calculated by iterating through the string which takes O(n) time. Therefore, the overall time complexity is O(26^n * n).

Space Complexity

O(2^N) – The algorithm generates all possible strings of length N. In the worst-case scenario, if we are considering strings consisting of two characters, the number of possible strings is 2^N. The algorithm stores each generated string to calculate its cost, thus requiring space proportional to the number of possible strings, 2^N. Since the algorithm stores one minimum cost string, it's space complexity will be dominated by the space required to hold all possible strings of length N for computation, resulting in O(2^N) space. Therefore, the auxiliary space complexity is O(2^N).

Optimal Solution

Approach

The goal is to build a string in such a way that minimizes the cost of adding characters. The key idea is to recognize repeating patterns and reuse them instead of building them from scratch each time. This involves identifying common substrings and efficiently incorporating them into the final string.

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

First, identify the repeating portions of the string that you want to build.
Calculate the cost to build each repeating portion from scratch.
Consider how much it costs to build the repeating portion just once, and then reuse it later by adding only the characters needed to connect the repeating sections.
Choose the most cost-effective method, either building from scratch each time, or building once and reusing.
Combine all portions together while ensuring you're making optimal decisions at each step to reduce the overall string build cost.

Code Implementation

def construct_string_min_cost(base_cost, append_cost, target_string):
    string_length = len(target_string)
    min_cost = float('inf')

    for substring_length in range(1, string_length + 1):
        # Iterate through possible substring lengths
        substring = target_string[:substring_length]
        substring_cost = base_cost + (substring_length - 1) * append_cost
        current_cost = substring_cost
        current_string = substring
        
        # Try reusing the substring to build the target string
        index = substring_length
        while index < string_length:
            if target_string[index:index + substring_length] == substring:
                current_cost += append_cost  # Cost to reuse substring
                index += substring_length
                current_string += substring
            else:
                current_cost += append_cost  # Cost to add a character
                current_string += target_string[index]
                index += 1

        # We only want the cost of building the target string
        if len(current_string) > string_length:
            cost_difference = len(current_string) - string_length
            current_cost -= cost_difference * append_cost

        # Compare current cost to minimum cost
        min_cost = min(min_cost, current_cost)

    # Now calculate cost if building without repeats
    no_repeat_cost = base_cost + (string_length - 1) * append_cost

    # Compare building with or without repeats to find minimum cost
    min_cost = min(min_cost, no_repeat_cost)

    return min_cost

def construct_string_min_cost_optimized(base_cost, append_cost, target_string):
    string_length = len(target_string)
    minimum_cost = base_cost + (string_length - 1) * append_cost

    # Iterate through possible substring lengths
    for substring_length in range(1, string_length // 2 + 1):
        substring = target_string[:substring_length]
        substring_cost = base_cost + (substring_length - 1) * append_cost
        
        # Find the number of times the substring can be reused
        reuse_count = 0
        index = substring_length
        while index + substring_length <= string_length and target_string[index:index + substring_length] == substring:
            reuse_count += 1
            index += substring_length

        # Calculate cost using repeats
        remaining_length = string_length - (substring_length + reuse_count * substring_length)
        
        # Calculate the cost of the rest of the string
        repeat_cost = substring_cost + reuse_count * append_cost + remaining_length * append_cost

        # Update the minimum cost
        minimum_cost = min(minimum_cost, repeat_cost)

    return minimum_cost

# Example usage:
# base_cost = 5
# append_cost = 3
# target_string = "abcabcabc"

# result = construct_string_min_cost_optimized(base_cost, append_cost, target_string)
# print(result)

def constructString(add, copy, s):
    n = len(s)
    dp = [0] * (n + 1)
    dp[1] = add
    
    for i in range(2, n + 1):
        dp[i] = dp[i - 1] + add
        # Check for substrings to copy
        for j in range(1, i // 2 + 1):

            # Check if the substring can be copied
            if s[i-j:i] == s[i-2*j:i-j]:
                dp[i] = min(dp[i], dp[i - j] + copy)

    return dp[n]

Big(O) Analysis

Time Complexity

O(n^2) – The dominant operation is identifying and evaluating repeating substrings. We iterate through all possible substring lengths and starting positions, which involves checking substrings of length 1 up to n/2, from each of the n possible starting positions. For each substring, we compare it to other substrings within the main string to find matches. Comparing strings takes O(n) in the worst case, and checking all possible combinations of substrings accounts for the O(n^2) time complexity. Therefore, the overall runtime is bounded by O(n^2).

Space Complexity

O(N) – The space complexity is determined by the size of the repeating portions identified and stored for potential reuse. In the worst-case scenario, the algorithm might need to store substrings whose combined length could approach the length of the input string N, where N is the length of the string to be built. This is because the identified repeating portions could theoretically cover almost the entire string. Therefore, the auxiliary space used to store these substrings grows linearly with the input size N. The cost to build each repeating portion or the connection between them does not contribute to additional space because once built, it will be combined.

Edge Cases

Case	How to Handle
Null or empty input string	Return an empty string or throw an IllegalArgumentException, depending on problem requirements and language conventions.
String of length 1	If the cost of adding a character is greater than 0, simply return the input string, otherwise if cost is zero and removal is cheaper return empty string.
String with all identical characters	The optimal solution might involve removing most or all characters, so consider costs carefully.
Costs are all zero	If all costs are zero, the solution should consider whether to return the original string or an empty string based on problem constraints.
Integer overflow when calculating the cost	Use appropriate data types (e.g., long in Java or C++) to prevent integer overflow during cost calculation.
Maximum length string exceeding memory limits	Consider using dynamic programming approaches with optimized memory usage or streaming algorithms if the string is extremely large.
Costs include negative values	The algorithm should handle negative costs correctly, potentially favoring adding characters when the cost is negative.
Multiple valid solutions exist with the same minimum cost	The problem description should specify if any valid solution is acceptable or if a specific one (e.g., lexicographically smallest) is required.

Null or empty input string

How to Handle:

Return an empty string or throw an IllegalArgumentException, depending on problem requirements and language conventions.

String of length 1

How to Handle:

If the cost of adding a character is greater than 0, simply return the input string, otherwise if cost is zero and removal is cheaper return empty string.

String with all identical characters

How to Handle:

The optimal solution might involve removing most or all characters, so consider costs carefully.

Costs are all zero

How to Handle:

If all costs are zero, the solution should consider whether to return the original string or an empty string based on problem constraints.

Integer overflow when calculating the cost

How to Handle:

Use appropriate data types (e.g., long in Java or C++) to prevent integer overflow during cost calculation.

Maximum length string exceeding memory limits

How to Handle:

Consider using dynamic programming approaches with optimized memory usage or streaming algorithms if the string is extremely large.

Costs include negative values

How to Handle:

The algorithm should handle negative costs correctly, potentially favoring adding characters when the cost is negative.

Multiple valid solutions exist with the same minimum cost

How to Handle:

The problem description should specify if any valid solution is acceptable or if a specific one (e.g., lexicographically smallest) is required.