#2746

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Decremental String Concatenation

Solve the Decremental String Concatenation problem by applying dynamic programming to minimize string length after concatenations.

array string dynamic programming

Problem Statement

You are given a 0-indexed array words containing n strings. The goal is to minimize the length of the concatenated string formed by joining the elements of the array. The join operation between two strings x and y is defined as concatenating them into xy, but if the last character of x is equal to the first character of y, one of them is deleted.

For example, the join operation between "ab" and "ba" will result in "aba" since the last character of the first string matches the first character of the second. On the other hand, joining "ab" and "cde" results in "abcde" without any reduction in length. Your task is to find the minimum possible length of the final concatenated string formed from the given words array.

Examples

Example 1

Input: words = ["aa","ab","bc"]

Output: 4

In this example, we can perform join operations in the following order to minimize the length of str2: str0 = "aa" str1 = join(str0, "ab") = "aab" str2 = join(str1, "bc") = "aabc" It can be shown that the minimum possible length of str2 is 4.

Example 2

Input: words = ["ab","b"]

Output: 2

In this example, str0 = "ab", there are two ways to get str1: join(str0, "b") = "ab" or join("b", str0) = "bab". The first string, "ab", has the minimum length. Hence, the answer is 2.

Example 3

Input: words = ["aaa","c","aba"]

Output: 6

In this example, we can perform join operations in the following order to minimize the length of str2: str0 = "aaa" str1 = join(str0, "c") = "aaac" str2 = join("aba", str1) = "abaaac" It can be shown that the minimum possible length of str2 is 6.

Constraints

1 <= words.length <= 1000
1 <= words[i].length <= 50
Each character in words[i] is an English lowercase letter

Solution Approach

Dynamic Programming with Memoization

The problem can be approached using dynamic programming with memoization. We can define a DP state dp[i] where i is the index in the array, and dp[i] represents the minimum length of the string obtained after performing join operations starting from words[i]. By recursively solving subproblems and utilizing memoization, we can efficiently find the optimal order of joins.

State Transitions Between Words

The key idea is to examine the transitions between each pair of words. We need to check if joining two strings results in a reduced length, based on whether the last character of the first string matches the first character of the second. This allows us to update the DP state accordingly and propagate the minimum length of concatenated strings throughout the array.

Bottom-Up Approach for Optimal Performance

We can optimize the solution by iterating over the array in a bottom-up manner. By starting from the last word and working backward, we can progressively compute the minimum length for each position in the array. This approach ensures that the solution is computed efficiently and reduces redundant computations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the final implementation. A brute force approach might result in exponential time complexity, while dynamic programming with memoization reduces this to O(n^2) where n is the number of words. The space complexity is O(n) due to the storage requirements for the DP states and memoization.

What Interviewers Usually Probe

The candidate demonstrates understanding of dynamic programming and its application to string manipulation.
The candidate correctly identifies the state transition process and efficiently computes the minimum length.
The candidate uses memoization to avoid redundant calculations, ensuring the solution is optimized.

Common Pitfalls or Variants

Common pitfalls

Failing to account for the deletion of characters when joining strings, leading to incorrect length calculations.
Using a brute force approach instead of dynamic programming, resulting in inefficient solutions for larger input sizes.
Not properly applying memoization, leading to redundant calculations and performance issues.

Follow-up variants

Extend the problem to handle larger character sets or longer strings.
Introduce additional constraints, such as limiting the number of allowed join operations.
Modify the problem to allow for multiple deletions when joining strings with matching characters.

FAQ

How do I solve the Decremental String Concatenation problem?

Use dynamic programming with memoization to minimize the string length after performing join operations between words. Ensure you account for character deletions when the last character of one string matches the first character of the next.

What is the main challenge in the Decremental String Concatenation problem?

The main challenge lies in efficiently minimizing the concatenated string length by correctly handling the join operation and optimizing the order of operations using dynamic programming.

How does dynamic programming help in solving the Decremental String Concatenation problem?

Dynamic programming helps by storing the results of subproblems (minimum string lengths) and avoiding redundant calculations, thus improving the solution's efficiency.

What is memoization in the context of this problem?

Memoization involves storing the results of previously computed DP states so that they can be reused in later calculations, avoiding redundant work and reducing time complexity.

What are some common mistakes to avoid when solving the Decremental String Concatenation problem?

Common mistakes include not handling string deletions correctly, using inefficient brute force approaches, and failing to apply memoization, leading to redundant calculations.

terminal

Solution

Solution 1: Memoization Search

We notice that when concatenating strings, the first and last characters of the string will affect the length of the concatenated string. Therefore, we design a function $dfs(i, a, b)$, which represents the minimum length of the concatenated string starting from the $i$-th string, and the first character of the previously concatenated string is $a$, and the last character is $b$.

class Solution:
    def minimizeConcatenatedLength(self, words: List[str]) -> int:
        @cache
        def dfs(i: int, a: str, b: str) -> int:
            if i >= len(words):
                return 0
            s = words[i]
            x = dfs(i + 1, a, s[-1]) - int(s[0] == b)
            y = dfs(i + 1, s[0], b) - int(s[-1] == a)
            return len(s) + min(x, y)

        return len(words[0]) + dfs(1, words[0][0], words[0][-1])

Continue Practicing

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