How do you solve Decremental String Concatenation on LeetCode?

Decremental String Concatenation (LeetCode #2746) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Only the first and last characters of the strings matter for determining the join length.

What is the brute force solution for Decremental String Concatenation?

The brute force approach for Decremental String Concatenation is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach tries all possible combinations of joining the strings to find the minimum length. It evaluates both ways of joining each string at every step, leading to a lot of redundant calculations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Decremental String Concatenation?

Decremental String Concatenation uses the following concepts: Array, String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Greedy Algorithms.

What are the interview tips for Decremental String Concatenation?

Always clarify the problem requirements and constraints before diving into coding. Think about edge cases, such as strings with the same characters or very short strings. Explain your thought process as you code; it shows your understanding and helps the interviewer follow along.

What are common mistakes when solving Decremental String Concatenation?

Not considering both join orders properly, leading to incorrect minimum lengths. Failing to track the lengths correctly during the join operations.

#2746

Decremental String Concatenation

Medium

ArrayStringDynamic ProgrammingDynamic ProgrammingGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses dynamic programming to store the minimum lengths for each string combination, avoiding redundant calculations. We only need to track the first and last characters of each string to determine the join length.

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Algorithm

3 steps

1Step 1: Create a DP array to store the minimum lengths for each string combination.
2Step 2: Initialize the first element of the DP array with the length of the first string.
3Step 3: For each subsequent string, calculate the minimum length by considering both join orders and update the DP array.

solution.py8 lines

1def minLength(words):
2    n = len(words)
3    dp = [0] * n
4    dp[0] = len(words[0])
5    for i in range(1, n):
6        dp[i] = min(dp[i-1] + len(words[i]) - (1 if words[i-1][-1] == words[i][0] else 0),
7                    len(words[i]) + len(words[i-1]) - (1 if words[i][0] == words[i-1][-1] else 0))
8    return dp[-1]

ℹ

Complexity note: The time complexity is O(n) because we only traverse the list of words once, and the space complexity is O(n) due to the DP array storing the minimum lengths.

1Only the first and last characters of the strings matter for determining the join length.
2Dynamic programming can significantly reduce redundant calculations by storing intermediate results.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.