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LeetCode 2746: Decremental String Concatenation Solution

Master LeetCode problem 2746 (Decremental String Concatenation), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2746. Decremental String Concatenation

Medium
ArrayStringDynamic Programming

Problem Explanation

Explanation

To solve this problem, we can use a dynamic programming approach. We will iterate through the array of words and for each word, we will consider two possibilities: either append the word to the current result or prepend the word to the current result. We will keep track of the minimum length achieved at each step. The key insight is that we need to consider the cases where the last character of the current result matches the first character of the word we are considering.

At the end of the iteration, the minimum possible length of the final concatenated string will be the answer.

Algorithm:

  1. Initialize a variable result to store the current result as the first word in the input array.
  2. Iterate through the array of words from index 1 to n-1.
  3. For each word, consider two possibilities: append the word to the result or prepend the word to the result.
  4. Calculate the length of the concatenated string after each operation and update result to the one with the minimum length.
  5. Return the length of the final concatenated string as the answer.

Time Complexity:

The time complexity of this approach is O(n**2) where n is the number of words in the input array.

Space Complexity:

The space complexity is O(n) to store the current result.

Solution Code

class Solution {
    public int minConcatenationLength(String[] words) {
        String result = words[0];
        
        for (int i = 1; i < words.length; i++) {
            String option1 = words[i] + result;
            String option2 = result + words[i];
            
            result = (option1.length() <= option2.length()) ? option1 : option2;
        }
        
        return result.length();
    }
}

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