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LeetCode 2573: Find the String with LCP Solution

Master LeetCode problem 2573 (Find the String with LCP), a hard 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.

Problem Explanation

Explanation

To solve this problem, we need to find the lexicographically smallest string that corresponds to the given lcp matrix. We can achieve this by iteratively constructing the string from left to right based on the LCP values.

  1. Start by creating a character array of size n to store the final string.
  2. Initialize the first character of the string as 'a'.
  3. Iterate from the second character to the last character of the string.
  4. For each character position, set the character to 'a' if the LCP value with the previous character is 0.
  5. Otherwise, set the character to the smallest character that has not been used in the prefix so far.
  6. If at any point we encounter a conflict where we cannot determine the correct character, return an empty string.
  7. Return the constructed string at the end.

This approach ensures that we construct the lexicographically smallest string based on the given LCP matrix.

Time Complexity

The time complexity of this approach is O(n^2) where n is the size of the LCP matrix.

Space Complexity

The space complexity is O(n) due to the character array used to store the final string.

Solution Code

class Solution {
    public String findTheStringWithLCP(int[][] lcp) {
        int n = lcp.length;
        char[] result = new char[n];
        result[0] = 'a';

        for (int i = 1; i < n; i++) {
            int prevLCP = lcp[i - 1][i];
            char prevChar = result[i - 1];
            char currentChar = 'a';

            while (currentChar <= prevChar) {
                currentChar++;
            }

            if (prevLCP == 0) {
                result[i] = 'a';
            } else if (currentChar - 'a' == prevLCP) {
                result[i] = currentChar;
            } else {
                return "";
            }
        }

        return String.valueOf(result);
    }
}

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