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Step-by-Step Explanation for Finding the k-th Lexicographical Number

This README will explain the core logic behind finding the k-th lexicographically smallest number between 1 and n for C++, Java, JavaScript, Python, and Go implementations.

General Approach

The problem revolves around lexicographical order (dictionary-like order) for numbers. We aim to find the k-th smallest number in that order. The solution uses a prefix tree traversal approach, where we explore numbers level by level (digit by digit) as they grow.

At each step, we either:

  1. Dive deeper into the current prefix (e.g., from 1 to 10, then to 100, etc.) if the k-th number exists within the subtree of the current prefix.
  2. Move to the next sibling prefix (e.g., from 1 to 2) if the k-th number doesn't lie within the current subtree.

We start at the smallest prefix (1) and count how many numbers exist under each prefix to determine whether we should skip the prefix or explore it further.

Step-by-Step Breakdown (Applicable to All Languages)

Step 1: Initialize Variables

  • curr: Start with the first lexicographical prefix, which is 1.
  • k: Decrement k by 1 because we already start at the first number (which is 1).

Step 2: Count Steps for a Prefix

To determine whether the k-th number lies under the current prefix, we need to calculate how many numbers exist that begin with this prefix. The logic is as follows:

  1. Initialize steps = 0: This keeps track of how many valid numbers start with the current prefix.
  2. Set bounds:
    • first = curr: The smallest number starting with the current prefix.
    • last = curr: The largest number starting with the current prefix.
  3. Expand the range:
    • Move to the next level by multiplying first and last by 10 (e.g., from 1 to 10, 19, 100, etc.).
    • At each level, count the valid numbers between first and last (ensuring not to exceed n).

Step 3: Compare Steps and k

Now that we know how many numbers exist under the current prefix, we compare:

  • If steps <= k: The k-th number is not within the subtree of this prefix.

    • Action: Move to the next sibling prefix (curr++).
    • Update k: Subtract the number of steps because we skip over these numbers.

  • If steps > k: The k-th number lies within this subtree.

    • Action: Dive deeper into the current prefix (curr *= 10).
    • Update k: Decrement by 1 because we're now moving to the next level.

Step 4: Return the Result

Once k reaches 0, the current prefix (curr) is the k-th lexicographical number, and we return it.

C++ Explanation

  • countSteps Function: A helper function that counts how many numbers exist in the lexicographical range starting with curr up to n.
  • findKthNumber Function: The main function that iterates through prefixes, comparing steps with k, and either dives deeper into the current prefix or moves to the next prefix until k reaches 0.

Java Explanation

  • Similar logic is implemented with helper and main functions:
    • countSteps: Calculates steps from the current prefix.
    • findKthNumber: Traverses the tree-like structure of prefixes and returns the result.

JavaScript Explanation

  • countSteps: Calculates the number of numbers with the current prefix.
  • findKthNumber: Finds the k-th number using the same prefix comparison strategy.

Python Explanation

  • countSteps: Returns the number of numbers that start with the current prefix.
  • findKthNumber: Implements the prefix traversal logic to determine the k-th lexicographical number.

Go Explanation

  • countSteps: Calculates the total number of numbers for a given prefix.
  • findKthNumber: Implements the main logic to explore and find the k-th smallest number lexicographically.

Conclusion

The logic behind finding the k-th lexicographical number is consistent across all languages, revolving around counting how many numbers exist under each prefix and either diving deeper into the prefix or skipping to the next one based on the comparison with k. The efficiency comes from the fact that we don’t need to generate all numbers, but rather, we traverse them systematically like a prefix tree.