How do you solve Kth Smallest Instructions on LeetCode?

Kth Smallest Instructions (LeetCode #1643) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Understanding combinatorial counting can greatly reduce the complexity of problems involving sequences.

What is the brute force solution for Kth Smallest Instructions?

The brute force approach for Kth Smallest Instructions is Brute Force, with O(n²) time complexity and O(n!) space complexity. The brute force approach generates all possible instruction sequences to reach the destination and sorts them to find the k-th smallest. This is straightforward but inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Kth Smallest Instructions?

Kth Smallest Instructions uses the following concepts: Array, Math, Dynamic Programming, Combinatorics. The recommended patterns to study are: Dynamic Programming, Combinatorics.

What are the interview tips for Kth Smallest Instructions?

Always start with a brute force solution to understand the problem better. Discuss your thought process with the interviewer; they may provide hints or guidance. Practice combinatorial problems to become comfortable with counting techniques.

What are common mistakes when solving Kth Smallest Instructions?

Failing to recognize the need for combinatorial counting and attempting to generate all sequences. Not considering edge cases, such as when either row or column is zero.

#1643

Kth Smallest Instructions

Hard

ArrayMathDynamic ProgrammingCombinatoricsDynamic ProgrammingCombinatorics

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of generating all combinations, we can use combinatorial counting to directly find the k-th instruction. This approach is efficient and avoids unnecessary computations.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty result string and set the remaining row and column moves.
2Step 2: For each step, calculate how many valid sequences start with 'H' and 'V'.
3Step 3: If the number of sequences starting with 'H' is greater than or equal to k, append 'H' to the result; otherwise, append 'V' and adjust k accordingly.
4Step 4: Repeat until all moves are used.

solution.py19 lines

1# Full working Python code
2from math import comb
3
4def kthSmallestInstructions(destination, k):
5    row, column = destination
6    result = []
7    while row > 0 or column > 0:
8        if column > 0:
9            count_H = comb(row + column - 1, row)
10        else:
11            count_H = 0
12        if k <= count_H:
13            result.append('H')
14            column -= 1
15        else:
16            result.append('V')
17            k -= count_H
18            row -= 1
19    return ''.join(result)

ℹ

Complexity note: The time complexity is O(n) since we only iterate through the moves, and the space complexity is O(1) as we use a constant amount of space.

1Understanding combinatorial counting can greatly reduce the complexity of problems involving sequences.
2Recognizing patterns in the problem (like the number of valid sequences) helps in making efficient decisions.

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