LeetCode Problem Workspace
Kth Smallest Instructions
Find the kth smallest lexicographic instruction sequence for reaching a destination in a grid using state transition dynamic programming.
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Practice Focus
Hard · State transition dynamic programming
Answer-first summary
Find the kth smallest lexicographic instruction sequence for reaching a destination in a grid using state transition dynamic programming.
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The Kth Smallest Instructions problem asks for the kth lexicographically smallest instruction sequence to travel in a grid from (0, 0) to a destination. By applying state transition dynamic programming, this problem can be efficiently solved by tracking the number of possible paths at each step and making decisions based on combinations, ensuring the correct instruction sequence is generated.
Problem Statement
Bob is at the top-left corner of a grid and wants to move to a destination at (row, column). He can only move right ('H') or down ('V'). There are multiple ways to reach the destination, and the goal is to determine the kth smallest lexicographical instruction sequence that leads Bob to the destination.
Each instruction sequence consists of exactly row 'H' moves and column 'V' moves. The number of such sequences is given by nCr(row + column, row). The challenge is to compute the kth sequence efficiently using a combination of dynamic programming and combinatorics, while maintaining the correct order of movements.
Examples
Example 1
Input: destination = [2,3], k = 1
Output: "HHHVV"
All the instructions that reach (2, 3) in lexicographic order are as follows: ["HHHVV", "HHVHV", "HHVVH", "HVHHV", "HVHVH", "HVVHH", "VHHHV", "VHHVH", "VHVHH", "VVHHH"].
Example 2
Input: destination = [2,3], k = 2
Output: "HHVHV"
Example details omitted.
Example 3
Input: destination = [2,3], k = 3
Output: "HHVVH"
Example details omitted.
Constraints
- destination.length == 2
- 1 <= row, column <= 15
- 1 <= k <= nCr(row + column, row), where nCr(a, b) denotes a choose b.
Solution Approach
State Transition Dynamic Programming
The problem can be modeled as a state transition where at each step, you decide whether to move right ('H') or down ('V'). By using dynamic programming, we track the number of possible paths from the current state to the destination and make decisions based on lexicographical order. The DP table stores how many ways there are to reach each intermediate point, allowing us to generate the kth sequence without having to enumerate all paths.
Combinatorics for Path Count Calculation
At each step, we need to decide whether moving right or down yields the kth sequence. The number of possible paths from a given point can be determined using combinatorics. Specifically, the number of paths from (x, y) to (row, column) is calculated using the binomial coefficient, nCr(row - x + column - y, row - x). This allows us to efficiently determine which move corresponds to the kth sequence.
Efficient Calculation of the kth Sequence
Instead of generating all possible paths, we can directly compute the kth sequence by progressively deciding whether to move right or down. For each decision point, we use the precomputed DP table and binomial coefficients to choose the appropriate move. This avoids the need to generate and sort all paths, making the solution efficient even for larger grid sizes.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time complexity of this approach is dominated by the computation of binomial coefficients and dynamic programming state transitions. Since there are row + column steps, the time complexity is O(row * column), and each decision step involves calculating binomial coefficients in constant time. The space complexity is O(row * column) due to the DP table used to store intermediate results.
What Interviewers Usually Probe
- The candidate demonstrates understanding of dynamic programming and combinatorics.
- The candidate efficiently applies binomial coefficients to reduce the problem complexity.
- The candidate navigates the state transition approach to solve the problem without brute-forcing all paths.
Common Pitfalls or Variants
Common pitfalls
- Forgetting to consider the lexicographical order of instructions, leading to incorrect kth sequences.
- Miscomputing the binomial coefficients, resulting in an inefficient or incorrect path count.
- Not using dynamic programming to optimize the calculation of intermediate results, leading to unnecessary recomputation.
Follow-up variants
- Given a different destination (row, column), how would the approach change?
- How would you modify the approach for finding the largest lexicographical instruction sequence?
- What happens if there are constraints on the number of right or down moves allowed?
FAQ
What is the main approach for solving the Kth Smallest Instructions problem?
The main approach involves using state transition dynamic programming to track the number of possible paths, combined with combinatorial calculations to directly compute the kth smallest lexicographic sequence.
How do you calculate the number of possible paths from a point to the destination?
The number of possible paths is calculated using binomial coefficients, specifically nCr(row - x + column - y, row - x), where (x, y) is the current point and (row, column) is the destination.
Why is dynamic programming important in solving this problem?
Dynamic programming helps by storing intermediate results, which avoids recomputing the number of paths multiple times, thus optimizing the solution and making it feasible for larger grids.
What are the constraints for the Kth Smallest Instructions problem?
The constraints are that destination.length == 2, 1 <= row, column <= 15, and 1 <= k <= nCr(row + column, row), where nCr is the binomial coefficient.
What happens if k exceeds the total number of possible paths?
If k exceeds the total number of paths from (0, 0) to (row, column), the problem would not have a valid kth sequence, and the answer should handle such edge cases.
Solution
Solution 1
#### Python3
class Solution:
def kthSmallestPath(self, destination: List[int], k: int) -> str:
v, h = destination
ans = []
for _ in range(h + v):
if h == 0:
ans.append("V")
else:
x = comb(h + v - 1, h - 1)
if k > x:
ans.append("V")
v -= 1
k -= x
else:
ans.append("H")
h -= 1
return "".join(ans)Continue Topic
array
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