LeetCode Problem Workspace
Manhattan Distances of All Arrangements of Pieces
This problem challenges you to calculate Manhattan distances in all valid arrangements of identical pieces on a grid.
2
Topics
0
Code langs
3
Related
Practice Focus
Hard · Math plus Combinatorics
Answer-first summary
This problem challenges you to calculate Manhattan distances in all valid arrangements of identical pieces on a grid.
Ace coding interviews with Interview AiBoxInterview AiBox guidance for Math plus Combinatorics
The problem requires calculating the sum of Manhattan distances for all valid placements of identical pieces on a grid. Key to the solution is leveraging combinatorics and mathematical reasoning. By analyzing the grid structure and using optimization techniques, an efficient approach is possible to avoid brute force methods.
Problem Statement
You are given three integers, m, n, and k. These represent the dimensions of a rectangular grid of size m × n, which contains k identical pieces. Your goal is to return the sum of Manhattan distances between every pair of pieces over all valid arrangements of these pieces on the grid.
A valid arrangement consists of placing all k pieces on the grid, with each piece occupying a unique cell. Manhattan distance is the sum of the horizontal and vertical distances between any two pieces. The problem requires finding the total Manhattan distance across all possible valid configurations of the pieces.
Examples
Example 1
Input: m = 2, n = 2, k = 2
Output: 8
The valid arrangements of pieces on the board are:
Thus, the total Manhattan distance across all valid arrangements is 1 + 1 + 1 + 1 + 2 + 2 = 8 .
Example 2
Input: m = 1, n = 4, k = 3
Output: 20
The valid arrangements of pieces on the board are:
The total Manhattan distance between all pairs of pieces across all arrangements is 4 + 6 + 6 + 4 = 20 .
Constraints
- 1 <= m, n <= 105
- 2 <= m * n <= 105
- 2 <= k <= m * n
Solution Approach
Mathematical Insight into Combinatorics
A key approach is leveraging combinatorial analysis and mathematical insights into grid symmetry to calculate distances without brute force. By reducing the problem to counting valid placements and distances systematically, this allows for a faster solution.
Fixing Pieces for Optimization
The problem can be simplified by fixing two pieces in specific locations and calculating the number of valid boards where this can happen. This reduces the computational complexity and provides a more efficient path to the solution.
Efficient Summation Techniques
Summing Manhattan distances across multiple valid grid placements is the heart of the problem. Efficient summation techniques, such as pre-computing possible distances for grid configurations, allow for a solution that scales well even for large grids.
Complexity Analysis
| Metric | Value |
|---|---|
| Time | Depends on the final approach |
| Space | Depends on the final approach |
The time and space complexity depend on the chosen approach. Brute force would be inefficient, with an expected time complexity of O(m * n choose k). However, using combinatorics and fixing pieces optimizes the process, potentially reducing time complexity to O(k * (m + n)). Memory usage can be minimized with mathematical pre-computation and optimized storage.
What Interviewers Usually Probe
- Looking for understanding of combinatorial principles in grid-based problems.
- Checking ability to optimize algorithms beyond brute force approaches.
- Evaluating problem-solving skills using mathematical reductions to simplify large grid problems.
Common Pitfalls or Variants
Common pitfalls
- Overcomplicating the problem with brute force approaches that do not scale for larger values of m and n.
- Failing to realize that symmetry and combinatorial techniques can drastically reduce the problem space.
- Misunderstanding Manhattan distance calculations or how to sum distances efficiently across multiple grid placements.
Follow-up variants
- Extending the problem to larger grids or with additional constraints on piece placements.
- Considering variations where the number of pieces is dynamic, and other objects must also be placed.
- Modifying the grid to have varying distances or non-rectangular shapes, introducing more complex distance calculations.
FAQ
What is the main challenge of the 'Manhattan Distances of All Arrangements of Pieces' problem?
The main challenge is calculating the sum of Manhattan distances across all valid placements of identical pieces on a grid. This requires optimizing the approach to avoid brute-force methods.
How can I optimize my solution for this problem?
Optimization is achieved by using combinatorial insights, fixing pieces in specific locations, and calculating valid placements systematically to avoid brute-force enumeration of all configurations.
What is the best approach for dealing with large grid sizes?
For large grid sizes, using mathematical combinatorics and symmetry principles is crucial to reduce the problem size and avoid unnecessary calculations.
How does Manhattan distance apply in this problem?
Manhattan distance measures the sum of the horizontal and vertical distances between two points on the grid. This problem requires summing these distances for all valid pairings of pieces.
What combinatorics techniques should I know for this problem?
You should understand how to calculate the number of valid placements of pieces on a grid and how to apply symmetry and pre-computation to reduce the complexity of summing distances across placements.
Solution
Solution 1
#### Python3
Continue Topic
math
Practice more edge cases under the same topic.
arrow_forwardauto_awesomeContinue Pattern
Math plus Combinatorics
Expand the same solving frame across more problems.
arrow_forwardsignal_cellular_altSame Difficulty Track
Hard
Stay on this level to stabilize interview delivery.
arrow_forward