Knight On Chess Board
Logical Breakdown
- [ ] Subproblem Identification:
- [ ] Optimal Substructure:
- [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
- [ ] Optimization: (Matrix Exponentiation if 'A' is huge)
Mathematical Rigor
State Definition
Let \(dp[i]\) be the state for the \(i\)-th subproblem.
Recurrence Relation
Visualization
Complexity Analysis
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Iterative DP | \(O(N)\) | \(O(1)\) |
Code Reference
package com.dsa.graphs;
import java.util.*;
/*
* ============================================================================================
* PROBLEM: Knight On Chess Board
* ============================================================================================
*
* ---------------------
* 1. Problem Description
* ---------------------
* Given a chessboard of size A x B. Find the minimum steps a Knight takes to move from
* source (C, D) to destination (E, F).
* - A Knight moves in an "L" shape: 2 steps in one cardinal direction and 1 step
* perpendicularly.
* - Return -1 if the destination is unreachable.
*
* ---------------------
* 2. Logical Breakdown & Mathematical Formulation
* ---------------------
* This is a Shortest Path problem in an unweighted grid, which is best solved using BFS.
*
* Step 1: Define Knight Moves
* A knight has 8 possible moves from (x, y):
* (x+2, y+1), (x+2, y-1), (x-2, y+1), (x-2, y-1),
* (x+1, y+2), (x+1, y-2), (x-1, y+2), (x-1, y-2)
*
* Step 2: BFS Implementation
* 1. Initialization:
* - Use a 2D visited array or dist array initialized to -1.
* - Convert 1-based coordinates to 0-based: (C-1, D-1) and (E-1, F-1).
* - Push the starting point into a Queue and set dist[start] = 0.
* 2. Traversal:
* - While Queue is not empty:
* - Pop current (currX, currY).
* - If current == destination, return dist[currX][currY].
* - Explore all 8 knight moves.
* - For each move (nextX, nextY):
* - If it's within bounds (0 <= nextX < A, 0 <= nextY < B) and NOT visited:
* - Set dist[nextX][nextY] = dist[currX][currY] + 1.
* - Enqueue (nextX, nextY).
*
* ---------------------
* 3. Constraints & Complexity Analysis
* ---------------------
* Time Complexity: O(A * B)
* - In the worst case, every cell of the board is visited once.
* Space Complexity: O(A * B)
* - To store the distance/visited matrix and the queue.
*
* ============================================================================================
/
public class KnightOnChessBoard {
public int solve(int A) {
// TODO: Implement solution
return 0;
}
public int knight(int i, int i1, int i2, int i3, int i4, int i5) {
return 1;
}
}