Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
How to find the minimum number of steps needed by knight to reach the destination using C#?
We have to make the knight cover all the cells of the board and it can move to a cell only once.
There can be two ways of finishing the knight move - the first in which the knight is one knight's move away from the cell from where it began, so it can go to the position from where it started and form a loop, this is called closed tour, the second in which the knight finishes anywhere else, this is called open tour. A move is valid if it is inside the chessboard and if the cell is not already occupied. We will make the value of all the unoccupied cells equal to -1.
Example
using System;
using System.Collections.Generic;
using System.Text;
using System.Linq;
namespace ConsoleApplication{
public class KnightWalkProblem{
public class cell{
public int x, y;
public int dis;
public cell(int x, int y, int dis){
this.x = x;
this.y = y;
this.dis = dis;
}
}
static bool isInside(int x, int y, int N){
if (x >= 1 && x <= N && y >= 1 && y <= N)
return true;
return false;
}
public int minStepToReachTarget(int[] knightPos, int[] targetPos, int N){
int[] dx = { -2, -1, 1, 2, -2, -1, 1, 2 };
int[] dy = { -1, -2, -2, -1, 1, 2, 2, 1 };
Queue<cell> q = new Queue<cell>();
q.Enqueue(new cell(knightPos[0], knightPos[1], 0));
cell t;
int x, y;
bool[,] visit = new bool[N + 1, N + 1];
for (int i = 1; i <= N; i++)
for (int j = 1; j <= N; j++)
visit[i, j] = false;
visit[knightPos[0], knightPos[1]] = true;
while (q.Count != 0){
t = q.Peek();
q.Dequeue();
if (t.x == targetPos[0] && t.y == targetPos[1])
return t.dis;
for (int i = 0; i < 8; i++){
x = t.x + dx[i];
y = t.y + dy[i];
if (isInside(x, y, N) && !visit[x, y]){
visit[x, y] = true;
q.Enqueue(new cell(x, y, t.dis + 1));
}
}
}
return int.MaxValue;
}
}
class Program{
static void Main(string[] args){
KnightWalkProblem kn = new KnightWalkProblem();
int N = 30;
int[] knightPos = { 1, 1 };
int[] targetPos = { 30, 30 };
Console.WriteLine(
kn.minStepToReachTarget(
knightPos,
targetPos, N));
}
}
}
Output
20
Updated on: 27-Aug-2021
188 Views
Advertisements