Minimum steps to reach target by a Knight | Set 1

Given a square chessboard of N x N size, the position of Knight and position of a target is given. We need to find out minimum steps a Knight will take to reach the target position.

Examples:



In above diagram Knight takes 3 step to reach 
from (4, 5) to (1, 1) (4, 5) -> (5, 3) -> (3, 2) 
-> (1, 1)  as shown in diagram

This problem can be seen as shortest path in unweighted graph. Therefore we use BFS to solve this problem. We try all 8 possible positions where a Knight can reach from its position. If reachable position is not already visited and is inside the board, we push this state into queue with distance 1 more than its parent state. Finally we return distance of target position, when it gets pop out from queue.
Below code implements BFS for searching through cells, where each cell contains its coordinate and distance from starting node. In worst case, below code visits all cells of board, making worst-case time complexity as O(N^2)

C++

// C++ program to find minimum steps to reach to 
// specific cell in minimum moves by Knight 
#include <bits/stdc++.h> 
using namespace std; 
  
// structure for storing a cell's data 
struct cell 
{ 
    int x, y; 
    int dis; 
    cell() {} 
    cell(int x, int y, int dis) : x(x), y(y), dis(dis) {} 
}; 
  
// Utility method returns true if (x, y) lies 
// inside Board 
bool isInside(int x, int y, int N) 
{ 
    if (x >= 1 && x <= N && y >= 1 && y <= N) 
        return true; 
    return false; 
} 
  
// Method returns minimum step to reach target position 
int minStepToReachTarget(int knightPos[], int targetPos[], 
                                                int N) 
{ 
    // x and y direction, where a knight can move 
    int dx[] = {-2, -1, 1, 2, -2, -1, 1, 2}; 
    int dy[] = {-1, -2, -2, -1, 1, 2, 2, 1}; 
  
    // queue for storing states of knight in board 
    queue<cell> q; 
  
    // push starting position of knight with 0 distance 
    q.push(cell(knightPos[0], knightPos[1], 0)); 
  
    cell t; 
    int x, y; 
    bool visit[N + 1][N + 1]; 
  
    // make all cell unvisited 
    for (int i = 1; i <= N; i++) 
        for (int j = 1; j <= N; j++) 
            visit[i][j] = false; 
  
    // visit starting state 
    visit[knightPos[0]][knightPos[1]] = true; 
  
    // loop untill we have one element in queue 
    while (!q.empty()) 
    { 
        t = q.front(); 
        q.pop(); 
  
        // if current cell is equal to target cell, 
        // return its distance 
        if (t.x == targetPos[0] && t.y == targetPos[1]) 
            return t.dis; 
  
        // loop for all reachable states 
        for (int i = 0; i < 8; i++) 
        { 
            x = t.x + dx[i]; 
            y = t.y + dy[i]; 
  
            // If reachable state is not yet visited and 
            // inside board, push that state into queue 
            if (isInside(x, y, N) && !visit[x][y]) { 
                visit[x][y] = true; 
                q.push(cell(x, y, t.dis + 1)); 
            } 
        } 
    } 
} 
  
// Driver code to test above methods 
int main() 
{ 
    int N = 30; 
    int knightPos[] = {1, 1}; 
    int targetPos[] = {30, 30}; 
    cout << minStepToReachTarget(knightPos, targetPos, N); 
    return 0; 
} 

/div>

Java

import java.util.*; 
  
// Java program to find minimum steps to reach to  
// specific cell in minimum moves by Knight  
class GFG 
{ 
  
// Class for storing a cell's data  
static class cell  
{  
    int x, y;  
    int dis;  
  
        public cell(int x, int y, int dis) 
        { 
            this.x = x; 
            this.y = y; 
            this.dis = dis; 
        } 
      
      
} 
  
// Utility method returns true if (x, y) lies  
// inside Board  
static boolean isInside(int x, int y, int N)  
{  
    if (x >= 1 && x <= N && y >= 1 && y <= N)  
        return true;  
    return false;  
}  
  
// Method returns minimum step  
// to reach target position  
static int minStepToReachTarget(int knightPos[], int targetPos[],  
                                                int N)  
{  
    // x and y direction, where a knight can move  
    int dx[] = {-2, -1, 1, 2, -2, -1, 1, 2};  
    int dy[] = {-1, -2, -2, -1, 1, 2, 2, 1};  
  
    // queue for storing states of knight in board  
    Vector<cell> q = new Vector<>();  
  
    // push starting position of knight with 0 distance  
    q.add(new cell(knightPos[0], knightPos[1], 0));  
  
    cell t ;  
    int x, y;  
    boolean visit[][] = new boolean[N + 1][N + 1];  
  
    // make all cell unvisited  
    for (int i = 1; i <= N; i++)  
        for (int j = 1; j <= N; j++)  
            visit[i][j] = false;  
  
    // visit starting state  
    visit[knightPos[0]][knightPos[1]] = true;  
  
    // loop untill we have one element in queue  
    while (!q.isEmpty())  
    {  
        t = q.firstElement();  
        q.remove(0);  
  
        // if current cell is equal to target cell,  
        // return its distance  
        if (t.x == targetPos[0] && t.y == targetPos[1])  
            return t.dis;  
  
        // loop for all reachable states  
        for (int i = 0; i < 8; i++)  
        {  
            x = t.x + dx[i];  
            y = t.y + dy[i];  
  
            // If reachable state is not yet visited and  
            // inside board, push that state into queue  
            if (isInside(x, y, N) && !visit[x][y]) 
            {  
                visit[x][y] = true;  
                q.add(new cell(x, y, t.dis + 1));  
            }  
        }  
    }  
        return Integer.MAX_VALUE; 
}  
  
// Driver code 
public static void main(String[] args)  
{ 
    int N = 30;  
    int knightPos[] = {1, 1};  
    int targetPos[] = {30, 30}; 
        System.out.println(minStepToReachTarget(knightPos, targetPos, N)); 
    } 
} 
  
// This code contributed by Rajput-Ji 

Python3

# Python3 code to find minimum steps to reach  
# to specific cell in minimum moves by Knight  
class cell: 
      
    def __init__(self, x = 0, y = 0, dist = 0): 
        self.x = x 
        self.y = y 
        self.dist = dist 
          
# checks whether given position is  
# inside the board 
def isInside(x, y, N): 
    if (x >= 1 and x <= N and 
        y >= 1 and y <= N):  
        return True
    return False
      
# Method returns minimum step to reach 
# target position  
def minStepToReachTarget(knightpos,  
                         targetpos, N): 
      
    #all possible movments for the knight 
    dx = [2, 2, -2, -2, 1, 1, -1, -1] 
    dy = [1, -1, 1, -1, 2, -2, 2, -2] 
      
    queue = [] 
      
    # push starting position of knight 
    # with 0 distance 
    queue.append(cell(knightpos[0], knightpos[1], 0)) 
      
    # make all cell unvisited  
    visited = [[False for i in range(N + 1)]  
                      for j in range(N + 1)] 
      
    # visit starting state 
    visited[knightpos[0]][knightpos[1]] = True
      
    # loop untill we have one element in queue  
    while(len(queue) > 0): 
          
        t = queue[0] 
        queue.pop(0) 
          
        # if current cell is equal to target  
        # cell, return its distance  
        if(t.x == targetpos[0] and 
           t.y == targetpos[1]): 
            return t.dist 
              
        # iterate for all reachable states  
        for i in range(8): 
              
            x = t.x + dx[i] 
            y = t.y + dy[i] 
              
            if(isInside(x, y, N) and not visited[x][y]): 
                visited[x][y] = True
                queue.append(cell(x, y, t.dist + 1)) 
  
# Driver Code      
if __name__=='__main__':  
    N = 30
    knightpos = [1, 1] 
    targetpos = [30, 30] 
    print(minStepToReachTarget(knightpos, 
                               targetpos, N)) 
      
# This code is contributed by  
# Kaustav kumar Chanda 

Output:

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Minimum steps to reach target by a Knight | Set 1

C++

Java

Python3

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