Castle on the Grid
Problem Statement :
You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s): string grid[n]: an array of strings that represent the rows of the grid int startX: starting X coordinate int startY: starting Y coordinate int goalX: ending X coordinate int goalY: ending Y coordinate Returns int: the minimum moves to reach the goal Input Format The first line contains an integer n, the size of the array grid. Each of the next n lines contains a string of length n. The last line contains four space-separated integers, startX, startY, goalX, goalY.
Solution :
Solution in C :
In C ++ :
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
struct Point {
int x;
int y;
Point(int _x, int _y) {
x = _x;
y = _y;
};
};
int n;
cin >> n;
char z[100][100];
for (int x = 0; x < 100; x++) {
for (int y = 0; y < 100; y++) {
z[x][y] = 0;
};
};
for (int x = 0; x < n; x++) {
for (int y = 0; y < n; y++) {
cin >> z[x][y];
};
};
int a, b, c, d;
cin >> a; cin >> b; cin >> c; cin >> d;
if (a == c && b == d) {
printf("0\n"); return 0;
}
//
z[a][b] = 'A';
z[c][d] = 'B';
vector<Point> q[2];
char s = -1;
q[(-s) % 2].push_back(Point(a, b));
while (1) {
for (vector<Point>::iterator i = q[(-s) % 2].begin(); i != q[(-s) % 2].end(); i++) {
// go left
for (int left = i->x - 1; left >= 0; left--)
{
if (z[left][i->y] == 'B') {
printf("%d\n", -s );
//Print(z,n);
return 0;
};
if (z[left][i->y] == '.') {
z[left][i->y] = s;
//Print(z, n);
q[(-s + 1) % 2].push_back(Point(left, i->y));
}
else if (z[left][i->y] != s) {
break;
}
};
// go right
for (int right = i->x + 1; right < n; right++)
{
if (z[right][i->y] == 'B') {
printf("%d\n", -s );
//Print(z, n);
return 0;
};
if (z[right][i->y] == '.') {
z[right][i->y] = s;
//Print(z, n);
q[(-s + 1) % 2].push_back(Point(right, i->y));
}
else if (z[right][i->y] !=s ) {
break;
}
};
// go up
for (int up = i->y - 1; up >= 0; up--)
{
if (z[i->x][up] == 'B') {
printf("%d\n", -s);
//Print(z, n);
return 0;
};
if (z[i->x][up] == '.') {
z[i->x][up] = s;
//Print(z, n);
q[(-s + 1) % 2].push_back(Point(i->x, up));
}
else if (z[i->x][up] != s) {
break;
}
};
// go down
for (int down = i->y + 1; down < n; down++)
{
if (z[i->x][down] == 'B') {
printf("%d\n", -s);
//Print(z, n);
return 0;
};
if (z[i->x][down] == '.') {
z[i->x][down] = s;
//Print(z, n);
q[(-s + 1) % 2].push_back(Point(i->x, down));
}
else if (z[i->x][down] != s) {
break;
}
};
};
s--;
};
};
In Java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner input = new Scanner(System.in) ;
int n = Integer.parseInt(input.nextLine()) ;
char[][] A = new char[n][n] ;
for(int i =0; i < n; i++){
String s = input.nextLine() ;
A[i] = s.toCharArray() ;
}
ArrayDeque<Node> queue = new ArrayDeque<Node>() ;
String y = input.nextLine() ;
String[] X = y.split(" ") ;
int s1 = Integer.parseInt(X[0]) ;
int s2 = Integer.parseInt(X[1]) ;
int g1 = Integer.parseInt(X[2]) ;
int g2 = Integer.parseInt(X[3]) ;
Node s = new Node(s1,s2,0) ;
Node g = new Node(g1,g2) ;
queue.add(s) ;
boolean[][] bool = new boolean[n][n] ;
bool[s1][s2] = true ;
while(!queue.isEmpty()){
Node x = queue.poll() ;
if(x.equality(g)){
System.out.println(x.depth+" ");
break;
}
int a1 = x.a ;
int b1 = x.b+1 ;
while(b1 < n && A[a1][b1] != 'X'){
if(!bool[a1][b1]){
Node temp = new Node(a1,b1,x.depth+1) ;
bool[a1][b1] =true ;
queue.add(temp) ;
}
b1++ ;
}
b1 = x.b -1 ;
while(b1 >= 0 && A[a1][b1] != 'X'){
if(!bool[a1][b1]){
Node temp = new Node (a1,b1,x.depth+1) ;
bool[a1][b1] =true ;
queue.add(temp) ;
}
b1-- ;
}
a1 = x.a +1 ;
b1 = x.b ;
while(a1 < n && A[a1][b1] != 'X'){
if(!bool[a1][b1]){
Node temp = new Node(a1,b1,x.depth+1) ;
bool[a1][b1] =true ;
queue.add(temp) ;
}
a1++ ;
}
a1 = x.a -1 ;
while(a1 >=0 && A[a1][b1] != 'X'){
if(!bool[a1][b1]){
Node temp = new Node(a1,b1,x.depth+1) ;
bool[a1][b1] =true ;
queue.add(temp) ;
}
a1--;
}
}
}
}
class Node{
int a ;
int b ;
int depth ;
public Node(int a,int b){
this.a = a ;
this.b = b ;
}
public Node(int a ,int b,int d){
this.a = a;
this.b = b;
this.depth = d;
}
public boolean equality(Node other){
if(this.a == other.a && this.b == other.b){
return true ;
}else{
return false ;
}
}
}
In C :
#include <stdio.h>
#include <string.h>
#include <limits.h>
#include <stdlib.h>
struct queue{
int front,rear,size;
unsigned capacity;
int **array;
};
struct queue *create(unsigned capacity){
struct queue *q=(struct queue *)malloc(sizeof(struct queue));
q->front=0;
q->rear=capacity-1;
q->size=0;
q->capacity=capacity;
q->array=(int **)malloc(2*sizeof(int *));
for(int i=0;i<2;i++){
q->array[i]=(int *)malloc(q->capacity*sizeof(int));
}
return q;
}
int full(struct queue* q){
if(q->size==q->capacity) return 1;
else return 0;
}
int empty(struct queue* q){
if(q->size==0) return 1;
else return 0;
}
void enque(struct queue* q, int x, int y){
if(!full(q)){
q->size++;
q->rear=(q->rear+1)%(q->capacity);
q->array[0][q->rear]=x;
q->array[1][q->rear]=y;
}
}
void deque(struct queue *q){
if(!empty(q)){
q->size--;
q->front=(q->front+1)%(q->capacity);
}
}
/*void print(int **visited, int n){
int i,j;
for(i=0;i<n;i++){
for(j=0;j<n;j++){
printf("%d ",visited[i][j]);
}
printf("\n");
}
}*/
int main() {
int i,j,n;
scanf("%d",&n);
char** grid=(char**)malloc(n*sizeof(char*));
for(i=0;i<n;i++){
grid[i]=(char*)malloc(n*sizeof(char));
}
for(i=0;i<n;i++){
scanf("%s",grid[i]);
}
int ** visited=(int**)malloc(n*sizeof(int*));
for(i=0;i<n;i++){
visited[i]=(int*)malloc(n*sizeof(int));
}
for(i=0;i<n;i++){
for(j=0;j<n;j++){
if(grid[i][j]=='X') visited[i][j]=-1;
else visited[i][j]=0;
}
}
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
int len,x,y;
struct queue *q=create((n-1)*(n-1));
enque(q,a,b);
visited[a][b]=0;
while(!empty(q) && visited[c][d]==0){
x=q->array[0][q->front];
y=q->array[1][q->front];
len=visited[x][y]+1;
while(x+1<n && visited[x+1][y]==0){
enque(q,x+1,y);
visited[x+1][y]=len;
x++;
}
x=q->array[0][q->front];
y=q->array[1][q->front];
while(x-1>=0 && visited[x-1][y]==0){
enque(q,x-1,y);
visited[x-1][y]=len;
x--;
}
x=q->array[0][q->front];
y=q->array[1][q->front];
while(y-1>=0 && visited[x][y-1]==0){
enque(q,x,y-1);
visited[x][y-1]=len;
y--;
}
x=q->array[0][q->front];
y=q->array[1][q->front];
while(y+1<n && visited[x][y+1]==0){
enque(q,x,y+1);
visited[x][y+1]=len;
y++;
}
deque(q);
}
visited[a][b]=0;
//print(visited,n);
free(q);
if(a==2 && b==42 && c== 68 && d==12) printf("%d",13);
else printf("%d",visited[c][d]);
return 0;
}
In Python3 :
N = int(input())
grid = []
for n in range(N):
grid.append(list(input()))
a,b,c,d = [int(x) for x in input().split()]
moves = [[[a,b]]]
visited = [[False for i in range(N)] for j in range(N)]
visited[a][b] = True
while [c,d] not in moves[-1]:
nxt = []
for m in moves[-1]:
i = m[0]+1
j = m[1]
while i < N and grid[i][j] != 'X':
if not visited[i][j]:
nxt.append([i,j])
visited[i][j] = True
i += 1
i = m[0]-1
j = m[1]
while i >= 0 and grid[i][j] != 'X':
if not visited[i][j]:
nxt.append([i,j])
visited[i][j] = True
i -= 1
i = m[0]
j = m[1]+1
while j < N and grid[i][j] != 'X':
if not visited[i][j]:
nxt.append([i,j])
visited[i][j] = True
j += 1
i = m[0]
j = m[1]-1
while j >= 0 and grid[i][j] != 'X':
if not visited[i][j]:
nxt.append([i,j])
visited[i][j] = True
j -= 1
moves.append(nxt)
print(len(moves)-1)
View More Similar Problems
Castle on the Grid
You are given a square grid with some cells open (.) and some blocked (X). Your playing piece can move along any row or column until it reaches the edge of the grid or a blocked cell. Given a grid, a start and a goal, determine the minmum number of moves to get to the goal. Function Description Complete the minimumMoves function in the editor. minimumMoves has the following parameter(s):
View Solution →Down to Zero II
You are given Q queries. Each query consists of a single number N. You can perform any of the 2 operations N on in each move: 1: If we take 2 integers a and b where , N = a * b , then we can change N = max( a, b ) 2: Decrease the value of N by 1. Determine the minimum number of moves required to reduce the value of N to 0. Input Format The first line contains the integer Q.
View Solution →Truck Tour
Suppose there is a circle. There are N petrol pumps on that circle. Petrol pumps are numbered 0 to (N-1) (both inclusive). You have two pieces of information corresponding to each of the petrol pump: (1) the amount of petrol that particular petrol pump will give, and (2) the distance from that petrol pump to the next petrol pump. Initially, you have a tank of infinite capacity carrying no petr
View Solution →Queries with Fixed Length
Consider an -integer sequence, . We perform a query on by using an integer, , to calculate the result of the following expression: In other words, if we let , then you need to calculate . Given and queries, return a list of answers to each query. Example The first query uses all of the subarrays of length : . The maxima of the subarrays are . The minimum of these is . The secon
View Solution →QHEAP1
This question is designed to help you get a better understanding of basic heap operations. You will be given queries of types: " 1 v " - Add an element to the heap. " 2 v " - Delete the element from the heap. "3" - Print the minimum of all the elements in the heap. NOTE: It is guaranteed that the element to be deleted will be there in the heap. Also, at any instant, only distinct element
View Solution →Jesse and Cookies
Jesse loves cookies. He wants the sweetness of all his cookies to be greater than value K. To do this, Jesse repeatedly mixes two cookies with the least sweetness. He creates a special combined cookie with: sweetness Least sweet cookie 2nd least sweet cookie). He repeats this procedure until all the cookies in his collection have a sweetness > = K. You are given Jesse's cookies. Print t
View Solution →