In this HackerRank Queen’s Attack II problem You will be given a square chessboard with one queen and a number of obstacles placed on it. Determine how many squares the queen can attack.
Problem solution in Python programming.
n, k = map(int, input().split())
rq, cq = map(int, input().split())
moves = {'n': n - rq, 'nw': min(n - rq, cq-1), 'ne': min(n - rq, n - cq),
's': rq-1, 'sw': min(rq-1, cq-1), 'se': min(rq-1, n - cq), 'w': cq-1, 'e': n - cq}
for _ in range(k):
r, c = map(int, input().split())
if c == cq:
if r > rq:
moves['n'] = min(r-rq-1, moves['n'])
else:
moves['s'] = min(rq-r-1, moves['s'])
elif r == rq:
if c > cq:
moves['e'] = min(c-cq-1, moves['e'])
else:
moves['w'] = min(cq-c-1, moves['w'])
elif c - r == cq - rq:
if c < cq and r < rq:
moves['sw'] = min(min(cq-c-1, rq-r-1), moves['sw'])
elif c > cq and r > rq:
moves['ne'] = min(min(c-cq-1, r-rq-1), moves['ne'])
elif c + r == cq + rq:
if c < cq and r > rq:
moves['nw'] = min(min(r-rq-1, cq-c-1), moves['nw'])
elif c > cq and r < rq:
moves['se'] = min(min(rq-r-1, c-cq-1), moves['se'])
print(sum(moves.values()))
Problem solution in Java Programming.
import java.util.*;
public final class Solution {
public static final void main(String[] args) {
int n, rq, cq;
Set<Long> o;
try (Scanner in = new Scanner(System.in)) {
n = in.nextInt();
int k = in.nextInt();
rq = in.nextInt();
cq = in.nextInt();
o = new HashSet<>(k);
while (k --> 0) {
int ro = in.nextInt(), co = in.nextInt();
o.add((long)ro << 32 | co);
}
}
int t = 0;
for (int d[] : new int[][] {{-1, -1}, {-1, 0}, {-1, +1},
{ 0, -1}, { 0, +1},
{+1, -1}, {+1, 0}, {+1, +1}}) {
for (int r = rq + d[0], c = cq + d[1];
1 <= r && r <= n && 1 <= c && c <= n && !o.contains((long)r << 32 | c);
r += d[0], c += d[1]) {
t++;
}
}
System.out.println(t);
}
}
Problem solution in C++ programming.
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
int n;
using namespace std;
map<pair<int,int>,int> q;
int dl(int i1,int i2){
int ret=0;
while(i1<n && i2>1){
i1++;
i2--;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int up(int i1,int i2){
int ret=0;
while(i1>1){
i1--;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int dw(int i1,int i2){
int ret=0;
while(i1<n){
i1++;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int lf(int i1,int i2){
int ret=0;
while(i2>1){
i2--;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int rg(int i1,int i2){
int ret=0;
while(i2<n){
i2++;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int ur(int i1,int i2){
int ret=0;
while(i1>1 && i2<n){
i1--;
i2++;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int ul(int i1,int i2){
int ret=0;
while(i1>1 && i2>1){
i1--;
i2--;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int dr(int i1,int i2){
int ret=0;
while(i1<n && i2<n){
i1++;
i2++;
if(q[{i1,i2}]) break;
ret++;
}
return ret;
}
int main(){
int k;
cin >> n >> k;
int f;
int g;
cin >> f >> g;
for(int a0 = 0; a0 < k; a0++){
int rObstacle;
int cObstacle;
cin >> rObstacle >> cObstacle;
q[{rObstacle,cObstacle}]=1;
}
cout<<up(f,g)+dw(f,g)+lf(f,g)+rg(f,g)+ur(f,g)+ul(f,g)+dl(f,g)+dr(f,g);
return 0;
}
Problem solution in C programming.
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
#define min(a,b) ((a) < (b) ? (a) : (b))
#define max(a,b) ((a) > (b) ? (a) : (b))
int rQueen;
int cQueen;
int B[8] = {0, 0, 0, 0, 0, 0, 0 , 0};
enum { U, D, L, R, UR, DR, DL, UL };
int blocks(int r, int c) {
if (r == rQueen) return c > cQueen ? R : L;
if (c == cQueen) return r > rQueen ? U : D;
if (abs(r - rQueen) == abs(c - cQueen)) {
if (r < rQueen && c < cQueen) return DL;
if (r < rQueen && c > cQueen) return DR;
if (r > rQueen && c < cQueen) return UL;
return UR;
}
return -1;
}
int main(){
int n;
int k;
scanf("%d %d",&n,&k);
scanf("%d %d",&rQueen,&cQueen);
B[D] = rQueen - 1;
B[U] = n - rQueen;
B[L] = cQueen - 1;
B[R] = n - cQueen;
B[UL] = min(B[U], B[L]);
B[UR] = min(B[U], B[R]);
B[DL] = min(B[D], B[L]);
B[DR] = min(B[D], B[R]);
for(int a0 = 0; a0 < k; a0++){
int rObstacle;
int cObstacle;
scanf("%d %d",&rObstacle,&cObstacle);
int b = blocks(rObstacle, cObstacle);
if (b != -1) {
int nv = max(abs(rObstacle-rQueen), abs(cObstacle-cQueen)) - 1;
if (nv < B[b]) B[b] = nv;
}
}
int s = 0;
for (int i = 0; i < 8; s += B[i++]) {}
printf("%dn", s);
return 0;
}
Problem solution in JavaScript programming.
process.stdin.resume();
process.stdin.setEncoding('ascii');
var input_stdin = "";
var input_stdin_array = "";
var input_currentline = 0;
process.stdin.on('data', function (data) {
input_stdin += data;
});
process.stdin.on('end', function () {
input_stdin_array = input_stdin.split("n");
main();
});
function readLine() {
return input_stdin_array[input_currentline++];
}
/////////////// ignore above this line ////////////////////
function main() {
var n_temp = readLine().split(' ');
var n = parseInt(n_temp[0]);
var k = parseInt(n_temp[1]);
var rQueen_temp = readLine().split(' ');
var rQueen = parseInt(rQueen_temp[0]);
var cQueen = parseInt(rQueen_temp[1]);
var left = cQueen - 1;
var right = n - cQueen;
var above = n - rQueen;
var below = rQueen - 1;
var above_left = Math.min(above, left);
var below_left = Math.min(below, left);
var above_right = Math.min(above, right);
var below_right = Math.min(below, right);
for(var a0 = 0; a0 < k; a0++){
var rObstacle_temp = readLine().split(' ');
var rObstacle = parseInt(rObstacle_temp[0]);
var cObstacle = parseInt(rObstacle_temp[1]);
if (rObstacle === rQueen) {
if (cObstacle < cQueen) {
left = Math.min(left, cQueen - cObstacle - 1);
} else {
right = Math.min(right, cObstacle - cQueen - 1);
}
} else if (cObstacle === cQueen) {
if (rObstacle < rQueen) {
below = Math.min(below, rQueen - rObstacle - 1);
} else {
above = Math.min(above, rObstacle - rQueen - 1);
}
} else if (rObstacle - cObstacle === rQueen - cQueen) {
if (cObstacle < cQueen) {
below_left = Math.min(below_left, cQueen - cObstacle - 1);
} else {
above_right = Math.min(above_right, cObstacle - cQueen - 1);
}
} else if (rObstacle + cObstacle === rQueen + cQueen) {
if (cObstacle < cQueen) {
above_left = Math.min(above_left, cQueen - cObstacle - 1);
} else {
below_right = Math.min(below_right, cObstacle - cQueen - 1);
}
}
}
console.log(left + right + above + below + above_left + below_left + above_right + below_right);
}