# Number of cells a queen can move with obstacles on the chessborad

Consider a N X N chessboard with a Queen and K obstacles. The Queen cannot pass through obstacles. Given the position (x, y) of Queen, the task is to find the number of cells the queen can move.

Examples:

```Input : N = 8, x = 4, y = 4,
K = 0
Output : 27

Input : N = 8, x = 4, y = 4,
K = 1, kx1 = 3, ky1 = 5
Output : 24

```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1:
The idea is to iterate over the cells the queen can attack and stop until there is an obstacle or end of the board. To do that, we need to iterate horizontally, vertically and diagonally. The moves from position (x, y) can be:
(x+1, y): one step horizontal move to the right.
(x-1, y): one step horizontal move to the left.
(x+1, y+1): one step diagonal move up-right.
(x-1, y-1): one step diagonal move down-left.
(x-1, y+1): one step diagonal move left-up.
(x+1, y-1): one step diagonal move right-down.
(x, y+1): one step downward.
(x, y-1): one step upward.

Below is C++ implementation of this approach:

 `// C++ program to find number of cells a queen can move  ` `// with obstacles on the chessborad ` `#include ` `using` `namespace` `std; ` ` `  `// Return if position is valid on chessboard ` `int` `range(``int` `n, ``int` `x, ``int` `y) ` `{ ` `  ``return` `(x <= n && x > 0 && y <= n && y > 0); ` `} ` ` `  `// Return the number of moves with a given direction ` `int` `check(``int` `n, ``int` `x, ``int` `y, ``int` `xx, ``int` `yy,  ` `                  ``map , ``int``> mp) ` `{ ` `  ``int` `ans = 0; ` `   `  `  ``// Checking valid move of Queen in a direction. ` `  ``while` `(range(n, x, y) && ! mp[{x, y}]) ` `  ``{ ` `    ``x += xx; ` `    ``y += yy; ` `    ``ans++; ` `  ``} ` `   `  `  ``return` `ans; ` `} ` ` `  `// Return the number of position a Queen can move. ` `int` `numberofPosition(``int` `n, ``int` `k, ``int` `x, ``int` `y,  ` `                  ``int` `obstPosx[], ``int` `obstPosy[]) ` `{ ` `  ``int` `x1, y1, ans = 0; ` `  ``map , ``int``> mp; ` `   `  `  ``// Mapping each obstacle's position ` `  ``while``(k--) ` `  ``{ ` `    ``x1 = obstPosx[k]; ` `    ``y1 = obstPosy[k]; ` `     `  `    ``mp[{x1, y1}] = 1; ` `  ``} ` `   `  `  ``// Fetching number of position a queen can ` `  ``// move in each direction. ` `  ``ans += check(n, x + 1, y, 1, 0, mp); ` `  ``ans += check(n, x-1, y, -1, 0, mp); ` `  ``ans += check(n, x, y + 1, 0, 1, mp); ` `  ``ans += check(n, x, y-1, 0, -1, mp); ` `  ``ans += check(n, x + 1, y + 1, 1, 1, mp); ` `  ``ans += check(n, x + 1, y-1, 1, -1, mp); ` `  ``ans += check(n, x-1, y + 1, -1, 1, mp); ` `  ``ans += check(n, x-1, y-1, -1, -1, mp); ` `   `  `  ``return` `ans; ` `} ` ` `  `// Driven Program ` `int` `main() ` `{ ` `  ``int` `n = 8;  ``// Chessboard size ` `  ``int` `k = 1;  ``// Number of obstacles ` `  ``int` `Qposx = 4; ``// Queen x position ` `  ``int` `Qposy = 4; ``// Queen y position ` `  ``int` `obstPosx[] = { 3 };  ``// x position of obstacles ` `  ``int` `obstPosy[] = { 5 };  ``// y position of obstacles ` `   `  `  ``cout << numberofPosition(n, k, Qposx, Qposy,  ` `                   ``obstPosx, obstPosy) << endl; ` `  ``return` `0; ` `} `

Output:

```24
```

Method 2:
The idea is to iterate over the obstacles and for those who are in the queen’s path, we calculate the free cells upto that obstacle. If there is no obstacle in the path we have to calculate the number of free cells upto end of board in that direction.
For any (x1, y1) and (x2, y2):

• If they are horizontally at same level: abs(x1 – x2 – 1)
• If they are vertically at same level: abs(y1 – y2 – 1) is the number of free cells between.
• If they are diagonal: both abs(x1 – x2 – 1) or abs(y1 – y2 – 1) is the number of free cells between.

Below is the implementation of this approach:

## C++

 `// C++ program to find number of cells a queen can move ` `// with obstacles on the chessborad ` `#include ` `using` `namespace` `std; ` ` `  `// Return the number of position a Queen can move. ` `int` `numberofPosition(``int` `n, ``int` `k, ``int` `x, ``int` `y, ` `                    ``int` `obstPosx[], ``int` `obstPosy[]) ` `{ ` `    ``// d11, d12, d21, d22 are for diagnoal distances. ` `    ``// r1, r2 are for vertical distance. ` `    ``// c1, c2 are for horizontal distance. ` `    ``int` `d11, d12, d21, d22, r1, r2, c1, c2; ` ` `  `    ``// Initialise the distance to end of the board. ` `    ``d11 = min( x-1, y-1 ); ` `    ``d12 = min( n-x, n-y ); ` `    ``d21 = min( n-x, y-1 ); ` `    ``d22 = min( x-1, n-y ); ` ` `  `    ``r1 = y-1; ` `    ``r2 = n-y; ` `    ``c1 = x-1; ` `    ``c2 = n-x; ` ` `  `    ``// For each obstacle find the minimum distance. ` `    ``// If obstacle is present in any direction, ` `    ``// distance will be updated. ` `    ``for` `(``int` `i = 0; i < k; i++) ` `    ``{ ` `        ``if` `( x > obstPosx[i] && y > obstPosy[i] && ` `                 ``x-obstPosx[i] == y-obstPosy[i] ) ` `            ``d11 = min(d11, x-obstPosx[i]-1); ` ` `  `        ``if` `( obstPosx[i] > x && obstPosy[i] > y && ` `                  ``obstPosx[i]-x == obstPosy[i]-y ) ` `            ``d12 = min( d12, obstPosx[i]-x-1); ` ` `  `        ``if` `( obstPosx[i] > x && y > obstPosy[i] && ` `                   ``obstPosx[i]-x == y-obstPosy[i] ) ` `            ``d21 = min(d21, obstPosx[i]-x-1); ` ` `  `        ``if` `( x > obstPosx[i] && obstPosy[i] > y && ` `                    ``x-obstPosx[i] == obstPosy[i]-y ) ` `            ``d22 = min(d22, x-obstPosx[i]-1); ` ` `  `        ``if` `( x == obstPosx[i] && obstPosy[i] < y ) ` `            ``r1 = min(r1, y-obstPosy[i]-1); ` ` `  `        ``if` `( x == obstPosx[i] && obstPosy[i] > y ) ` `            ``r2 = min(r2, obstPosy[i]-y-1); ` ` `  `        ``if` `( y == obstPosy[i] && obstPosx[i] < x ) ` `            ``c1 = min(c1, x-obstPosx[i]-1); ` ` `  `        ``if` `( y == obstPosy[i] && obstPosx[i] > x ) ` `            ``c2 = min(c2, obstPosx[i]-x-1); ` `    ``} ` ` `  `    ``return` `d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; ` `} ` ` `  `// Driver code ` `int` `main(``void``) ` `{ ` `    ``int` `n = 8;  ``// Chessboard size ` `    ``int` `k = 1;  ``// number of obstacles ` `    ``int` `Qposx = 4; ``// Queen x position ` `    ``int` `Qposy = 4; ``// Queen y position ` `    ``int` `obstPosx[] = { 3 };  ``// x position of obstacles ` `    ``int` `obstPosy[] = { 5 };  ``// y position of obstacles ` ` `  `    ``cout << numberofPosition(n, k, Qposx, Qposy, ` `                             ``obstPosx, obstPosy); ` `    ``return` `0; ` `} `

## Java

 `// Java program to find number of cells a  ` `// queen can move with obstacles on the  ` `// chessborad ` `import` `java.io.*; ` ` `  `class` `GFG { ` ` `  `    ``// Return the number of position a Queen ` `    ``// can move. ` `    ``static` `int` `numberofPosition(``int` `n, ``int` `k, ``int` `x, ` `                 ``int` `y, ``int` `obstPosx[], ``int` `obstPosy[]) ` `    ``{ ` `         `  `        ``// d11, d12, d21, d22 are for diagnoal distances. ` `        ``// r1, r2 are for vertical distance. ` `        ``// c1, c2 are for horizontal distance. ` `        ``int` `d11, d12, d21, d22, r1, r2, c1, c2; ` `     `  `        ``// Initialise the distance to end of the board. ` `        ``d11 = Math.min( x-``1``, y-``1` `); ` `        ``d12 = Math.min( n-x, n-y ); ` `        ``d21 = Math.min( n-x, y-``1` `); ` `        ``d22 = Math.min( x-``1``, n-y ); ` `     `  `        ``r1 = y-``1``; ` `        ``r2 = n-y; ` `        ``c1 = x-``1``; ` `        ``c2 = n-x; ` `     `  `        ``// For each obstacle find the minimum distance. ` `        ``// If obstacle is present in any direction, ` `        ``// distance will be updated. ` `        ``for` `(``int` `i = ``0``; i < k; i++) ` `        ``{ ` `            ``if` `( x > obstPosx[i] && y > obstPosy[i] && ` `                    ``x-obstPosx[i] == y-obstPosy[i] ) ` `                ``d11 = Math.min(d11, x-obstPosx[i]-``1``); ` `     `  `            ``if` `( obstPosx[i] > x && obstPosy[i] > y && ` `                    ``obstPosx[i]-x == obstPosy[i]-y ) ` `                ``d12 = Math.min( d12, obstPosx[i]-x-``1``); ` `     `  `            ``if` `( obstPosx[i] > x && y > obstPosy[i] && ` `                    ``obstPosx[i]-x == y-obstPosy[i] ) ` `                ``d21 = Math.min(d21, obstPosx[i]-x-``1``); ` `     `  `            ``if` `( x > obstPosx[i] && obstPosy[i] > y && ` `                        ``x-obstPosx[i] == obstPosy[i]-y ) ` `                ``d22 = Math.min(d22, x-obstPosx[i]-``1``); ` `     `  `            ``if` `( x == obstPosx[i] && obstPosy[i] < y ) ` `                ``r1 = Math.min(r1, y-obstPosy[i]-``1``); ` `     `  `            ``if` `( x == obstPosx[i] && obstPosy[i] > y ) ` `                ``r2 = Math.min(r2, obstPosy[i]-y-``1``); ` `     `  `            ``if` `( y == obstPosy[i] && obstPosx[i] < x ) ` `                ``c1 = Math.min(c1, x-obstPosx[i]-``1``); ` `     `  `            ``if` `( y == obstPosy[i] && obstPosx[i] > x ) ` `                ``c2 = Math.min(c2, obstPosx[i]-x-``1``); ` `        ``} ` `     `  `        ``return` `d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `main (String[] args) { ` `    ``int` `n = ``8``; ``// Chessboard size ` `    ``int` `k = ``1``; ``// number of obstacles ` `    ``int` `Qposx = ``4``; ``// Queen x position ` `    ``int` `Qposy = ``4``; ``// Queen y position ` `    ``int` `obstPosx[] = { ``3` `}; ``// x position of obstacles ` `    ``int` `obstPosy[] = { ``5` `}; ``// y position of obstacles ` ` `  `    ``System.out.println(numberofPosition(n, k, Qposx, ` `                          ``Qposy, obstPosx, obstPosy)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## C#

 `// C# program to find number of cells a  ` `// queen can move with obstacles on the  ` `// chessborad ` `using System; ` ` `  `class` `GFG { ` ` `  `    ``// Return the number of position a Queen ` `    ``// can move. ` `    ``static` `int` `numberofPosition(``int` `n, ``int` `k, ``int` `x, ` `                ``int` `y, ``int` `[]obstPosx, ``int` `[]obstPosy) ` `    ``{ ` `         `  `        ``// d11, d12, d21, d22 are for diagnoal distances. ` `        ``// r1, r2 are for vertical distance. ` `        ``// c1, c2 are for horizontal distance. ` `        ``int` `d11, d12, d21, d22, r1, r2, c1, c2; ` `     `  `        ``// Initialise the distance to end of the board. ` `        ``d11 = Math.Min( x-``1``, y-``1` `); ` `        ``d12 = Math.Min( n-x, n-y ); ` `        ``d21 = Math.Min( n-x, y-``1` `); ` `        ``d22 = Math.Min( x-``1``, n-y ); ` `     `  `        ``r1 = y-``1``; ` `        ``r2 = n-y; ` `        ``c1 = x-``1``; ` `        ``c2 = n-x; ` `     `  `        ``// For each obstacle find the Minimum distance. ` `        ``// If obstacle is present in any direction, ` `        ``// distance will be updated. ` `        ``for` `(``int` `i = ``0``; i < k; i++) ` `        ``{ ` `            ``if` `( x > obstPosx[i] && y > obstPosy[i] && ` `                    ``x-obstPosx[i] == y-obstPosy[i] ) ` `                ``d11 = Math.Min(d11, x-obstPosx[i]-``1``); ` `     `  `            ``if` `( obstPosx[i] > x && obstPosy[i] > y && ` `                    ``obstPosx[i]-x == obstPosy[i]-y ) ` `                ``d12 = Math.Min( d12, obstPosx[i]-x-``1``); ` `     `  `            ``if` `( obstPosx[i] > x && y > obstPosy[i] && ` `                    ``obstPosx[i]-x == y-obstPosy[i] ) ` `                ``d21 = Math.Min(d21, obstPosx[i]-x-``1``); ` `     `  `            ``if` `( x > obstPosx[i] && obstPosy[i] > y && ` `                        ``x-obstPosx[i] == obstPosy[i]-y) ` `                ``d22 = Math.Min(d22, x-obstPosx[i]-``1``); ` `     `  `            ``if` `( x == obstPosx[i] && obstPosy[i] < y ) ` `                ``r1 = Math.Min(r1, y-obstPosy[i]-``1``); ` `     `  `            ``if` `( x == obstPosx[i] && obstPosy[i] > y ) ` `                ``r2 = Math.Min(r2, obstPosy[i]-y-``1``); ` `     `  `            ``if` `( y == obstPosy[i] && obstPosx[i] < x ) ` `                ``c1 = Math.Min(c1, x-obstPosx[i]-``1``); ` `     `  `            ``if` `( y == obstPosy[i] && obstPosx[i] > x ) ` `                ``c2 = Math.Min(c2, obstPosx[i]-x-``1``); ` `        ``} ` `     `  `        ``return` `d11 + d12 + d21 + d22 + r1 + r2 + c1 + c2; ` `    ``} ` `     `  `    ``// Driver code ` `    ``public` `static` `void` `Main ()  ` `    ``{ ` `        ``int` `n = ``8``; ``// Chessboard size ` `        ``int` `k = ``1``; ``// number of obstacles ` `        ``int` `Qposx = ``4``; ``// Queen x position ` `        ``int` `Qposy = ``4``; ``// Queen y position ` `        ``int` `[]obstPosx = { ``3` `}; ``// x position of obstacles ` `        ``int` `[]obstPosy = { ``5` `}; ``// y position of obstacles ` `     `  `        ``Console.WriteLine(numberofPosition(n, k, Qposx, ` `                            ``Qposy, obstPosx, obstPosy)); ` `    ``} ` `} ` ` `  `// This code is contributed by anuj_67. `

## tags:

Matrix chessboard-problems Matrix