Scale Engineer

Number of Valid Move Combinations On Chessboard

Hard
#1872Time: O(M * n^2 * d), where `M` is the total number of move combinations, `n` is the number of pieces, and `d` is the maximum travel distance on the board (which is 7). The complexity is dominated by validating every single combination.Space: O(M * n). The primary memory cost is storing the list of all `M` combinations, where each combination stores `n` coordinates. This can easily lead to out-of-memory errors.1 company
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Prompt

There is an 8 x 8 chessboard containing n pieces (rooks, queens, or bishops). You are given a string array pieces of length n, where pieces[i] describes the type (rook, queen, or bishop) of the ith piece. In addition, you are given a 2D integer array positions also of length n, where positions[i] = [ri, ci] indicates that the ith piece is currently at the 1-based coordinate (ri, ci) on the chessboard.

When making a move for a piece, you choose a destination square that the piece will travel toward and stop on.

  • A rook can only travel horizontally or vertically from (r, c) to the direction of (r+1, c), (r-1, c), (r, c+1), or (r, c-1).
  • A queen can only travel horizontally, vertically, or diagonally from (r, c) to the direction of (r+1, c), (r-1, c), (r, c+1), (r, c-1), (r+1, c+1), (r+1, c-1), (r-1, c+1), (r-1, c-1).
  • A bishop can only travel diagonally from (r, c) to the direction of (r+1, c+1), (r+1, c-1), (r-1, c+1), (r-1, c-1).

You must make a move for every piece on the board simultaneously. A move combination consists of all the moves performed on all the given pieces. Every second, each piece will instantaneously travel one square towards their destination if they are not already at it. All pieces start traveling at the 0th second. A move combination is invalid if, at a given time, two or more pieces occupy the same square.

Return the number of valid move combinations​​​​​.

Notes:

  • No two pieces will start in the same square.
  • You may choose the square a piece is already on as its destination.
  • If two pieces are directly adjacent to each other, it is valid for them to move past each other and swap positions in one second.

 

Example 1:

Input: pieces = ["rook"], positions = [[1,1]]
Output: 15
Explanation: The image above shows the possible squares the piece can move to.

Example 2:

Input: pieces = ["queen"], positions = [[1,1]]
Output: 22
Explanation: The image above shows the possible squares the piece can move to.

Example 3:

Input: pieces = ["bishop"], positions = [[4,3]]
Output: 12
Explanation: The image above shows the possible squares the piece can move to.

 

Constraints:

  • n == pieces.length
  • n == positions.length
  • 1 <= n <= 4
  • pieces only contains the strings "rook", "queen", and "bishop".
  • There will be at most one queen on the chessboard.
  • 1 <= ri, ci <= 8
  • Each positions[i] is distinct.

Approaches

2 approaches with complexity analysis and trade-offs.

A straightforward approach that first generates all possible move combinations and then iterates through them to validate each one.

Algorithm

  • For each piece i, generate a list of all its possible destination squares, Moves_i.
  • Generate the Cartesian product of Moves_0, Moves_1, ..., Moves_{n-1} to get a list of all move combinations.
  • Initialize valid_count = 0.
  • For each combination in the list of all move combinations:
    • If is_valid(combination) is true:
      • valid_count++.
  • Return valid_count.
  • The is_valid function simulates movement second by second and checks for collisions.

Walkthrough

This method breaks the problem into two main parts. First, we determine every possible destination for each piece. Then, we create the Cartesian product of these move sets to get a complete list of all move combinations. For example, if piece 1 has 15 moves and piece 2 has 22 moves, we generate all 15 * 22 = 330 combinations. After generating this list, we iterate through it. For each combination, we simulate the pieces' movements second by second. If at any point in time two pieces occupy the same square, the combination is marked invalid. If all pieces reach their destinations without any collisions along their paths, the combination is valid, and we increment our counter. While simple to understand, this approach is highly inefficient in terms of memory because it requires storing all combinations, which can be a very large number.

Complexity

Time

O(M * n^2 * d), where `M` is the total number of move combinations, `n` is the number of pieces, and `d` is the maximum travel distance on the board (which is 7). The complexity is dominated by validating every single combination.

Space

O(M * n). The primary memory cost is storing the list of all `M` combinations, where each combination stores `n` coordinates. This can easily lead to out-of-memory errors.

Trade-offs

Pros

  • Conceptually simple and easy to separate the logic for generation and validation.

Cons

  • Extremely high space complexity, making it impractical for the given constraints.

  • Can be slow due to the overhead of creating and managing a large data structure for all combinations.

Solutions

class Solution { int n , m = 9 , ans ; int [][][] dist ; int [][] end ; String [] pieces ; int [][] positions ; int [][] rookDirs = { { 1 , 0 }, {- 1 , 0 }, { 0 , 1 }, { 0 , - 1 } }; int [][] bishopDirs = { { 1 , 1 }, { 1 , - 1 }, {- 1 , 1 }, {- 1 , - 1 } }; int [][] queenDirs = { { 1 , 0 }, {- 1 , 0 }, { 0 , 1 }, { 0 , - 1 }, { 1 , 1 }, { 1 , - 1 }, {- 1 , 1 }, {- 1 , - 1 } }; public int countCombinations ( String [] pieces , int [][] positions ) { n = pieces . length ; dist = new int [ n ][ m ][ m ]; end = new int [ n ][ 3 ]; ans = 0 ; this . pieces = pieces ; this . positions = positions ; dfs ( 0 ); return ans ; } private void dfs ( int i ) { if ( i >= n ) { ans ++; return ; } int x = positions [ i ][ 0 ], y = positions [ i ][ 1 ]; resetDist ( i ); dist [ i ][ x ][ y ] = 0 ; end [ i ] = new int [] { x , y , 0 }; if ( checkStop ( i , x , y , 0 )) { dfs ( i + 1 ); } int [][] dirs = getDirs ( pieces [ i ]); for ( int [] dir : dirs ) { resetDist ( i ); dist [ i ][ x ][ y ] = 0 ; int nx = x + dir [ 0 ], ny = y + dir [ 1 ], nt = 1 ; while ( isValid ( nx , ny ) && checkPass ( i , nx , ny , nt )) { dist [ i ][ nx ][ ny ] = nt ; end [ i ] = new int [] { nx , ny , nt }; if ( checkStop ( i , nx , ny , nt )) { dfs ( i + 1 ); } nx += dir [ 0 ]; ny += dir [ 1 ]; nt ++; } } } private void resetDist ( int i ) { for ( int j = 0 ; j < m ; j ++) { for ( int k = 0 ; k < m ; k ++) { dist [ i ][ j ][ k ] = - 1 ; } } } private boolean checkStop ( int i , int x , int y , int t ) { for ( int j = 0 ; j < i ; j ++) { if ( dist [ j ][ x ][ y ] >= t ) { return false ; } } return true ; } private boolean checkPass ( int i , int x , int y , int t ) { for ( int j = 0 ; j < i ; j ++) { if ( dist [ j ][ x ][ y ] == t ) { return false ; } if ( end [ j ][ 0 ] == x && end [ j ][ 1 ] == y && end [ j ][ 2 ] <= t ) { return false ; } } return true ; } private boolean isValid ( int x , int y ) { return x >= 1 && x < m && y >= 1 && y < m ; } private int [][] getDirs ( String piece ) { char c = piece . charAt ( 0 ); return switch ( c ) { case 'r' -> rookDirs ; case 'b' -> bishopDirs ; default -> queenDirs ; }; } }

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