Available Captures for Rook

Intuition

• The brute force method of checking up down right left starting from the rook positions seems to be the best solution here.

Complexity

Runtime

$O(MN)$ to initially find the position of the rook.

Space

$O(1)$

Code

class Solution:
    def numRookCaptures(self, board: List[List[str]]) -> int:
        pawn_positions = []
        bishop_positions = []
        rook_r, rook_c = -1, -1
        broken = True
        for r in range(8):
            for c in range(8):
                if board[r][c] == 'R':
                    rook_r, rook_c = r, c
        
        res = 0
 
        i = rook_r - 1
        # Go up from rook
        while i >= 0:
            if board[i][rook_c] == 'B':
                break
            elif board[i][rook_c] == 'p':
                res += 1
                break
            i -= 1
        
        # Go down from rook
        i = rook_r + 1
        while i < 8:
            if board[i][rook_c] == 'B':
                break
            elif board[i][rook_c] == 'p':
                res += 1
                break
            i += 1
        
        # Go left from rook
        i = rook_c - 1
        while i >= 0:
            if board[rook_r][i] == 'B':
                break
            elif board[rook_r][i] == 'p':
                res += 1
                break
            i -= 1
        
        # Go right from rook
        i = rook_c + 1
        while i < 8:
            if board[rook_r][i] == 'B':
                break
            elif board[rook_r][i] == 'p':
                res += 1
                break
            i += 1
        
        return res

Notes

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References

Available Captures for Rook - LeetCode