Number of Valid Move Combinations On Chessboard - Problem

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 of the i^th piece. In addition, you are given a 2D integer array positions also of length n, where positions[i] = [ri, ci] indicates that the i^th 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.

Note: 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.

Input & Output

Example 1 — Basic Two Pieces

$ Input: pieces = ["rook","bishop"], positions = [[1,1],[1,8]]

› Output: 15

💡 Note: Rook at (1,1) can move horizontally/vertically, bishop at (1,8) can move diagonally. They have multiple valid combinations without collision.

Example 2 — Single Piece

$ Input: pieces = ["queen"], positions = [[1,1]]

› Output: 22

💡 Note: A single queen at (1,1) can stay in place or move to any of 21 valid squares on the 8x8 board.

Example 3 — Adjacent Pieces

$ Input: pieces = ["rook","rook"], positions = [[1,1],[1,2]]

› Output: 49

💡 Note: Two rooks adjacent horizontally can move independently in many directions without collision.

Constraints

1 ≤ pieces.length ≤ 4
pieces[i] is either "rook", "queen", or "bishop"
There will be at most one piece of each type on the board
1 ≤ x_i, y_i ≤ 8
All pieces are on distinct squares

Visualization

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Asked in

G Google 12 f Facebook 8 M Microsoft 6

The key insight is to use backtracking with simulation to check all possible move combinations while pruning invalid paths early. Best approach is optimized backtracking that generates moves piece by piece and detects collisions during path simulation. Time: O(8^n), Space: O(n × max_distance)

Common Approaches

✓ Brute Force with Full Simulation

⏱️ Time: O(8^n × max_distance) Space: O(n × max_distance)

For each piece, generate all possible valid destinations based on its type. Then try every combination of moves and simulate the complete movement to check for collisions at any time step.

Optimized Backtracking with Pruning

⏱️ Time: O(8^n) with pruning Space: O(n × max_distance)

Generate moves for pieces one by one using backtracking. At each step, check if the current partial combination could lead to collisions, and prune impossible branches early to avoid unnecessary computation.

Brute Force with Full Simulation — Algorithm Steps

Generate all possible moves for each piece type
Try every combination of moves
Simulate movement step by step for collision detection

Visualization

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Step-by-Step Walkthrough

Generate Moves

List all valid destinations for each piece

Try Combinations

Test every combination of moves

Simulate

Check for collisions during movement

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MAX_PIECES 10
#define MAX_MOVES 64
#define MAX_PATH 16

static int allDestinations[MAX_PIECES][MAX_MOVES][2];
static int destCounts[MAX_PIECES];
static int currentMoves[MAX_PIECES][2];
static int positions[MAX_PIECES][2];
static char pieces[MAX_PIECES][10];
static int numPieces;

void getMoves(char* piece, int pos[2], int moves[][2], int* count) {
    *count = 0;
    moves[(*count)][0] = pos[0];
    moves[(*count)][1] = pos[1];
    (*count)++;
    
    int r = pos[0], c = pos[1];
    int directions[8][2];
    int numDirs = 0;
    
    if (strcmp(piece, "rook") == 0) {
        directions[0][0] = 0; directions[0][1] = 1;
        directions[1][0] = 0; directions[1][1] = -1;
        directions[2][0] = 1; directions[2][1] = 0;
        directions[3][0] = -1; directions[3][1] = 0;
        numDirs = 4;
    } else if (strcmp(piece, "bishop") == 0) {
        directions[0][0] = 1; directions[0][1] = 1;
        directions[1][0] = 1; directions[1][1] = -1;
        directions[2][0] = -1; directions[2][1] = 1;
        directions[3][0] = -1; directions[3][1] = -1;
        numDirs = 4;
    } else { // queen
        directions[0][0] = 0; directions[0][1] = 1;
        directions[1][0] = 0; directions[1][1] = -1;
        directions[2][0] = 1; directions[2][1] = 0;
        directions[3][0] = -1; directions[3][1] = 0;
        directions[4][0] = 1; directions[4][1] = 1;
        directions[5][0] = 1; directions[5][1] = -1;
        directions[6][0] = -1; directions[6][1] = 1;
        directions[7][0] = -1; directions[7][1] = -1;
        numDirs = 8;
    }
    
    for (int d = 0; d < numDirs; d++) {
        int dr = directions[d][0], dc = directions[d][1];
        int nr = r + dr, nc = c + dc;
        while (nr >= 1 && nr <= 8 && nc >= 1 && nc <= 8) {
            moves[*count][0] = nr;
            moves[*count][1] = nc;
            (*count)++;
            nr += dr;
            nc += dc;
        }
    }
}

int getPath(int start[2], int end[2], int path[][2]) {
    int pathLen = 0;
    if (start[0] == end[0] && start[1] == end[1]) {
        path[0][0] = start[0];
        path[0][1] = start[1];
        return 1;
    }
    
    path[pathLen][0] = start[0];
    path[pathLen][1] = start[1];
    pathLen++;
    
    int r = start[0], c = start[1];
    int er = end[0], ec = end[1];
    
    // Calculate direction
    int dr = (r == er) ? 0 : ((er > r) ? 1 : -1);
    int dc = (c == ec) ? 0 : ((ec > c) ? 1 : -1);
    
    // Generate path
    while (r != er || c != ec) {
        r += dr;
        c += dc;
        path[pathLen][0] = r;
        path[pathLen][1] = c;
        pathLen++;
    }
    
    return pathLen;
}

int isValidCombination(int moves[][2], int numMoves) {
    static int paths[MAX_PIECES][MAX_PATH][2];
    static int pathLens[MAX_PIECES];
    int maxSteps = 0;
    
    // Generate paths for all pieces
    for (int i = 0; i < numMoves; i++) {
        pathLens[i] = getPath(positions[i], moves[i], paths[i]);
        maxSteps = (pathLens[i] - 1 > maxSteps) ? pathLens[i] - 1 : maxSteps;
    }
    
    // Check for collisions at each time step
    for (int t = 0; t <= maxSteps; t++) {
        static char occupied[MAX_PIECES * MAX_PATH][10];
        int occupiedCount = 0;
        
        for (int i = 0; i < numMoves; i++) {
            // Get position at time t
            int posIdx = (t < pathLens[i]) ? t : pathLens[i] - 1;
            int pos[2] = {paths[i][posIdx][0], paths[i][posIdx][1]};
            
            char posKey[10];
            sprintf(posKey, "%d,%d", pos[0], pos[1]);
            
            // Check if already occupied
            for (int j = 0; j < occupiedCount; j++) {
                if (strcmp(occupied[j], posKey) == 0) {
                    return 0;
                }
            }
            strcpy(occupied[occupiedCount], posKey);
            occupiedCount++;
        }
    }
    
    return 1;
}

int backtrack(int index) {
    if (index == numPieces) {
        return isValidCombination(currentMoves, numPieces) ? 1 : 0;
    }
    
    int count = 0;
    for (int i = 0; i < destCounts[index]; i++) {
        currentMoves[index][0] = allDestinations[index][i][0];
        currentMoves[index][1] = allDestinations[index][i][1];
        count += backtrack(index + 1);
    }
    
    return count;
}

int solution() {
    // Generate all possible destinations for each piece
    for (int i = 0; i < numPieces; i++) {
        getMoves(pieces[i], positions[i], allDestinations[i], &destCounts[i]);
    }
    
    return backtrack(0);
}

void parsePieces(char* line) {
    numPieces = 0;
    char* ptr = line + 1; // Skip [
    char* start = NULL;
    
    while (*ptr) {
        if (*ptr == '"') {
            if (start == NULL) {
                start = ptr + 1;
            } else {
                int len = ptr - start;
                strncpy(pieces[numPieces], start, len);
                pieces[numPieces][len] = '\0';
                numPieces++;
                start = NULL;
            }
        }
        ptr++;
    }
}

void parsePositions(char* line) {
    int pieceIdx = 0;
    char* ptr = line + 1; // Skip first [
    
    while (*ptr && pieceIdx < numPieces) {
        if (*ptr == '[') {
            ptr++; // Skip [
            char numStr[10];
            int numIdx = 0;
            int coordIdx = 0;
            
            while (*ptr && *ptr != ']') {
                if (*ptr >= '0' && *ptr <= '9') {
                    numStr[numIdx++] = *ptr;
                } else if (*ptr == ',' && numIdx > 0) {
                    numStr[numIdx] = '\0';
                    positions[pieceIdx][coordIdx++] = atoi(numStr);
                    numIdx = 0;
                }
                ptr++;
            }
            if (numIdx > 0) {
                numStr[numIdx] = '\0';
                positions[pieceIdx][coordIdx] = atoi(numStr);
            }
            pieceIdx++;
        }
        ptr++;
    }
}

int main() {
    char line1[1000], line2[1000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    // Remove newlines
    line1[strcspn(line1, "\n")] = 0;
    line2[strcspn(line2, "\n")] = 0;
    
    parsePieces(line1);
    parsePositions(line2);
    
    printf("%d\n", solution());
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(8^n × max_distance)

Each piece has up to 8 directions with multiple distances, creating exponential combinations

✓ Linear Growth

Space Complexity

O(n × max_distance)

Store all possible moves and simulation states

⚡ Linearithmic Space

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Number of Valid Move Combinations On Chessboard - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force with Full Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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