LeetCode #2056 — HARD

Number of Valid Move Combinations On Chessboard

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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 i^th piece. In addition, you are given a 2D integer array positions also of length n, where positions[i] = [r_i, c_i] indicates that the i^th piece is currently at the 1-based coordinate (r_i, c_i) 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 0^th 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 <= r_i, c_i <= 8
Each positions[i] is distinct.

Step 01

Brute Force Baseline

Problem summary: 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,

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Backtracking

Example 1

["rook"]
[[1,1]]

Example 2

["queen"]
[[1,1]]

Example 3

["bishop"]
[[4,3]]

Step 02

Core Insight

What unlocks the optimal approach

N is small, we can generate all possible move combinations.
For each possible move combination, determine which ones are valid.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2056: Number of Valid Move Combinations On Chessboard
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;
        };
    }
}

// Accepted solution for LeetCode #2056: Number of Valid Move Combinations On Chessboard
func countCombinations(pieces []string, positions [][]int) (ans int) {
	n := len(pieces)
	m := 9
	dist := make([][][]int, n)
	for i := range dist {
		dist[i] = make([][]int, m)
		for j := range dist[i] {
			dist[i][j] = make([]int, m)
		}
	}

	end := make([][3]int, n)

	rookDirs := [][2]int{{1, 0}, {-1, 0}, {0, 1}, {0, -1}}
	bishopDirs := [][2]int{{1, 1}, {1, -1}, {-1, 1}, {-1, -1}}
	queenDirs := [][2]int{{1, 0}, {-1, 0}, {0, 1}, {0, -1}, {1, 1}, {1, -1}, {-1, 1}, {-1, -1}}

	resetDist := func(i int) {
		for j := 0; j < m; j++ {
			for k := 0; k < m; k++ {
				dist[i][j][k] = -1
			}
		}
	}

	checkStop := func(i, x, y, t int) bool {
		for j := 0; j < i; j++ {
			if dist[j][x][y] >= t {
				return false
			}
		}
		return true
	}

	checkPass := func(i, x, y, t int) bool {
		for 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
	}

	isValid := func(x, y int) bool {
		return x >= 1 && x < m && y >= 1 && y < m
	}

	getDirs := func(piece string) [][2]int {
		switch piece[0] {
		case 'r':
			return rookDirs
		case 'b':
			return bishopDirs
		default:
			return queenDirs
		}
	}

	var dfs func(i int)
	dfs = func(i int) {
		if i >= n {
			ans++
			return
		}

		x, y := positions[i][0], positions[i][1]
		resetDist(i)
		dist[i][x][y] = 0
		end[i] = [3]int{x, y, 0}

		if checkStop(i, x, y, 0) {
			dfs(i + 1)
		}

		dirs := getDirs(pieces[i])
		for _, dir := range dirs {
			resetDist(i)
			dist[i][x][y] = 0
			nx, ny, nt := x+dir[0], y+dir[1], 1

			for isValid(nx, ny) && checkPass(i, nx, ny, nt) {
				dist[i][nx][ny] = nt
				end[i] = [3]int{nx, ny, nt}
				if checkStop(i, nx, ny, nt) {
					dfs(i + 1)
				}
				nx += dir[0]
				ny += dir[1]
				nt++
			}
		}
	}

	dfs(0)
	return
}

# Accepted solution for LeetCode #2056: Number of Valid Move Combinations On Chessboard
rook_dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)]
bishop_dirs = [(1, 1), (1, -1), (-1, 1), (-1, -1)]
queue_dirs = rook_dirs + bishop_dirs


def get_dirs(piece: str) -> List[Tuple[int, int]]:
    match piece[0]:
        case "r":
            return rook_dirs
        case "b":
            return bishop_dirs
        case _:
            return queue_dirs


class Solution:
    def countCombinations(self, pieces: List[str], positions: List[List[int]]) -> int:
        def check_stop(i: int, x: int, y: int, t: int) -> bool:
            return all(dist[j][x][y] < t for j in range(i))

        def check_pass(i: int, x: int, y: int, t: int) -> bool:
            for j in range(i):
                if dist[j][x][y] == t:
                    return False
                if end[j][0] == x and end[j][1] == y and end[j][2] <= t:
                    return False
            return True

        def dfs(i: int) -> None:
            if i >= n:
                nonlocal ans
                ans += 1
                return
            x, y = positions[i]
            dist[i][:] = [[-1] * m for _ in range(m)]
            dist[i][x][y] = 0
            end[i] = (x, y, 0)
            if check_stop(i, x, y, 0):
                dfs(i + 1)
            dirs = get_dirs(pieces[i])
            for dx, dy in dirs:
                dist[i][:] = [[-1] * m for _ in range(m)]
                dist[i][x][y] = 0
                nx, ny, nt = x + dx, y + dy, 1
                while 1 <= nx < m and 1 <= ny < m and check_pass(i, nx, ny, nt):
                    dist[i][nx][ny] = nt
                    end[i] = (nx, ny, nt)
                    if check_stop(i, nx, ny, nt):
                        dfs(i + 1)
                    nx += dx
                    ny += dy
                    nt += 1

        n = len(pieces)
        m = 9
        dist = [[[-1] * m for _ in range(m)] for _ in range(n)]
        end = [(0, 0, 0) for _ in range(n)]
        ans = 0
        dfs(0)
        return ans

// Accepted solution for LeetCode #2056: Number of Valid Move Combinations On Chessboard
/**
 * [2056] Number of Valid Move Combinations On Chessboard
 *
 * 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 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 0^th 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:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/09/23/a1.png" style="width: 215px; height: 215px;" />
 * Input: pieces = ["rook"], positions = [[1,1]]
 * Output: 15
 * Explanation: The image above shows the possible squares the piece can move to.
 *
 * Example 2:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/09/23/a2.png" style="width: 215px; height: 215px;" />
 * Input: pieces = ["queen"], positions = [[1,1]]
 * Output: 22
 * Explanation: The image above shows the possible squares the piece can move to.
 *
 * Example 3:
 * <img alt="" src="https://assets.leetcode.com/uploads/2021/09/23/a3.png" style="width: 214px; height: 215px;" />
 * 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.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-valid-move-combinations-on-chessboard/
// discuss: https://leetcode.com/problems/number-of-valid-move-combinations-on-chessboard/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn count_combinations(pieces: Vec<String>, positions: Vec<Vec<i32>>) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2056_example_1() {
        let pieces = vec_string!["rook"];
        let positions = vec![vec![1, 1]];

        let result = 15;

        assert_eq!(Solution::count_combinations(pieces, positions), result);
    }

    #[test]
    #[ignore]
    fn test_2056_example_2() {
        let pieces = vec_string!["queen"];
        let positions = vec![vec![1, 1]];

        let result = 22;

        assert_eq!(Solution::count_combinations(pieces, positions), result);
    }

    #[test]
    #[ignore]
    fn test_2056_example_3() {
        let pieces = vec_string!["bishop"];
        let positions = vec![vec![4, 3]];

        let result = 12;

        assert_eq!(Solution::count_combinations(pieces, positions), result);
    }
}

// Accepted solution for LeetCode #2056: Number of Valid Move Combinations On Chessboard
const rookDirs: [number, number][] = [
    [1, 0],
    [-1, 0],
    [0, 1],
    [0, -1],
];
const bishopDirs: [number, number][] = [
    [1, 1],
    [1, -1],
    [-1, 1],
    [-1, -1],
];
const queenDirs = [...rookDirs, ...bishopDirs];

function countCombinations(pieces: string[], positions: number[][]): number {
    const n = pieces.length;
    const m = 9;
    let ans = 0;

    const dist = Array.from({ length: n }, () =>
        Array.from({ length: m }, () => Array(m).fill(-1)),
    );

    const end: [number, number, number][] = Array(n).fill([0, 0, 0]);

    const resetDist = (i: number) => {
        for (let j = 0; j < m; j++) {
            for (let k = 0; k < m; k++) {
                dist[i][j][k] = -1;
            }
        }
    };

    const checkStop = (i: number, x: number, y: number, t: number): boolean => {
        for (let j = 0; j < i; j++) {
            if (dist[j][x][y] >= t) {
                return false;
            }
        }
        return true;
    };

    const checkPass = (i: number, x: number, y: number, t: number): boolean => {
        for (let 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;
    };

    const isValid = (x: number, y: number): boolean => {
        return x >= 1 && x < m && y >= 1 && y < m;
    };

    const getDirs = (piece: string): [number, number][] => {
        switch (piece[0]) {
            case 'r':
                return rookDirs;
            case 'b':
                return bishopDirs;
            default:
                return queenDirs;
        }
    };

    const dfs = (i: number) => {
        if (i >= n) {
            ans++;
            return;
        }

        const [x, y] = positions[i];
        resetDist(i);
        dist[i][x][y] = 0;
        end[i] = [x, y, 0];

        if (checkStop(i, x, y, 0)) {
            dfs(i + 1);
        }

        const dirs = getDirs(pieces[i]);
        for (const [dx, dy] of dirs) {
            resetDist(i);
            dist[i][x][y] = 0;
            let nx = x + dx,
                ny = y + dy,
                nt = 1;

            while (isValid(nx, ny) && checkPass(i, nx, ny, nt)) {
                dist[i][nx][ny] = nt;
                end[i] = [nx, ny, nt];
                if (checkStop(i, nx, ny, nt)) {
                    dfs(i + 1);
                }
                nx += dx;
                ny += dy;
                nt++;
            }
        }
    };

    dfs(0);
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n × M)

Space

O(n × M)

Approach Breakdown

EXHAUSTIVE

O(nⁿ) time

O(n) space

Generate every possible combination without any filtering. At each of n positions we choose from up to n options, giving nⁿ total candidates. Each candidate takes O(n) to validate. No pruning means we waste time on clearly invalid partial solutions.

BACKTRACKING + PRUNING

O(n!) time

O(n) space

Backtracking explores a decision tree, but prunes branches that violate constraints early. Worst case is still factorial or exponential, but pruning dramatically reduces the constant factor in practice. Space is the recursion depth (usually O(n) for n-level decisions).

Shortcut: Backtracking time = size of the pruned search tree. Focus on proving your pruning eliminates most branches.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Missing undo step on backtrack

Wrong move: Mutable state leaks between branches.

Usually fails on: Later branches inherit selections from earlier branches.

Fix: Always revert state changes immediately after recursive call.