What is the brute force solution for Minimum Moves to Move a Box to Their Target Location?

The brute force approach for Minimum Moves to Move a Box to Their Target Location is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible move for the player and the box until the box reaches the target. This can be visualized as exploring all paths in the grid, which is simple but inefficient. The optimal Optimal Solution approach improves this to O(n).

What are the interview tips for Minimum Moves to Move a Box to Their Target Location?

Clarify the problem requirements and constraints before diving into coding. Discuss your thought process and approach with the interviewer. Test edge cases and validate your solution with different grid configurations.

What are common mistakes when solving Minimum Moves to Move a Box to Their Target Location?

Not checking if the player can reach the box before attempting to push it. Failing to account for walls and obstacles in the grid.

#1263

Minimum Moves to Move a Box to Their Target Location

Hard

ArrayBreadth-First SearchHeap (Priority Queue)MatrixBreadth-First SearchState Space Exploration

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses BFS to find the shortest path for the player while ensuring the box can be pushed to the target. This method efficiently tracks the states and avoids unnecessary explorations.

⚙️

Algorithm

3 steps

1Step 1: Initialize BFS with the player's position and the box's position.
2Step 2: For each state, check if the box can be pushed in any direction and if the player can reach the new box position.
3Step 3: Continue until the box reaches the target or all possibilities are exhausted.

solution.py30 lines

1from collections import deque
2
3def minPushBox(grid):
4    m, n = len(grid), len(grid[0])
5    directions = [(0, 1), (1, 0), (0, -1), (-1, 0)]
6    start = (0, 0, 0, 0)
7    for i in range(m):
8        for j in range(n):
9            if grid[i][j] == 'S':
10                start = (i, j)
11            elif grid[i][j] == 'B':
12                box = (i, j)
13            elif grid[i][j] == 'T':
14                target = (i, j)
15    queue = deque([(start[0], start[1], box[0], box[1], 0)])
16    visited = set()
17    visited.add((start[0], start[1], box[0], box[1]))
18
19    while queue:
20        px, py, bx, by, pushes = queue.popleft()
21        if (bx, by) == target:
22            return pushes
23        for dx, dy in directions:
24            nbx, nby = bx + dx, by + dy
25            if 0 <= nbx < m and 0 <= nby < n and grid[nbx][nby] != '#':
26                if (px, py) == (bx - dx, by - dy):
27                    if (px, py) != (bx, by):
28                        queue.append((bx, by, nbx, nby, pushes + 1))
29                        visited.add((bx, by, nbx, nby))
30    return -1

ℹ

Complexity note: The time complexity is O(n) because we only explore valid states of the player and box, avoiding unnecessary checks. The space complexity is O(n) due to the queue and visited set storing states.

1Understanding the grid structure is crucial for navigating the player and box.
2Identifying valid moves and ensuring the player can reach the box before pushing it is key.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.