LeetCode Problem 1368

Difficulty: 🔴 Hard
Topics: Array, Breadth-First Search, Graph Theory, Heap (Priority Queue), Matrix, Shortest Path

Solution

LeetCode 1368 - Minimum Cost to Make at Least One Valid Path in a Grid

Problem Understanding

This problem gives us a two dimensional grid where every cell contains a directional sign. The sign tells us which neighboring cell we should move to next. The four possible directions are:

• 1 means move right
• 2 means move left
• 3 means move down
• 4 means move up

We always begin at the top-left corner (0, 0) and want to eventually reach the bottom-right corner (m - 1, n - 1).

The important detail is that we are allowed to modify the direction of any cell, and each modification costs 1. A cell can only be modified once, but since we only care about constructing a single valid path, that restriction does not create additional complications.

The goal is to determine the minimum total modification cost needed so that at least one valid path exists from the start cell to the destination cell.

A move is considered free if we follow the current arrow direction of the cell. If we want to move in a different direction, we must pay a cost of 1 to change that arrow.

This naturally becomes a shortest path problem:

• Every grid cell is a node

• Moving from one cell to a neighboring cell is an edge

• The edge cost is:

• 0 if the movement follows the existing arrow

• 1 if we must modify the arrow

We must find the minimum total cost path from (0, 0) to (m - 1, n - 1).

The constraints are important:

• m, n <= 100
• Maximum number of cells is 10,000

A brute-force search over all possible arrow modifications would be completely infeasible because each cell has multiple possible states. We need an efficient graph shortest path algorithm.

Several edge cases are important:

• A valid path may already exist, answer is 0
• Arrows may point outside the grid
• The optimal path may revisit regions conceptually, but shortest path algorithms prevent unnecessary repeated work
• A 1 x 1 grid already starts at the destination, so answer is 0

Approaches

Brute Force Approach

A brute-force solution would attempt all possible combinations of arrow modifications and check whether a valid path exists.

For each cell, we could theoretically choose one of four directions. Since there are m * n cells, the number of possible configurations becomes:

$$4^{m \times n}$$

For every configuration, we would simulate whether a valid path exists from the top-left to the bottom-right cell.

This approach is correct because it exhaustively explores all possible arrow assignments, guaranteeing that the minimum-cost valid configuration will eventually be found.

However, it is computationally impossible for grids up to 100 x 100. Even much smaller grids would already produce an astronomical number of states.

Optimal Approach, 0-1 BFS

The key insight is that this is not really a grid modification problem. It is a shortest path problem with only two possible edge weights:

• 0 if we follow the current arrow
• 1 if we change the arrow

Whenever all edge weights are either 0 or 1, we can use a specialized shortest path algorithm called 0-1 BFS.

The idea behind 0-1 BFS is:

• Use a deque instead of a priority queue
• If moving to a neighbor costs 0, push it to the front
• If moving to a neighbor costs 1, push it to the back

This guarantees that cells are processed in increasing order of total cost, similar to Dijkstra’s algorithm, but more efficiently.

Each cell is treated as a graph node, and each neighboring movement represents an edge with cost 0 or 1.

Since each node and edge is processed only a limited number of times, the algorithm runs efficiently even for the largest constraints.

Approach Comparison

Approach	Time Complexity	Space Complexity	Notes
Brute Force	O(4^(m*n))	O(m*n)	Tries all possible arrow modifications
Optimal, 0-1 BFS	O(m*n)	O(m*n)	Uses deque with edge weights 0 or 1

Algorithm Walkthrough

Step 1: Define Direction Mapping

We map each direction number to its corresponding movement:

Direction	Movement
1	Right (0, 1)
2	Left (0, -1)
3	Down (1, 0)
4	Up (-1, 0)

This lets us easily compare whether a move follows the existing arrow.

Step 2: Initialize Distance Matrix

We create a distance matrix where:

• distance[r][c] stores the minimum cost needed to reach cell (r, c)

Initially:

• All values are infinity
• distance[0][0] = 0

This matrix prevents unnecessary revisits with worse costs.

Step 3: Initialize the Deque

We use a deque for 0-1 BFS.

The deque initially contains:

(0, 0)

representing the starting cell.

Step 4: Process Cells from the Deque

While the deque is not empty:

1. Remove the front cell
2. Explore all four neighboring directions
3. Determine movement cost:

• 0 if the move matches the current arrow
• 1 otherwise

4. Compute new total cost
5. If the new cost improves the neighbor's distance:

• Update the distance

• Push to:

• front if cost is 0

• back if cost is 1

Step 5: Continue Until Destination is Reached

Because 0-1 BFS processes states in increasing cost order, the first time we finalize the destination cost, it is guaranteed to be optimal.

The final answer is:

distance[m - 1][n - 1]

Why it works

The algorithm models the grid as a graph where every move has weight 0 or 1. 0-1 BFS is specifically designed for such graphs and guarantees that nodes are processed in nondecreasing order of shortest distance.

Whenever a better path to a cell is found, the deque placement ensures that cheaper states are processed earlier. Therefore, once the algorithm completes, the recorded distance for every cell is the minimum possible modification cost.

Python Solution

from collections import deque
from typing import List

class Solution:
    def minCost(self, grid: List[List[int]]) -> int:
        rows = len(grid)
        cols = len(grid[0])

        directions = {
            1: (0, 1),    # right
            2: (0, -1),   # left
            3: (1, 0),    # down
            4: (-1, 0)    # up
        }

        distance = [[float('inf')] * cols for _ in range(rows)]
        distance[0][0] = 0

        deque_01 = deque([(0, 0)])

        while deque_01:
            row, col = deque_01.popleft()

            current_cost = distance[row][col]

            for direction_value, (dr, dc) in directions.items():
                next_row = row + dr
                next_col = col + dc

                if not (0 <= next_row < rows and 0 <= next_col < cols):
                    continue

                extra_cost = 0 if grid[row][col] == direction_value else 1

                new_cost = current_cost + extra_cost

                if new_cost < distance[next_row][next_col]:
                    distance[next_row][next_col] = new_cost

                    if extra_cost == 0:
                        deque_01.appendleft((next_row, next_col))
                    else:
                        deque_01.append((next_row, next_col))

        return distance[rows - 1][cols - 1]

The implementation begins by storing the grid dimensions and constructing a direction mapping. This mapping converts each arrow value into its corresponding row and column movement.

The distance matrix stores the minimum modification cost needed to reach every cell. Initially, all cells are set to infinity because they are unreachable until proven otherwise.

The deque is the central structure of 0-1 BFS. Cells reached with no additional cost are pushed to the front, while cells requiring a modification are pushed to the back. This ordering ensures that lower-cost paths are always explored first.

For every popped cell, the algorithm checks all four possible directions. If the chosen movement matches the current cell’s arrow, the move costs 0. Otherwise, the move costs 1 because we must modify the sign.

Whenever a cheaper path to a neighbor is found, the distance is updated and the neighbor is inserted into the deque in the appropriate position.

Finally, the algorithm returns the minimum cost stored for the bottom-right cell.

Go Solution

package main

import "container/list"

func minCost(grid [][]int) int {
	rows := len(grid)
	cols := len(grid[0])

	directions := [][]int{
		{},
		{0, 1},   // right
		{0, -1},  // left
		{1, 0},   // down
		{-1, 0},  // up
	}

	const INF = int(1e9)

	distance := make([][]int, rows)
	for i := range distance {
		distance[i] = make([]int, cols)
		for j := range distance[i] {
			distance[i][j] = INF
		}
	}

	type State struct {
		row int
		col int
	}

	deque := list.New()

	distance[0][0] = 0
	deque.PushFront(State{0, 0})

	for deque.Len() > 0 {
		front := deque.Front()
		current := front.Value.(State)
		deque.Remove(front)

		row := current.row
		col := current.col

		currentCost := distance[row][col]

		for direction := 1; direction <= 4; direction++ {
			dr := directions[direction][0]
			dc := directions[direction][1]

			nextRow := row + dr
			nextCol := col + dc

			if nextRow < 0 || nextRow >= rows || nextCol < 0 || nextCol >= cols {
				continue
			}

			extraCost := 1
			if grid[row][col] == direction {
				extraCost = 0
			}

			newCost := currentCost + extraCost

			if newCost < distance[nextRow][nextCol] {
				distance[nextRow][nextCol] = newCost

				if extraCost == 0 {
					deque.PushFront(State{nextRow, nextCol})
				} else {
					deque.PushBack(State{nextRow, nextCol})
				}
			}
		}
	}

	return distance[rows-1][cols-1]
}

The Go implementation follows the exact same logic as the Python version, but uses Go’s container/list package to implement the deque required for 0-1 BFS.

A custom State struct stores row and column coordinates. The distance matrix is initialized using nested slices.

Since Go does not provide infinity values for integers directly, a large constant INF is used instead.

The rest of the implementation mirrors the Python solution closely.

Worked Examples

Example 1

grid =
[
  [1,1,1,1],
  [2,2,2,2],
  [1,1,1,1],
  [2,2,2,2]
]

Initial state:

Cell	Distance
(0,0)	0
others	inf

Starting from (0,0):

• Right move matches arrow 1
• Cost remains 0

Deque state:

Operation	Deque
Start	[(0,0)]
Move right	[(0,1)]
Move right	[(0,2)]
Move right	[(0,3)]

At (0,3):

• Moving down requires changing arrow
• Extra cost = 1

Distance update:

Cell	New Cost
(1,3)	1

Continue similarly:

• Row transition at (1,0) costs another 1
• Row transition at (2,3) costs another 1

Final minimum cost:

3

Example 2

grid =
[
  [1,1,3],
  [3,2,2],
  [1,1,4]
]

Path followed naturally:

(0,0) -> (0,1) -> (0,2)
                     |
                     v
                  (1,2)
                  <- (1,1)
                     |
                     v
                  (2,1) -> (2,2)

Every movement already matches the arrows.

Total cost:

0

Example 3

grid =
[
  [1,2],
  [4,3]
]

Starting at (0,0):

• Right move to (0,1) costs 0

At (0,1):

• Arrow points left
• Need to move down instead
• Modification cost = 1

Total minimum cost:

1

Complexity Analysis

Measure	Complexity	Explanation
Time	O(m*n)	Each cell is processed efficiently through 0-1 BFS
Space	O(m*n)	Distance matrix and deque storage

The grid contains at most 10,000 cells. Every cell may be inserted into the deque multiple times, but only when a strictly better distance is discovered. Since each edge has weight 0 or 1, 0-1 BFS processes the graph very efficiently, giving linear complexity relative to the number of nodes and edges.

The space usage comes primarily from the distance matrix and deque.

Test Cases

from typing import List

class Solution:
    def minCost(self, grid: List[List[int]]) -> int:
        from collections import deque

        rows = len(grid)
        cols = len(grid[0])

        directions = {
            1: (0, 1),
            2: (0, -1),
            3: (1, 0),
            4: (-1, 0)
        }

        distance = [[float('inf')] * cols for _ in range(rows)]
        distance[0][0] = 0

        dq = deque([(0, 0)])

        while dq:
            row, col = dq.popleft()

            for direction, (dr, dc) in directions.items():
                nr = row + dr
                nc = col + dc

                if not (0 <= nr < rows and 0 <= nc < cols):
                    continue

                extra = 0 if grid[row][col] == direction else 1
                new_cost = distance[row][col] + extra

                if new_cost < distance[nr][nc]:
                    distance[nr][nc] = new_cost

                    if extra == 0:
                        dq.appendleft((nr, nc))
                    else:
                        dq.append((nr, nc))

        return distance[-1][-1]

solution = Solution()

assert solution.minCost([
    [1,1,1,1],
    [2,2,2,2],
    [1,1,1,1],
    [2,2,2,2]
]) == 3  # Provided example with multiple required changes

assert solution.minCost([
    [1,1,3],
    [3,2,2],
    [1,1,4]
]) == 0  # Already has a valid path

assert solution.minCost([
    [1,2],
    [4,3]
]) == 1  # Single modification required

assert solution.minCost([
    [1]
]) == 0  # Single cell grid

assert solution.minCost([
    [2,2],
    [2,2]
]) == 1  # Must modify to move toward destination

assert solution.minCost([
    [4,4,4],
    [4,4,4],
    [4,4,4]
]) == 4  # All arrows initially unusable

assert solution.minCost([
    [1,1,1],
    [1,1,1],
    [1,1,1]
]) == 2  # Need downward transitions

assert solution.minCost([
    [3,3,3],
    [3,3,3],
    [3,3,3]
]) == 2  # Need rightward transitions

Test Summary

Test	Why
Example 1	Verifies multiple modifications across rows
Example 2	Confirms zero-cost valid path
Example 3	Tests single modification scenario
Single cell grid	Smallest possible input
All left arrows	Tests forced modifications
All up arrows	Tests arrows pointing outside grid
All right arrows	Requires vertical corrections
All down arrows	Requires horizontal corrections

Edge Cases

Single Cell Grid

A 1 x 1 grid is a special case because the start and destination are the same cell. No movement is required, so the answer must always be 0.

Naive implementations sometimes attempt unnecessary traversal logic or fail because there are no valid neighbors. This implementation handles the case naturally because the starting distance is initialized to 0, and the algorithm immediately returns that value.

Arrows Pointing Outside the Grid

Some cells may contain arrows that lead beyond the grid boundary. These directions are effectively useless for traversal.

A buggy implementation might attempt invalid array accesses. This solution safely checks bounds before exploring neighbors, preventing out-of-range errors.

Multiple Possible Paths

There may be many different ways to reach the destination with different modification costs.

A greedy strategy that always follows existing arrows could fail because a short-term expensive modification might enable a much cheaper overall route.

0-1 BFS correctly evaluates all reachable states in increasing cost order, guaranteeing that the globally minimum modification cost is found.