What is the time complexity of Minimum Cost to Make at Least One Valid Path in a Grid?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Minimum Cost to Make at Least One Valid Path in a Grid?

This problem uses Array, Breadth-First Search, Graph, Heap (Priority Queue), Matrix, Shortest Path. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimum Cost to Make at Least One Valid Path in a Grid?

Insight 1: The problem can be modeled as a shortest path problem where the cost of moving to a neighbor cell is either 0 (if we follow the direction in the current cell) or 1 (if we need to change the direction). Insight 2: Since the cost of edges are either 0 or 1, we can use a 0-1 BFS (Breadth-First Search) algorithm, which efficiently finds the shortest path in a graph with edges of these weights.

Which companies ask LeetCode 1368. Minimum Cost to Make at Least One Valid Path in a Grid in interviews?

This problem is frequently asked at Cleartrip.

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

Q: What is the space complexity of Minimum Cost to Make at Least One Valid Path in a Grid?

The space complexity is O(m*n).

Minimum Cost to Make at Least One Valid Path in a Grid - Complete Solution Guide

Minimum Cost to Make at Least One Valid Path in a Grid is LeetCode problem 1368, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

Given an m x n grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of grid[i][j] can be: 1 which means go to the cell to the right. (i.e go from grid[i][j] to grid[i][j + 1] ) 2 which means go to the cell to the left. (i.e go from grid[i][j] to grid[i][j - 1] ) 3 which means go to the lower cell. (i.e go from grid[i][j] to grid[i + 1][j] ) 4 which means go to the upper cell. (i.e go from grid[i][j] to grid[i - 1][j] ) Noti

Detailed Explanation

The problem asks us to find the minimum cost to make at least one valid path from the top-left cell (0, 0) to the bottom-right cell (m-1, n-1) in a given grid. Each cell in the grid contains a number from 1 to 4, representing the direction we should move to the next cell: 1 (right), 2 (left), 3 (down), or 4 (up). We start at (0, 0) and want to reach (m-1, n-1) following the directions specified in the grid. However, the provided grid might not have a valid path. We are allowed to change the direction of a cell with a cost of 1. We can change the sign on a cell at most once. The goal is to determine the minimum number of changes (cost) needed to ensure there's at least one path from the starting cell to the destination cell.

Solution Approach

The provided code uses a 0-1 BFS algorithm to find the minimum cost. The algorithm starts at the top-left cell (0, 0) and explores the grid, keeping track of the minimum cost to reach each cell. It uses a deque to prioritize cells that can be reached with a cost of 0 (i.e., following the given direction). When a cell is reached by following the correct direction, it's added to the front of the deque; otherwise, it's added to the rear. This helps to prioritize the exploration of paths with lower costs first. The algorithm continues until the bottom-right cell (m-1, n-1) is reached, at which point it returns the minimum cost.

Step-by-Step Algorithm

Step 1: Initialize a distance matrix `dist` with infinity for all cells, except for the starting cell (0, 0), which is initialized to 0. This matrix stores the minimum cost to reach each cell from the starting cell.
Step 2: Create a deque `dq` and add the starting cell (0, 0) to it.
Step 3: Define the possible moves (right, left, down, up) using a `moves` array.
Step 4: While the deque is not empty:
a. Remove the first cell (r, c) from the deque.
b. Get the current cost `d` to reach (r, c) from the `dist` matrix.
c. If (r, c) is the destination cell (m-1, n-1), return `d`.
d. For each possible move (dr, dc) in `moves`:
i. Calculate the new row `nr` and new column `nc`.
ii. Check if the new cell (nr, nc) is within the grid boundaries.
iii. Calculate the cost `cost` to move from (r, c) to (nr, nc). The cost is 0 if the direction in `grid[r][c]` matches the move, and 1 otherwise.
iv. If the new cost `d + cost` is less than the current cost `dist[nr][nc]` to reach (nr, nc), update `dist[nr][nc]` to `d + cost`.
v. If `cost` is 0, add (nr, nc) to the front of the deque; otherwise, add it to the rear. This prioritizes cells reachable with a cost of 0.
Step 5: If the destination cell is not reached, return -1 (which should be unreachable given the constraints).
Step 6: Memory management for C version, freeing the deque.

Key Insights

Insight 1: The problem can be modeled as a shortest path problem where the cost of moving to a neighbor cell is either 0 (if we follow the direction in the current cell) or 1 (if we need to change the direction).
Insight 2: Since the cost of edges are either 0 or 1, we can use a 0-1 BFS (Breadth-First Search) algorithm, which efficiently finds the shortest path in a graph with edges of these weights.
Insight 3: A standard BFS wouldn't work efficiently here. Using a deque allows prioritizing moves with cost 0 (adding to the front) and moves with cost 1 (adding to the rear), thus mimicking Dijkstra's algorithm for this special case, without using a priority queue.
Insight 4: Representing the grid with its directions makes it easy to determine the cost of moving from a cell to an adjacent cell. This representation simplifies the cost calculation.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Array, Breadth-First Search, Graph, Heap (Priority Queue), Matrix, Shortest Path.

Companies

Asked at: Cleartrip.

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

Hard

Array Breadth-First Search Graph Heap (Priority Queue)Matrix Shortest Path

Time: O(m*n)Space: O(m*n)

Solution Code

Available in:

import java.util.ArrayDeque;
import java.util.Deque;
import java.util.Arrays;

class Solution {
    public int minCost(int[][] grid) {
        int m = grid.length;
        int n = grid[0].length;

        int[][] dist = new int[m][n];
        for (int i = 0; i < m; i++) {
            Arrays.fill(dist[i], Integer.MAX_VALUE);
        }
        dist[0][0] = 0;

        Deque<int[]> dq = new ArrayDeque<>(); // Stores (row, col)
        dq.offer(new int[]{0, 0});

        // Directions: moves[0] -> right (sign 1), moves[1] -> left (sign 2), etc.
        int[][] moves = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        while (!dq.isEmpty()) {
            int[] curr = dq.poll();
            int r = curr[0];
            int c = curr[1];
            int d = dist[r][c];

            if (r == m - 1 && c == n - 1) {
                return d;
            }

            for (int i = 0; i < 4; i++) {
                int dr = moves[i][0];
                int dc = moves[i][1];
                int nr = r + dr;
                int nc = c + dc;

                if (0 <= nr && nr < m && 0 <= nc && nc < n) {
                    int cost = (grid[r][c] == i + 1) ? 0 : 1;

                    if (d + cost < dist[nr][nc]) {
                        dist[nr][nc] = d + cost;
                        if (cost == 0) {
                            dq.offerFirst(new int[]{nr, nc});
                        } else {
                            dq.offerLast(new int[]{nr, nc});
                        }
                    }
                }
            }
        }

        return -1; // Should be unreachable given problem constraints
    }
}

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Problem Info

Description

Problem 1368: Minimum Cost to Make at Least One Valid Path in a Grid - Detailed explanation and solution approaches.

Topics

Array Breadth-First Search Graph Heap (Priority Queue)Matrix Shortest Path

Companies

Cleartrip

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1368

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

15 steps

1
Step 1: Initialize a distance matrix `dist` with infinity for all cells, except for the starting cell (0, 0), which is initialized to 0. This matrix stores the minimum cost to reach each cell from the starting cell.
2
Step 2: Create a deque `dq` and add the starting cell (0, 0) to it.
3
Step 3: Define the possible moves (right, left, down, up) using a `moves` array.
4
Step 4: While the deque is not empty:
5
a. Remove the first cell (r, c) from the deque.
6
b. Get the current cost `d` to reach (r, c) from the `dist` matrix.
7
c. If (r, c) is the destination cell (m-1, n-1), return `d`.
8
d. For each possible move (dr, dc) in `moves`:
9
i. Calculate the new row `nr` and new column `nc`.
10
ii. Check if the new cell (nr, nc) is within the grid boundaries.
11
iii. Calculate the cost `cost` to move from (r, c) to (nr, nc). The cost is 0 if the direction in `grid[r][c]` matches the move, and 1 otherwise.
12
iv. If the new cost `d + cost` is less than the current cost `dist[nr][nc]` to reach (nr, nc), update `dist[nr][nc]` to `d + cost`.
13
v. If `cost` is 0, add (nr, nc) to the front of the deque; otherwise, add it to the rear. This prioritizes cells reachable with a cost of 0.
14
Step 5: If the destination cell is not reached, return -1 (which should be unreachable given the constraints).
15
Step 6: Memory management for C version, freeing the deque.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity is O(m*n) because each cell is visited at most once and processed in constant time. The space complexity is O(m*n) because the `dist` matrix stores the shortest distance to each cell, and the deque can hold up to m*n elements in the worst-case scenario.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dijkstra's Algorithm

Dijkstra's algorithm can also be used to solve this problem. Treat the grid as a graph, where each cell is a node, and the cost of moving between adjacent cells is either 0 or 1. Dijkstra's algorithm will find the shortest path from (0, 0) to (m-1, n-1).

Time: O(m*n*log(m*n)) using a priority queue implementation of Dijkstra.Space: O(m*n)

Trade-offs: While Dijkstra's algorithm is a general solution for shortest path problems, it's less efficient than 0-1 BFS when the edge weights are limited to 0 and 1. The 0-1 BFS is more efficient because it leverages the specific properties of the problem and avoids the overhead of a priority queue. For general shortest path problems with arbitrary edge weights, Dijkstra or Bellman-Ford are appropriate.

✨

Key Insights

1Insight 1: The problem can be modeled as a shortest path problem where the cost of moving to a neighbor cell is either 0 (if we follow the direction in the current cell) or 1 (if we need to change the direction).
2Insight 2: Since the cost of edges are either 0 or 1, we can use a 0-1 BFS (Breadth-First Search) algorithm, which efficiently finds the shortest path in a graph with edges of these weights.
3Insight 3: A standard BFS wouldn't work efficiently here. Using a deque allows prioritizing moves with cost 0 (adding to the front) and moves with cost 1 (adding to the rear), thus mimicking Dijkstra's algorithm for this special case, without using a priority queue.
4Insight 4: Representing the grid with its directions makes it easy to determine the cost of moving from a cell to an adjacent cell. This representation simplifies the cost calculation.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Graph

Nodes connected by edges. Like a social network or city map!

Edge Cases

Edge case 1: The grid already has a valid path from (0, 0) to (m-1, n-1). In this case, the algorithm will correctly return 0, as it will traverse the path without needing to change any directions.
Edge case 2: A grid where no path exists between (0, 0) and (m-1, n-1) without changing any directions. The algorithm will correctly find the minimum number of changes required to create a valid path.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using a regular BFS algorithm without considering the cost. This would not prioritize paths with fewer changes, resulting in a suboptimal solution.

✗Common mistake 2: Not handling the edge cases where the grid is empty or has only one cell. However, the problem constraints specify m, n >= 1.

✗Common mistake 3: Failing to check for valid indices when moving to the next cell, leading to array out-of-bounds errors.

✗Common mistake 4: Not understanding the difference between 'offerFirst' and 'offerLast' in the Java deque and the corresponding functionalities in Python and C++. This causes wrong order of deque and leads to wrong results

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Pathfinding in robotics, where the robot needs to navigate a grid-like environment with varying costs for different movements. For instance, some movements might require more energy or time, which can be modeled as costs.

🏦Real-world scenario 2: Network routing, where the goal is to find the shortest path between two nodes in a network with varying costs for different connections. For example, some connections might have higher latency or bandwidth limitations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 13

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Minimum Cost to Make at Least One Valid Path in a Grid"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether breadth-first search techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?