What is the time complexity of Unique Paths III?

The optimal solution has a time complexity of O(4^(m*n)).

What is the space complexity of Unique Paths III?

The space complexity is O(m*n).

What data structures are used to solve Unique Paths III?

This problem uses Array, Backtracking, Bit Manipulation, Matrix. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Unique Paths III?

Insight 1: Backtracking is a suitable approach since we need to explore all possible paths from the starting cell to the ending cell. Insight 2: To ensure each non-obstacle cell is visited exactly once, we can mark visited cells during the backtracking process and revert the marking when backtracking.

Which companies ask LeetCode 980. Unique Paths III in interviews?

This problem is frequently asked at Cruise, Pinterest.

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Unique Paths III - LeetCode 980 Solution

Unique Paths III - Complete Solution Guide

Unique Paths III is LeetCode problem 980, 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

You are given an m x n integer array grid where grid[i][j] could be: 1 representing the starting square. There is exactly one starting square. 2 representing the ending square. There is exactly one ending square. 0 representing empty squares we can walk over. -1 representing obstacles that we cannot walk over. Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once . Example 1: Input: grid = [[1,0,0,0],[0,0,0,0

Detailed Explanation

The problem asks us to find the number of unique paths from a starting cell (marked as 1) to an ending cell (marked as 2) in a grid. The grid also contains empty cells (marked as 0) and obstacles (marked as -1). The path must traverse every non-obstacle cell exactly once. We can only move in four directions: up, down, left, and right.

Solution Approach

The provided code employs a backtracking algorithm to explore all possible paths from the starting cell to the ending cell. It keeps track of visited cells by temporarily marking them as obstacles (-1). Once a path reaches the ending cell, it checks if all non-obstacle cells have been visited. If so, it increments the path count. After exploring each direction, the cell is marked back to its original value to allow other paths to explore it.

Step-by-Step Algorithm

Step 1: Initialize the number of non-obstacle cells and find the starting cell's coordinates.
Step 2: Define a recursive backtrack function with the current cell's coordinates and the number of visited cells as parameters.
Step 3: Base Case 1: If the current cell is out of bounds or is an obstacle, return 0 (invalid path).
Step 4: Base Case 2: If the current cell is the ending cell, check if the number of visited cells equals the total number of non-obstacle cells. If so, return 1 (valid path); otherwise, return 0 (invalid path).
Step 5: Mark the current cell as visited (set to -1).
Step 6: Recursively call the backtrack function for all four possible directions (up, down, left, right), incrementing the visited cells count.
Step 7: Unmark the current cell (set back to its original value) to allow other paths to explore it.
Step 8: Return the total number of valid paths found from the current cell.

Key Insights

Insight 1: Backtracking is a suitable approach since we need to explore all possible paths from the starting cell to the ending cell.
Insight 2: To ensure each non-obstacle cell is visited exactly once, we can mark visited cells during the backtracking process and revert the marking when backtracking.
Insight 3: The critical condition for a valid path is that we reach the ending cell after visiting all non-obstacle cells. We need to count the number of non-obstacle cells at the beginning and compare it with the number of visited cells when we reach the ending cell.

Complexity Analysis

Time Complexity: O(4^(m*n))

Space Complexity: O(m*n)

Topics

This problem involves: Array, Backtracking, Bit Manipulation, Matrix.

Companies

Asked at: Cruise, Pinterest.

Unique Paths III | LeetCode 980

Hard

Array Backtracking Bit Manipulation Matrix

Time: O(4^(m*n))Space: O(m*n)

Solution Code

Available in:

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

        int non_obstacles_count = 0;
        int start_r = -1, start_c = -1;

        for (int r = 0; r < m; r++) {
            for (int c = 0; c < n; c++) {
                if (grid[r][c] >= 0) {
                    non_obstacles_count++;
                }
                if (grid[r][c] == 1) {
                    start_r = r;
                    start_c = c;
                }
            }
        }

        return backtrack(grid, start_r, start_c, 1, non_obstacles_count);
    }

    private int backtrack(int[][] grid, int r, int c, int visited_count, int non_obstacles_count) {
        int m = grid.length;
        int n = grid[0].length;

        if (!(0 <= r && r < m && 0 <= c && c < n && grid[r][c] != -1)) {
            return 0;
        }

        if (grid[r][c] == 2) {
            if (visited_count == non_obstacles_count) {
                return 1;
            } else {
                return 0;
            }
        }

        int temp = grid[r][c];
        grid[r][c] = -1;

        int paths_found = 0;
        int[][] directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
        for (int[] direction : directions) {
            int dr = direction[0];
            int dc = direction[1];
            paths_found += backtrack(grid, r + dr, c + dc, visited_count + 1, non_obstacles_count);
        }

        grid[r][c] = temp;

        return paths_found;
    }
}

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Problem Info

Description

Problem 980: Unique Paths III - Detailed explanation and solution approaches.

Topics

Array Backtracking Bit Manipulation Matrix

Companies

CruisePinterest

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:980

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

8 steps

1
Step 1: Initialize the number of non-obstacle cells and find the starting cell's coordinates.
2
Step 2: Define a recursive backtrack function with the current cell's coordinates and the number of visited cells as parameters.
3
Step 3: Base Case 1: If the current cell is out of bounds or is an obstacle, return 0 (invalid path).
4
Step 4: Base Case 2: If the current cell is the ending cell, check if the number of visited cells equals the total number of non-obstacle cells. If so, return 1 (valid path); otherwise, return 0 (invalid path).
5
Step 5: Mark the current cell as visited (set to -1).
6
Step 6: Recursively call the backtrack function for all four possible directions (up, down, left, right), incrementing the visited cells count.
7
Step 7: Unmark the current cell (set back to its original value) to allow other paths to explore it.
8
Step 8: Return the total number of valid paths found from the current cell.

💡

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(4^(m*n))

💾

Space Complexity

O(m*n)

📝

Why?

The time complexity stems from the fact that each cell has up to 4 possible moves, leading to an exponential increase in paths as we explore the grid. The space complexity is due to the call stack depth during the recursive calls, proportional to the total number of cells.

📖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

Iterative Depth-First Search (DFS)

An iterative DFS approach using a stack could avoid recursion overhead. It mimics the backtracking process iteratively.

Time: O(4^(m*n))Space: O(m*n)

Trade-offs: Can be more memory-efficient than recursive DFS in some cases by controlling the stack size and might avoid stack overflow issues for very large grids. However, it might be less readable.

✨

Key Insights

1Insight 1: Backtracking is a suitable approach since we need to explore all possible paths from the starting cell to the ending cell.
2Insight 2: To ensure each non-obstacle cell is visited exactly once, we can mark visited cells during the backtracking process and revert the marking when backtracking.
3Insight 3: The critical condition for a valid path is that we reach the ending cell after visiting all non-obstacle cells. We need to count the number of non-obstacle cells at the beginning and compare it with the number of visited cells when we reach the ending cell.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: The grid has no path from start to end. The backtracking algorithm will naturally return 0 in this case, as it won't find any valid paths.
Edge case 2: The starting or ending cell is blocked (marked as -1). The initial checks in the backtrack function prevent any exploration from starting or ending at such cells, resulting in 0 valid paths.
Edge case 3: The grid has no 0 cells. In this case, the path from start to end must directly traverse the start and end cells. The visited count equals to the non_obstacle_count if start cell directly connects to end cell. Otherwise, there's no valid path.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to revert the grid cell value after backtracking. This prevents other paths from exploring that cell, leading to incorrect results.

✗Common mistake 2: Not correctly counting the non-obstacle cells initially. This leads to incorrect validation when reaching the ending cell.

✗Common mistake 3: Failing to check for out-of-bounds or obstacle cells at the beginning of the backtrack function. This can lead to errors and incorrect path exploration.

✗Common mistake 4: Not distinguishing between visiting every empty cell (0) and every non-obstacle cell. The prompt specifies visiting every non-obstacle cell exactly once. The check for 'visited_count' must take this into account.

✅

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 a maze. The algorithm can be adapted to find all possible paths through a maze, subject to certain constraints.

🏦Real-world scenario 2: Robotics navigation. A robot can use this algorithm to plan a route through an environment, avoiding obstacles and ensuring it covers all relevant areas.

🎮Real-world scenario 3: Game AI: Determining possible move sequences for a character in a game to achieve a certain objective while visiting all required locations.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: August 20, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

❓

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.

What is the difference between backtracking and brute force?

Backtracking prunes invalid paths early using constraints, while brute force explores all possibilities. In backtracking, you "choose" an option, "explore" it recursively, then "unchoose" (backtrack) if it doesn't lead to a solution. This eliminates entire branches of the search tree.

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 "Unique Paths III"?

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 backtracking techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?