What is the time complexity of Unique Paths?

The optimal solution has a time complexity of O(1).

What data structures are used to solve Unique Paths?

This problem uses Math, Dynamic Programming, Combinatorics. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Unique Paths?

Insight 1: The problem can be modeled as a combinatorics problem. Every path consists of (m-1) down moves and (n-1) right moves. Therefore, any unique path is a sequence of these moves. Insight 2: We can find the total number of moves required which is (m-1) + (n-1) = m + n - 2. The problem transforms to selecting (m-1) positions out of (m+n-2) total moves for the down moves, or equivalently (n-1) positions for right moves.

Which companies ask LeetCode 62. Unique Paths in interviews?

This problem is frequently asked at Accenture, Adobe, Amazon, Apple, Bloomberg and 15 more companies.

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Unique Paths - LeetCode 62 Solution

Unique Paths - Complete Solution Guide

Unique Paths is LeetCode problem 62, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a robot on an m x n grid. The robot is initially located at the top-left corner (i.e., grid[0][0] ). The robot tries to move to the bottom-right corner (i.e., grid[m - 1][n - 1] ). The robot can only move either down or right at any point in time. Given the two integers m and n , return the number of possible unique paths that the robot can take to reach the bottom-right corner . The test cases are generated so that the answer will be less than or equal to 2 * 10 9 . Example 1: Input: m

Detailed Explanation

The "Unique Paths" problem on LeetCode asks us to find the number of possible routes a robot can take to reach the bottom-right corner of an m x n grid. The robot starts at the top-left corner and can only move down or right at any given step. We need to return the total count of these unique paths, given the grid's dimensions (m and n). The answer is guaranteed to fit within a 32-bit integer.

Solution Approach

The provided code utilizes a combinatorics approach to solve the problem. It avoids creating an actual grid or using recursive methods that would lead to exponential time complexity. The core idea is recognizing that any valid path consists of `m-1` downward moves and `n-1` rightward moves. The code calculates the number of ways to choose the positions for these moves within the sequence of all moves using the combination formula. Python implementation leverages the built-in `math.comb` which optimizes for this calculation, while Java, C++ and C implementation calculates the combination using iterative multiplication and division in order to avoid overflow.

Step-by-Step Algorithm

Step 1: Calculate the total number of moves needed: `total_moves = m + n - 2`.
Step 2: Determine the number of down moves (or right moves): `down_moves = m - 1` or `right_moves = n - 1`. Both will yield the same result.
Step 3: Calculate the combination C(total_moves, down_moves). This can be directly computed using Python's `math.comb`.
Step 4: For Java, C++ and C, the Combination function is implemented using a for loop that iteratively calculates the value in order to avoid potential overflow.

Key Insights

Insight 1: The problem can be modeled as a combinatorics problem. Every path consists of (m-1) down moves and (n-1) right moves. Therefore, any unique path is a sequence of these moves.
Insight 2: We can find the total number of moves required which is (m-1) + (n-1) = m + n - 2. The problem transforms to selecting (m-1) positions out of (m+n-2) total moves for the down moves, or equivalently (n-1) positions for right moves.
Insight 3: The number of unique paths can be calculated as the combination C(m+n-2, m-1) or C(m+n-2, n-1), which is (m+n-2)! / ((m-1)! * (n-1)!).

Complexity Analysis

Time Complexity: O(1)

Space Complexity: O(1)

Topics

This problem involves: Math, Dynamic Programming, Combinatorics.

Companies

Asked at: Accenture, Adobe, Amazon, Apple, Bloomberg, ByteDance, Cisco, Coupang, DE Shaw, Goldman Sachs, Grammarly, Meta, Microsoft, Oracle, Roblox, TikTok, Uber, Yahoo, Zomato, tcs.

Unique Paths | LeetCode 62

Medium

Math Dynamic Programming Combinatorics

Time: O(1)Space: O(1)

Solution Code

Available in:

class Solution {
    public int uniquePaths(int m, int n) {
        // The problem can be solved using combinatorics.
        // To get from the top-left to the bottom-right corner, the robot must
        // make a total of (m-1) down moves and (n-1) right moves.
        // The total number of moves is (m-1) + (n-1) = m + n - 2.
        // The number of unique paths is the number of ways to choose the positions
        // for the (m-1) down moves (or (n-1) right moves) out of the total moves.
        // This is equivalent to calculating combinations "C(total_moves, down_moves)"
        // or "C(total_moves, right_moves)".
        //
        // C(n, k) = n! / (k! * (n-k)!)
        // We can use math.comb for a direct and efficient calculation.
        
        long totalMoves = m + n - 2;
        long downMoves = m - 1;

        return (int) combinations(totalMoves, downMoves);
    }

    private long combinations(long n, long k) {
        if (k < 0 || k > n) {
            return 0;
        }
        if (k == 0 || k == n) {
            return 1;
        }
        if (k > n / 2) {
            k = n - k;
        }

        long result = 1;
        for (long i = 1; i <= k; ++i) {
            result = result * (n - i + 1) / i;
        }
        return result;
    }
}

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Problem Info

Description

Problem 62: Unique Paths - Detailed explanation and solution approaches.

Topics

Math Dynamic Programming Combinatorics

Companies

AccentureAdobeAmazonAppleBloombergByteDanceCiscoCoupang

Stats

Difficulty:Medium

Languages:4

Companies:20

Problem #:62

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

4 steps

1
Step 1: Calculate the total number of moves needed: `total_moves = m + n - 2`.
2
Step 2: Determine the number of down moves (or right moves): `down_moves = m - 1` or `right_moves = n - 1`. Both will yield the same result.
3
Step 3: Calculate the combination C(total_moves, down_moves). This can be directly computed using Python's `math.comb`.
4
Step 4: For Java, C++ and C, the Combination function is implemented using a for loop that iteratively calculates the value in order to avoid potential overflow.

💡

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(1)

⚡ Constant time - Super fast! Same speed regardless of input size.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The Python version with `math.comb` will have a logarithmic time complexity in the arguments used in the combination formula in many implementations of the underlying algorithm. Other languages require an iterative calculation of the combination, where the number of iterations depends on the smaller of `m` and `n`, resulting in O(min(m, n)) complexity. No extra space is used (other than a few variables) regardless of m and n, giving O(1) space complexity.

📖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

Dynamic Programming

Create a 2D array `dp` of size `m x n`. `dp[i][j]` stores the number of unique paths to reach cell (i, j). Initialize `dp[0][j] = 1` and `dp[i][0] = 1` for all valid i and j. Then, `dp[i][j] = dp[i-1][j] + dp[i][j-1]`. The final answer is `dp[m-1][n-1]`.

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

Trade-offs: Dynamic programming has a higher time and space complexity than the combinatorial approach. However, it is easier to understand and implement for beginners. The DP approach requires more memory but is still viable, whereas the combinatorics solution can overflow with simple implementations.

Optimized Dynamic Programming

Similar to the dynamic programming approach, but optimize space by using only a 1D array. Instead of storing the whole `m x n` grid, store just the current row's results. Update this row iteratively based on the previous row's values.

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

Trade-offs: The time complexity is the same as regular DP but the space complexity is drastically improved.

✨

Key Insights

1Insight 1: The problem can be modeled as a combinatorics problem. Every path consists of (m-1) down moves and (n-1) right moves. Therefore, any unique path is a sequence of these moves.
2Insight 2: We can find the total number of moves required which is (m-1) + (n-1) = m + n - 2. The problem transforms to selecting (m-1) positions out of (m+n-2) total moves for the down moves, or equivalently (n-1) positions for right moves.
3Insight 3: The number of unique paths can be calculated as the combination C(m+n-2, m-1) or C(m+n-2, n-1), which is (m+n-2)! / ((m-1)! * (n-1)!).

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `m = 1`, or `n = 1`. In this scenario, there is only one possible path (either all right moves or all down moves). The code correctly handles these cases as it returns 1 due to the combination formula.
Edge case 2: `m = 100, n = 100`. These large inputs could lead to integer overflow when calculating factorials in C, C++, Java if naive implementation is used. The solutions avoid direct factorial calculation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Using a recursive approach without memoization leads to exponential time complexity and Time Limit Exceeded errors.

✗Common mistake 2: Incorrectly implementing the combination formula or factorial calculation, especially failing to address potential integer overflows.

✗Common mistake 3: Assuming that the robot can move diagonally.

✅

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 algorithms in robotics and game development. For example, planning a route for a delivery robot in a warehouse.

🏦Real-world scenario 2: Network routing. Determining the number of possible paths for data packets to travel from one node to another in a network.

💡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

Consider multiple approaches before coding. Mentioning alternatives shows breadth of knowledge.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

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 2, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

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

@CodeNinjaJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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"?

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

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