What is the time complexity of Manhattan Distances of All Arrangements of Pieces?

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

What is the space complexity of Manhattan Distances of All Arrangements of Pieces?

The space complexity is O(m*n).

What data structures are used to solve Manhattan Distances of All Arrangements of Pieces?

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

What are the key insights for solving Manhattan Distances of All Arrangements of Pieces?

Insight 1: The key is to avoid generating all possible arrangements and calculating the Manhattan distance for each. This would be computationally expensive for large values of m, n, and k. Insight 2: We can use combinatorics and mathematical reasoning to derive a formula for the total Manhattan distance across all arrangements. The problem simplifies to finding the sum of all pair-wise Manhattan distances in a fully occupied grid (m x n) and then multiplying it by the number of ways to choose k-2 pieces from the remaining N-2 spots, where N = m * n.

Which companies ask LeetCode 3426. Manhattan Distances of All Arrangements of Pieces in interviews?

This problem is frequently asked at Rubrik.

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Manhattan Distances of All Arrangements of Pieces - LeetCode 3426 Solution

Manhattan Distances of All Arrangements of Pieces - Complete Solution Guide

Manhattan Distances of All Arrangements of Pieces is LeetCode problem 3426, 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 three integers m , n , and k . There is a rectangular grid of size m × n containing k identical pieces. Return the sum of Manhattan distances between every pair of pieces over all valid arrangements of pieces. A valid arrangement is a placement of all k pieces on the grid with at most one piece per cell. Since the answer may be very large, return it modulo 10 9 + 7 . The Manhattan Distance between two cells (x i , y i ) and (x j , y j ) is |x i - x j | + |y i - y j | . Exampl

Detailed Explanation

The problem asks us to calculate the sum of Manhattan distances between all possible pairs of pieces in a rectangular grid (m x n) containing 'k' identical pieces. We need to consider all possible valid arrangements of these 'k' pieces, where each cell can have at most one piece. The Manhattan distance between two cells (x1, y1) and (x2, y2) is |x1 - x2| + |y1 - y2|. The final result must be returned modulo 10^9 + 7.

Solution Approach

The solution uses combinatorics to efficiently calculate the total Manhattan distance. It involves these steps: 1. Calculate the total number of possible arrangements using combinations (nCr). 2. Calculate the total Manhattan distances for all possible pairs in a grid. Use a closed-form expression for this to avoid iterating through all pairs. 3. Multiply the total distance from step 2 by the number of combinations that represent how many times each distance appears across all arrangements. This gives the desired result.

Step-by-Step Algorithm

Step 1: Calculate N = m * n, which is the total number of cells in the grid.
Step 2: Precompute factorials (fact) and their modular inverses (invfact) from 0 to N. These will be used to efficiently calculate nCr modulo MOD.
Step 3: Define a function nCr_mod(n, r) that calculates n choose r modulo MOD using the precomputed factorials and their inverses. This function returns 0 if r is invalid (r < 0 or r > n).
Step 4: Calculate the number of combinations of choosing k-2 cells from N-2 available cells, which is `comb_term = nCr_mod(N - 2, k - 2)`. This represents the number of times each pair of pieces will appear together in an arrangement.
Step 5: Calculate the total Manhattan distance between all possible pairs of cells if they were occupied. The formula used is (N * (N - 1) * (m + n) / 6) % MOD. We use modular inverse of 6 since we are doing division.
Step 6: Multiply `comb_term` with the `dist_sum_all_pairs` calculated in Step 5 and take the modulo to get the `total_sum`. This is the required answer.
Step 7: Return the integer value of `total_sum`.

Key Insights

Insight 1: The key is to avoid generating all possible arrangements and calculating the Manhattan distance for each. This would be computationally expensive for large values of m, n, and k.
Insight 2: We can use combinatorics and mathematical reasoning to derive a formula for the total Manhattan distance across all arrangements. The problem simplifies to finding the sum of all pair-wise Manhattan distances in a fully occupied grid (m x n) and then multiplying it by the number of ways to choose k-2 pieces from the remaining N-2 spots, where N = m * n.
Insight 3: Pre-calculating factorials and their modular inverses is crucial for efficient computation of combinations (nCr) under modulo. Also, the inverse of 6 is needed for a division in the formula, which is done via modular inverse.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(m*n)

Topics

This problem involves: Math, Combinatorics.

Companies

Asked at: Rubrik.

Manhattan Distances of All Arrangements of Pieces | LeetCode 3426

Hard

Math Combinatorics

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

Solution Code

Available in:

class Solution {
    public int distanceSum(int m, int n, int k) {
        int MOD = (int)1e9 + 7;
        int N = m * n;

        if (k < 2) {
            return 0;
        }

        long[] fact = new long[N + 1];
        long[] invfact = new long[N + 1];
        fact[0] = 1;
        invfact[0] = 1;
        for (int i = 1; i <= N; i++) {
            fact[i] = (fact[i - 1] * i) % MOD;
        }

        invfact[N] = power(fact[N], MOD - 2, MOD);
        for (int i = N - 1; i >= 0; i--) {
            invfact[i] = (invfact[i + 1] * (i + 1)) % MOD;
        }

        
        
        
        
        
        
        
        
        
        
        

        long comb_term = nCr_mod(N - 2, k - 2, fact, invfact, MOD);
        
        long inv6 = power(6, MOD - 2, MOD);
        
        long term1 = ((long)N * (N - 1)) % MOD;
        long term2 = (m + n) % MOD;
        
        long dist_sum_all_pairs = (term1 * term2) % MOD;
        dist_sum_all_pairs = (dist_sum_all_pairs * inv6) % MOD;

        long total_sum = (comb_term * dist_sum_all_pairs) % MOD;
        
        return (int)total_sum;
    }
    
    private long nCr_mod(int n_val, int r_val, long[] fact, long[] invfact, int MOD) {
        if (r_val < 0 || r_val > n_val) {
            return 0;
        }
        long num = fact[n_val];
        long den = (invfact[r_val] * invfact[n_val - r_val]) % MOD;
        return (num * den) % MOD;
    }
    
    private long power(long base, long exp, int MOD) {
        long res = 1;
        base = base % MOD;
        while (exp > 0) {
            if (exp % 2 == 1) res = (res * base) % MOD;
            exp = exp >> 1;
            base = (base * base) % MOD;
        }
        return res;
    }
}

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Problem Info

Description

Problem 3426: Manhattan Distances of All Arrangements of Pieces - Detailed explanation and solution approaches.

Topics

Math Combinatorics

Companies

Rubrik

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3426

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

7 steps

1
Step 1: Calculate N = m * n, which is the total number of cells in the grid.
2
Step 2: Precompute factorials (fact) and their modular inverses (invfact) from 0 to N. These will be used to efficiently calculate nCr modulo MOD.
3
Step 3: Define a function nCr_mod(n, r) that calculates n choose r modulo MOD using the precomputed factorials and their inverses. This function returns 0 if r is invalid (r < 0 or r > n).
4
Step 4: Calculate the number of combinations of choosing k-2 cells from N-2 available cells, which is `comb_term = nCr_mod(N - 2, k - 2)`. This represents the number of times each pair of pieces will appear together in an arrangement.
5
Step 5: Calculate the total Manhattan distance between all possible pairs of cells if they were occupied. The formula used is (N * (N - 1) * (m + n) / 6) % MOD. We use modular inverse of 6 since we are doing division.
6
Step 6: Multiply `comb_term` with the `dist_sum_all_pairs` calculated in Step 5 and take the modulo to get the `total_sum`. This is the required answer.
7
Step 7: Return the integer value of `total_sum`.

💡

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 dominant operation is the precomputation of factorials and their modular inverses. The loops in these computations run from 1 to N, resulting in O(N) time. The space complexity is O(N) because we are storing factorials and inverse factorials in arrays of size N+1.

📖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

Brute Force Approach

Generate all possible arrangements of k pieces and calculate the Manhattan distance for each pair within each arrangement. This approach is highly inefficient.

Time: O(C(m*n, k) * k^2) where C(m*n, k) represents the number of combinations to choose k cells among m*n. The k^2 factor comes from calculating all pairwise distances for each arrangement.Space: O(k) to store a single arrangement.

Trade-offs: This approach is simple to understand but is only feasible for very small values of m, n, and k. It will quickly exceed the time limit for the given constraints. The combinatorial complexity makes it impractical.

✨

Key Insights

1Insight 1: The key is to avoid generating all possible arrangements and calculating the Manhattan distance for each. This would be computationally expensive for large values of m, n, and k.
2Insight 2: We can use combinatorics and mathematical reasoning to derive a formula for the total Manhattan distance across all arrangements. The problem simplifies to finding the sum of all pair-wise Manhattan distances in a fully occupied grid (m x n) and then multiplying it by the number of ways to choose k-2 pieces from the remaining N-2 spots, where N = m * n.
3Insight 3: Pre-calculating factorials and their modular inverses is crucial for efficient computation of combinations (nCr) under modulo. Also, the inverse of 6 is needed for a division in the formula, which is done via modular inverse.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: k < 2. The problem statement implies that k >= 2. The code explicitly checks for k < 2 and returns 0 in this case, which is correct since we cannot have a pair to calculate distance if there are fewer than two pieces.
Edge case 2: Large values of m, n, and k that can lead to integer overflow. The code uses `long long` (or `long` in Java) and performs all operations modulo 10^9 + 7 (MOD) to prevent integer overflow and keep the intermediate and final results within the acceptable range.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling modulo operations correctly. When performing calculations involving factorials and combinations, the intermediate results can quickly become very large, leading to integer overflows. It's crucial to perform modulo operations after each multiplication and exponentiation.

✗Common mistake 2: Incorrect calculation of nCr (n choose r). Using the formula n! / (r! * (n-r)!) directly can lead to integer overflow. It is more efficient and safer to precompute factorials and their modular inverses, then use those for the calculation.

✗Common mistake 3: Not correctly accounting for all possible pairs. The formula used to calculate Manhattan distances relies on correct understanding of Combinatorial principles. Not using a correct and optimized formula may result in inefficient computations.

✅

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: Facility location. Consider optimizing the placement of service centers (e.g., hospitals, fire stations) to minimize the average distance to residents. This involves calculating Manhattan distances between potential facility locations and residential areas, similar to this problem.

🏦Real-world scenario 2: Network routing. In a grid-based network, finding efficient paths for data packets often involves minimizing the Manhattan distance. This algorithm could be adapted to optimize routing strategies by considering all possible paths and selecting the one with the minimum total distance.

💡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

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: June 7, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

❓

Frequently Asked Questions

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 "Manhattan Distances of All Arrangements of Pieces"?

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

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