LeetCode #3426 — HARD

Manhattan Distances of All Arrangements of Pieces

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

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 109 + 7.

The Manhattan Distance between two cells (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|.

Example 1:

Input: m = 2, n = 2, k = 2

Output: 8

Explanation:

The valid arrangements of pieces on the board are:

  • In the first 4 arrangements, the Manhattan distance between the two pieces is 1.
  • In the last 2 arrangements, the Manhattan distance between the two pieces is 2.

Thus, the total Manhattan distance across all valid arrangements is 1 + 1 + 1 + 1 + 2 + 2 = 8.

Example 2:

Input: m = 1, n = 4, k = 3

Output: 20

Explanation:

The valid arrangements of pieces on the board are:

  • The first and last arrangements have a total Manhattan distance of 1 + 1 + 2 = 4.
  • The middle two arrangements have a total Manhattan distance of 1 + 2 + 3 = 6.

The total Manhattan distance between all pairs of pieces across all arrangements is 4 + 6 + 6 + 4 = 20.

Constraints:

  • 1 <= m, n <= 105
  • 2 <= m * n <= 105
  • 2 <= k <= m * n

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: 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 109 + 7. The Manhattan Distance between two cells (xi, yi) and (xj, yj) is |xi - xj| + |yi - yj|.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

2
2
2

Example 2

1
4
3
Step 02

Core Insight

What unlocks the optimal approach

  • Fix two pieces in two specific locations and find the number of boards where this can happen.
  • A particular pair of positions will be counted exactly <code>C(m * n - 2, k - 2)</code> times. Calculate the total distance for all pairs of positions and multiply it with <code>C(m * n - 2, k - 2)</code>.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3426: Manhattan Distances of All Arrangements of Pieces
class Solution {
  public int distanceSum(int m, int n, int k) {
    final long rowContrib = 1L * n * n * (1L * m * m * m - m) / 6 % MOD;
    final long colContrib = 1L * m * m * (1L * n * n * n - n) / 6 % MOD;
    return (int) ((rowContrib + colContrib) % MOD * nCk(m * n - 2, k - 2) % MOD);
  }

  private static final int MOD = 1_000_000_007;

  private long nCk(int n, int k) {
    long res = 1;
    for (int i = 1; i <= k; ++i)
      // By Fermat's little theorem.
      res = res * (n - i + 1) % MOD * modPow(i, MOD - 2) % MOD;
    return res;
  }

  private long modPow(long x, long n) {
    if (n == 0)
      return 1;
    if (n % 2 == 1)
      return x * modPow(x, n - 1) % MOD;
    return modPow(x * x % MOD, n / 2);
  }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(log n)
Space
O(1)

Approach Breakdown

ITERATIVE
O(n) time
O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT
O(log n) time
O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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