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Manhattan Distances of All Arrangements of Pieces

Number: 3739

Difficulty: Hard

Paid? No

Companies: Google

Problem Description

Given an m×n grid and k identical pieces to be placed (with at most one piece per cell), we need to compute the sum over all valid placements of pieces of the Manhattan distances between every pair of pieces. A valid placement is any set of k distinct cells chosen from the grid, and the Manhattan distance between two cells (x1, y1) and (x2, y2) is defined as |x1 – x2| + |y1 – y2|. The answer should be returned modulo 10^9+7.

Key Insights

  • Instead of iterating over every valid arrangement (which is computationally infeasible), observe that each valid arrangement corresponds to choosing k cells out of m*n.
  • Each pair of distinct cells appears together in exactly C(m*n - 2, k - 2) arrangements.
  • The overall sum can be computed as: (Combination Factor) × (Sum of Manhattan distances over all pairs of distinct cells).
  • The Manhattan distance is separable into a row part and a column part. Compute the contribution from rows and columns independently:
    • Row contribution: n² * (Sum_{d=1}^{m-1} d * (m - d))
    • Column contribution: m² * (Sum_{d=1}^{n-1} d * (n - d))
  • Modular arithmetic needs be applied throughout since the numbers can be very large.

Space and Time Complexity

Time Complexity: O(mn_max + (m+n)) where mn_max is the cost for precomputing factorials up to mn (which is at most 10^5), and O(m+n) for computing the distance sums. Space Complexity: O(mn_max) for storing factorials and inverse factorials up to m*n.

Solution

The key is to decompose the problem into two parts:

  1. Compute the constant combination factor, which is C(total_cells - 2, k - 2) mod MOD where total_cells = m * n. This represents how many arrangements include a given pair of cells.
  2. Compute the overall Manhattan distance sum over all pairs of cells in the grid by separately summing over row differences and column differences:
    • The row part is: n² * (sum for all d=1 to m-1 of d*(m-d))
    • The column part is: m² * (sum for all d=1 to n-1 of d*(n-d))
  3. Multiply the computed Manhattan sum by the combination factor and return the result modulo MOD.

Data structures and techniques used include:

  • Precomputation of factorials and modular inverses for efficiently computing binomial coefficients.
  • Summing arithmetic series using simple loops (or formulas) because m and n are relatively small (≤ 10^5 cells, but m*n is at most 10^5).
  • Modular arithmetic to ensure that integers do not overflow.

Code Solutions

# Python solution with detailed comments

MOD = 10**9 + 7

# Function to precompute factorials and inverse factorials up to max_n
def precompute_factorials(max_n):
    factorial = [1] * (max_n + 1)
    inv_factorial = [1] * (max_n + 1)
    # Compute factorials modulo MOD
    for i in range(1, max_n + 1):
        factorial[i] = (factorial[i - 1] * i) % MOD
    # Compute inverse factorial using Fermat's little theorem
    inv_factorial[max_n] = pow(factorial[max_n], MOD - 2, MOD)
    for i in range(max_n, 0, -1):
        inv_factorial[i - 1] = (inv_factorial[i] * i) % MOD
    return factorial, inv_factorial

# Function to compute binomial coefficient C(n, k) modulo MOD
def comb(n, k, factorial, inv_factorial):
    if k < 0 or k > n:
        return 0
    return factorial[n] * inv_factorial[k] % MOD * inv_factorial[n - k] % MOD

def manhattanArrangement(m, n, k):
    total_cells = m * n
    # Precompute factorials up to total_cells
    factorial, inv_factorial = precompute_factorials(total_cells)
    
    # Combination factor: number of arrangements containing a specific pair of cells
    comb_factor = comb(total_cells - 2, k - 2, factorial, inv_factorial)
    
    # Compute row contribution: for rows 0 to m-1, the sum of differences is:
    # sum_{d=1}^{m-1} d * (m - d) and each difference is counted n^2 times.
    row_sum = 0
    for d in range(1, m):
        row_sum = (row_sum + d * (m - d)) % MOD
    row_sum = (row_sum * (n * n % MOD)) % MOD

    # Compute column contribution: each difference counted m^2 times.
    col_sum = 0
    for d in range(1, n):
        col_sum = (col_sum + d * (n - d)) % MOD
    col_sum = (col_sum * (m * m % MOD)) % MOD

    # Total Manhattan sum for all pairs of cells in one grid
    total_distance_sum = (row_sum + col_sum) % MOD

    # Multiply with combination factor to account for arrangements in which each pair is used
    answer = (comb_factor * total_distance_sum) % MOD
    return answer

# Example usage:
if __name__ == '__main__':
    # Test Example 1: m = 2, n = 2, k = 2, Expected output: 8
    print(manhattanArrangement(2, 2, 2))
    # Test Example 2: m = 1, n = 4, k = 3, Expected output: 20
    print(manhattanArrangement(1, 4, 3))
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