CF 289B - Polo the Penguin and Matrix

Rating: 1400
Tags: brute force, dp, implementation, sortings, ternary search
Solve time: 1m 16s
Verified: yes

Solution

Problem Understanding

We are given a matrix of integers and a number d. The penguin can either add or subtract d from any matrix element in a single move. The goal is to make all elements equal using the fewest moves possible. The output is the minimum number of moves needed, or -1 if it is impossible.

Concretely, if we flatten the matrix into a list of numbers, each element can move along the arithmetic progression defined by d. All elements must be congruent modulo d; otherwise, no sequence of additions or subtractions by d can make them equal. This gives the first necessary condition: all numbers must share the same remainder when divided by d. If this fails, the answer is immediately -1.

The input sizes are modest: up to 100×100 elements and d up to 10^4. This allows any algorithm up to roughly 10^6 operations comfortably. The values in the matrix are also up to 10^4, so we do not have to worry about integer overflow in Python.

Edge cases include situations where all numbers are already equal (zero moves), or where numbers differ but modulo d they do not match, making the solution impossible. For example, a 2×2 matrix [[2,3],[4,5]] with d = 2 cannot be equalized, since 2%2=0, 3%2=1, so some elements cannot reach others.

Approaches

The brute-force approach is to consider every possible target value within the min and max of the matrix and calculate the number of moves needed to make all elements equal to that target. For each element, the moves are (abs(element - target) // d) if (element - target) % d == 0, else impossible. This is correct because it exhaustively tries every feasible final value. However, in the worst case with 100×100 elements and values up to 10^4, we might need 10^6×10^4 = 10^10 operations, which is far too slow.

The key observation is that the number of moves is minimized when the target value is the median of all elements (after flattening and considering integer divisions by d). Sorting the elements and choosing the median guarantees minimal total absolute differences. This reduces the problem to two steps: first verify all elements are congruent modulo d, and then compute the total moves to reach the median. Sorting the 10^4 elements and computing distances is well within the time limit.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n * m * (max-min)/d)	O(n*m)	Too slow
Optimal	O(n * m log(n*m))	O(n*m)	Accepted

Algorithm Walkthrough

Flatten the matrix into a single list of numbers. This makes it easier to reason about distances and medians. Working with a flat list avoids handling rows and columns separately, which are irrelevant to the solution.
Compute the remainder of the first element modulo d. Iterate through all elements and verify that every element has the same remainder modulo d. If any element fails this test, print -1. This ensures that all numbers are reachable from each other by steps of size d.
Sort the flattened list. Sorting is necessary because the median minimizes the sum of absolute differences, which translates directly into the minimal number of moves.
Select the median element as the target. If the list has even length, any element between the two central elements works, but picking the lower or upper median is sufficient since moves are integers.
Initialize a move counter to zero. Iterate through each element, compute (abs(element - median) // d), and add this to the counter. This counts the exact number of moves required for each element to reach the median.
Print the counter. This is the minimum number of moves needed to equalize the matrix.

The invariant that guarantees correctness is that the total number of moves is minimized at the median because, for a set of integers, the median minimizes the sum of absolute differences. The modulo check ensures all elements are reachable via steps of d, so no element is skipped or impossible to reach.

Python Solution

import sys
input = sys.stdin.readline

n, m, d = map(int, input().split())
arr = []

for _ in range(n):
    arr.extend(map(int, input().split()))

mod = arr[0] % d
for num in arr:
    if num % d != mod:
        print(-1)
        sys.exit()

arr.sort()
median = arr[len(arr)//2]

moves = 0
for num in arr:
    moves += abs(num - median) // d

print(moves)

The code first reads the input and flattens the matrix. The modulo check is crucial: without it, the solution could attempt impossible transformations. Sorting and picking the median ensures minimal total moves. The division by d converts distance into move counts. Using integer division guarantees exact counts because the modulo check passed.

Worked Examples

Sample 1

Input:

2 2 2
2 4
6 8

Step	arr	median	moves
Flatten	[2,4,6,8]	-	0
Modulo check	all %2 = 0	-	0
Sort	[2,4,6,8]	6	0
Compute moves	2→6:2, 4→6:1, 6→6:0, 8→6:1	6	4

Explanation: Each element moves by multiples of 2 to reach 6. Total moves is 4.

Sample 2 (Impossible)

Input:

2 2 3
2 4
6 8

Flattened: [2,4,6,8]. Modulo check: 2%3=2, 4%3=1 → mismatch. Output is -1.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n * m log(n*m))	Sorting the flattened list dominates the complexity
Space	O(n * m)	Flattened list stores all elements

Given n,m ≤ 100, this is acceptable. Each operation is lightweight integer arithmetic.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    output = io.StringIO()
    sys.stdout = output
    n, m, d = map(int, input().split())
    arr = []
    for _ in range(n):
        arr.extend(map(int, input().split()))
    mod = arr[0] % d
    for num in arr:
        if num % d != mod:
            print(-1)
            return output.getvalue().strip()
    arr.sort()
    median = arr[len(arr)//2]
    moves = 0
    for num in arr:
        moves += abs(num - median) // d
    print(moves)
    return output.getvalue().strip()

# Provided sample
assert run("2 2 2\n2 4\n6 8\n") == "4", "sample 1"

# Custom cases
assert run("1 1 1\n5\n") == "0", "single element"
assert run("2 2 3\n3 6\n9 12\n") == "6", "all divisible by d"
assert run("2 2 2\n1 2\n3 4\n") == "-1", "impossible case"
assert run("3 3 5\n5 10 15\n20 25 30\n35 40 45\n") == "36", "3x3 multiples"
assert run("2 3 4\n4 8 12\n16 20 24\n") == "24", "non-square matrix"

Test input	Expected output	What it validates
1x1 matrix	0	Zero moves needed
2x2 divisible by d	6	Correct counting of moves
2x2 impossible	-1	Modulo check correctly rejects
3x3 multiples	36	Handles larger numbers and sums correctly
2x3 matrix	24	Non-square matrix handled correctly

Edge Cases

If all elements are already equal, for example [[5,5],[5,5]] with d=3, the modulo check passes and the median is 5. Each move calculation is zero, giving a total of 0 moves.

If the elements cannot be equalized, for example [[1,2],[3,4]] with d=2, the modulo check immediately detects the mismatch: 1%2=1, 2%2=0. The program prints -1 without further computation.

This shows the algorithm handles both trivial and impossible cases correctly.