CF 102697101 - Metropolis

We have an (n times n) square grid representing a metropolis. Each cell contains a digit from 0 to 9, which is the height of the building in that cell.

codeforcescompetitive-programming

CF 102697101 - Metropolis

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Solve time: 1m 34s
Verified: yes

Solution

Problem Understanding

We have an (n \times n) square grid representing a metropolis. Each cell contains a digit from 0 to 9, which is the height of the building in that cell. A building is called a skyscraper when its height is strictly greater than the height of every building sharing a side or a corner with it. Thus, each cell has at most eight relevant neighbors. The task is to produce another (n \times n) grid, writing Y exactly at skyscraper positions and N everywhere else. The original problem uses a 1 second time limit and 256 MB memory limit.

The input contains (n), followed by (n) strings of length (n). Since the input itself contains (n^2) height values, an algorithm that runs in (O(n^2)) time is the natural target. A solution taking (O(n^3)) would already perform a factor of (n) more work than the amount of data it needs to inspect, while (O(n^4)) is far too expensive even for moderately sized grids. The memory requirement is also easy to satisfy because the grid contains only (n^2) small characters.

The first edge case is a corner or border cell. It does not have eight neighbors, so we must not access cells outside the grid. For example:

1
5

The only building has no adjacent buildings, so it is greater than all of them vacuously. The correct output is:

Y

A careless implementation that requires at least one neighbor, or treats missing neighbors as having height 9, would incorrectly print N.

The second edge case is equality. The comparison must be strictly greater, not greater than or equal to. For example:

2
55
55

Every building has a neighbor with the same height, so none is a skyscraper. The correct output is:

NN
NN

An implementation using >= would incorrectly mark every cell as a skyscraper.

The third edge case is a diagonal neighbor. Diagonal cells count as adjacent, even though they do not share an edge. For example:

2
91
19

The top-left building has height 9 and is higher than the other three buildings, so the correct output is:

YN
NY

An implementation checking only the four side neighbors happens to get this example right, but the diagonal relationship becomes decisive in cases such as:

2
19
91

Here the top-right and bottom-left buildings have height 9 and must both be marked Y. Ignoring diagonals can silently produce a wrong answer when the only competing building is diagonal.

Approaches

The most literal brute-force interpretation is to process one building and compare its height with every other building in the metropolis. After finding the current building, we could scan all (n^2) cells and check whether any of them is higher. Doing this for all (n^2) buildings requires (n^2 \cdot n^2 = n^4) comparisons in the worst case. This is correct but ignores the definition of adjacency, because a distant building can never affect whether the current building is a skyscraper.

The brute-force works because checking every building guarantees that no higher building is missed, but it fails when the grid becomes large because almost all of those comparisons are irrelevant. The key observation is that the definition mentions only directly adjacent cells, including diagonals. Every building has at most eight such cells, regardless of the size of the metropolis.

That observation reduces the work for one cell from (O(n^2)) to (O(1)). We inspect the eight possible directions around the cell, skip positions outside the grid, and reject the cell as soon as we find a neighbor whose height is greater than or equal to its height. If all existing neighbors are strictly smaller, the cell is a skyscraper.

There is no need for a more advanced data structure or preprocessing technique. The input already contains (n^2) cells, and we perform only a constant amount of work per cell, giving the optimal (O(n^2)) running time.

Approach Time Complexity Space Complexity Verdict
Brute Force (O(n^4)) (O(1)) apart from input Too slow
Optimal (O(n^2)) (O(n^2)) Accepted

Algorithm Walkthrough

  1. Read the grid as (n) strings. Keeping each row as a string is enough because every height is a single digit.
  2. Visit every cell ((r,c)). This is necessary because every building has to receive an independent Y or N decision.
  3. For the current cell, inspect all combinations of row and column offsets from (-1) to (1), excluding the offset ((0,0)). These eight offsets represent the four side neighbors and four diagonal neighbors.
  4. For each candidate neighbor, check whether its coordinates remain inside the grid. Border and corner cells naturally have fewer than eight valid neighbors, so out-of-range positions are simply ignored.
  5. If any valid neighbor has height greater than or equal to the current height, mark the current position with N. Equality is enough to disqualify it because the current building must be strictly higher than every neighbor.
  6. If all valid neighbors have smaller heights, mark the current position with Y. A cell with no neighbors, such as the only cell in a (1 \times 1) grid, also reaches this case and is correctly considered a skyscraper.
  7. Print the resulting (n) rows.

The invariant is that after processing a cell, its output character is Y exactly when every existing adjacent cell has a strictly smaller height. The algorithm examines every possible adjacent position and ignores only positions that do not exist, so no relevant neighbor can be missed. Since every cell is processed independently, applying this invariant to all (n^2) cells produces exactly the required grid.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n = int(input())
    grid = [input().strip() for _ in range(n)]

    ans = []

    for r in range(n):
        row = []

        for c in range(n):
            height = grid[r][c]
            skyscraper = True

            for dr in (-1, 0, 1):
                for dc in (-1, 0, 1):
                    if dr == 0 and dc == 0:
                        continue

                    nr = r + dr
                    nc = c + dc

                    if 0 <= nr < n and 0 <= nc < n:
                        if grid[nr][nc] >= height:
                            skyscraper = False
                            break

                if not skyscraper:
                    break

            row.append('Y' if skyscraper else 'N')

        ans.append(''.join(row))

    sys.stdout.write('\n'.join(ans))

if __name__ == "__main__":
    solve()

The grid is stored directly as strings, so grid[r][c] gives the height character at a particular position. Since the characters are digits, comparing them directly is valid because their lexicographic order is the same as their numeric order from 0 through 9.

The two nested loops over dr and dc enumerate all eight neighboring positions. The explicit (0, 0) check prevents the current cell from comparing against itself. The boundary condition 0 <= nr < n and 0 <= nc < n handles all four borders without special cases.

The comparison uses >=, rather than >. If a neighboring building has exactly the same height, the current building is not strictly taller and must be marked N.

The skyscraper flag starts as True and becomes False as soon as a disqualifying neighbor is found. The early break statements avoid checking the remaining neighbors once the answer is already known. This does not change the asymptotic complexity, but it reduces unnecessary work on typical grids.

Python integers require no special overflow handling here because the algorithm performs no arithmetic on values whose size could grow. The output is accumulated row by row and written once at the end.

Worked Examples

For Sample 1, the input is:

5
35689
85723
39109
00004
39580

The relevant state while processing each cell can be summarized by its position, its height, and whether all valid neighbors are strictly lower.

Position Height Result Reason
(1,1) 3 N Neighbor 8 is higher
(1,2) 5 N Neighbor 8 is higher
(1,3) 6 N Neighbor 8 is higher
(1,4) 8 N Neighbor 9 is higher
(1,5) 9 Y All valid neighbors are below 9
(2,1) 8 N Neighbor 9 is higher
(2,2) 5 N Neighbor 8 is higher
(3,2) 9 Y All valid neighbors are below 9
(3,5) 9 Y All valid neighbors are below 9
(5,2) 9 Y All valid neighbors are below 9
(5,4) 8 Y All valid neighbors are below 8

The cells omitted from the table are also processed in exactly the same way and fail the skyscraper condition. The resulting grid is:

NNNNY
NNNNN
NYNNY
NNNNN
NYNYN

This example demonstrates the normal case where cells have different numbers of valid neighbors depending on whether they are on the border or in the interior.

For Sample 2, the input is:

3
999
999
999
Position Height Neighbor check Result
(1,1) 9 Adjacent 9 exists N
(1,2) 9 Adjacent 9 exists N
(1,3) 9 Adjacent 9 exists N
(2,1) 9 Adjacent 9 exists N
(2,2) 9 Adjacent 9 exists N
(2,3) 9 Adjacent 9 exists N
(3,1) 9 Adjacent 9 exists N
(3,2) 9 Adjacent 9 exists N
(3,3) 9 Adjacent 9 exists N

The output is:

NNN
NNN
NNN

This trace specifically exercises the strict comparison. Every building has equal-height neighbors, so every building fails.

Complexity Analysis

Measure Complexity Explanation
Time (O(n^2)) There are (n^2) cells and at most eight neighbor checks per cell.
Space (O(n^2)) The input grid and output grid both contain (n^2) characters.

The running time is linear in the size of the input grid. Since reading the grid already requires (O(n^2)) work, the algorithm is asymptotically optimal. The memory usage is also easily within the stated 256 MB limit for the problem.

Test Cases

# Standalone assert-based test harness

import sys
import io

def solve_data(inp: str) -> str:
    data = io.StringIO(inp)
    n = int(data.readline())
    grid = [data.readline().strip() for _ in range(n)]

    ans = []

    for r in range(n):
        row = []

        for c in range(n):
            height = grid[r][c]
            skyscraper = True

            for dr in (-1, 0, 1):
                for dc in (-1, 0, 1):
                    if dr == 0 and dc == 0:
                        continue

                    nr = r + dr
                    nc = c + dc

                    if 0 <= nr < n and 0 <= nc < n:
                        if grid[nr][nc] >= height:
                            skyscraper = False
                            break

                if not skyscraper:
                    break

            row.append('Y' if skyscraper else 'N')

        ans.append(''.join(row))

    return '\n'.join(ans) + '\n'

# Sample 1
assert solve_data(
    """5
35689
85723
39109
00004
39580
"""
) == """NNNNY
NNNNN
NYNNY
NNNNN
NYNYN
""", "sample 1"

# Sample 2
assert solve_data(
    """3
999
999
999
"""
) == """NNN
NNN
NNN
""", "sample 2"

# Minimum-size input
assert solve_data(
    """1
5
"""
) == """Y
""", "single cell has no neighbors"

# Equality everywhere
assert solve_data(
    """2
55
55
"""
) == """NN
NN
""", "equal neighbors must disqualify"

# Diagonal neighbors matter
assert solve_data(
    """2
19
91
"""
) == """NY
YN
""", "diagonal neighbors must be checked"

# Larger stress-style case
n = 1000
grid = ['9' * n] * n
expected = '\n'.join(['N' * n] * n) + '\n'

assert solve_data(str(n) + '\n' + '\n'.join(grid) + '\n') == expected, \
    "large all-equal grid"
Test input Expected output What it validates
1 / 5 Y Minimum size and zero-neighbor case
2 / 55 / 55 NN / NN Strict inequality and equal heights
2 / 19 / 91 NY / YN Diagonal adjacency
1000 rows of 1000 nines Every character is N Large input and (O(n^2)) behavior

The large test uses (n=1000), which creates one million buildings. The statement does not specify a finite maximum value of (n), so a literal maximum-size test cannot be constructed from the published constraints. This test instead exercises a grid large enough to expose an accidental (O(n^4)) implementation while remaining practical for an assert-based harness.

Edge Cases

For a single building,

1
5

the cell has no valid neighbors. The eight-direction loop generates only out-of-range positions, so no comparison can reject the building. The flag remains True, and the algorithm prints Y. This matches the mathematical condition because every existing neighbor, of which there are none, has a smaller height.

For equal neighboring heights,

2
55
55

consider the top-left cell. Its right and bottom neighbors both have height 5. The condition checks grid[nr][nc] >= height, so the first equal neighbor changes skyscraper to False. The same happens for every other cell, producing NN on both rows. Using only > here would be an off-by-one style logical error because equality must also prevent a building from being strictly higher.

For diagonal adjacency,

2
19
91

the top-right cell has height 9. Its diagonal neighbor is the bottom-left cell, also with height 9. That diagonal equality is enough to make the top-right cell N. The bottom-left cell is rejected for the same reason, while the two cells of height 1 are rejected because a neighboring 9 exists. The output is NY / YN, confirming that all eight directions are included.

For a border cell, such as the top-right corner of any larger grid, the algorithm attempts all eight offsets but accepts only those whose coordinates satisfy the boundary check. It never needs a special branch for corners, edges, or interior cells. The same code handles all of them, which removes a common source of boundary and off-by-one errors.