HARDAtCoderLiveCodeBench

LCB-05: Defect

Baseline: PASS
Augmented: PASS
Injection: 2,356 chars

The Task

There is a grid with H rows and W columns. Let (i, j) denote the square at the i-th row from the top and j-th column from the left of the grid. Each square of the grid is holed or not. There are exactly N holed squares: (a_1, b_1), (a_2, b_2), \dots, (a_N, b_N). When the triple of positive integers (i, j, n) satisfies the following condition, the square region whose top-left corner is (i, j) and whose bottom-right corner is (i + n - 1, j + n - 1) is called a holeless square. - i + n - 1 \leq H. - j + n - 1 \leq W. - For every pair of non-negative integers (k, l) such that 0 \leq k \leq n - 1, 0 \leq l \leq n - 1, square (i + k, j + l) is not holed. How many holeless squares are in the grid? Input The input is given from Standard Input in the following format: H W N a_1 b_1 a_2 b_2 \vdots a_N b_N Output Print the number of holeless squares. Constraints - 1 \leq H, W \leq 3000 - 0 \leq N \leq \min(H \times W, 10^5) - 1 \leq a_i \leq H - 1 \leq b_i \leq W - All (a_i, b_i) are pairwise different. - All input values are integers. Sample Input 1 2 3 1 2 3 Sample Output 1 6 There are six holeless squares, listed below. For the first five, n = 1, and the top-left and bottom-right corners are the same square. - The square region whose top-left and bottom-right corners are (1, 1). - The square region whose top-left and bottom-right corners are (1, 2). - The square region whose top-left and bottom-right corners are (1, 3). - The square region whose top-left and bottom-right corners are (2, 1). - The square region whose top-left and bottom-right corners are (2, 2). - The square region whose top-left corner is (1, 1) and whose bottom-right corner is (2, 2). Sample Input 2 3 2 6 1 1 1 2 2 1 2 2 3 1 3 2 Sample Output 2 0 There may be no holeless square. Sample Input 3 1 1 0 Sample Output 3 1 The whole grid may be a holeless square. Sample Input 4 3000 3000 0 Sample Output 4 9004500500

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The Code

Claude Opus 4.6 with max-effort extended thinking. Exact outputs. Side by side.

Baseline

PASS1131ch / 515s
import sys

def main():
    inp = sys.stdin.buffer.read().split()
    ptr = 0
    H = int(inp[ptr]); ptr += 1
    W = int(inp[ptr]); ptr += 1
    N = int(inp[ptr]); ptr += 1

    S = W + 2
    hole = bytearray(S * (H + 2))
    row_has_hole = bytearray(H + 2)
    for _ in range(N):
        a = int(inp[ptr]); ptr += 1
        b = int(inp[ptr]); ptr += 1
        hole[a * S + b] = 1
        row_has_hole[a] = 1

    nxt = [0] * S
    ans = 0

    for i in range(H, 0, -1):
        cur = [0] * S
        if row_has_hole[i]:
            off = i * S
            for j in range(W, 0, -1):
                if not hole[off + j]:
                    a = cur[j + 1]
                    b = nxt[j]
                    c = nxt[j + 1]
                    if b < a: a = b
                    if c < a: a = c
                    cur[j] = a + 1
        else:
            for j in range(W, 0, -1):
                a = cur[j + 1]
                b = nxt[j]
                c = nxt[j + 1]
                if b < a: a = b
                if c < a: a = c
                cur[j] = a + 1
        ans += sum(cur)
        nxt = cur

    print(ans)

main()

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Augmented

PASS906ch / 157s
import sys

def main():
    data = sys.stdin.buffer.read().split()
    idx = 0
    H = int(data[idx]); idx += 1
    W = int(data[idx]); idx += 1
    N = int(data[idx]); idx += 1

    holed = bytearray(H * W)
    for _ in range(N):
        a = int(data[idx]) - 1; idx += 1
        b = int(data[idx]) - 1; idx += 1
        holed[a * W + b] = 1

    prev = [0] * (W + 1)
    curr = [0] * (W + 1)
    ans = 0

    for i in range(H - 1, -1, -1):
        base = i * W
        for j in range(W - 1, -1, -1):
            if holed[base + j]:
                curr[j] = 0
            else:
                v = prev[j]
                c = curr[j + 1]
                p = prev[j + 1]
                if c < v:
                    v = c
                if p < v:
                    v = p
                v += 1
                curr[j] = v
                ans += v
        prev, curr = curr, prev

    print(ans)

main()

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Blind Evaluation

Independent evaluator. Randomized A/B labels. No knowledge of which solution used the injection.

Augmented 42, Baseline 38

DimensionBaselineAugmented
Correctness1010
Efficiency87
Code Structure68
Readability58
Robustness99
Blind Analysis

Both solutions implement the classic "maximal square" DP to count all holeless squares. The recurrence is: dp[i][j] = min(dp[neighbor1], dp[neighbor2], dp[diagonal]) + 1 when the cell has no hole, else 0. The total answer is the sum of all DP values, since a cell with value k contributes exactly k holeless squares (sizes 1 through k) anchored at that corner. Both are algorithmically correct with no edge-case bugs.

Blind Observation

The baseline reveals a performance-oriented mindset: flat memory layout, branch-based loop specialization, awareness that most rows are hole-free when N is small. These are the instincts of someone who has been burned by Python TLE verdicts and knows where CPython's interpreter overhead lives. The augmented solution reveals a clarity-oriented mindset: do the standard thing, do it cleanly, trust that the algorithm's O(HW) complexity is sufficient.

Source: LiveCodeBench Hard benchmark. Full code outputs: baseline · augmented · runner script & methodology

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