LCB-06: Sleep Log
The Task
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The Code
Claude Opus 4.6 with max-effort extended thinking. Exact outputs. Side by side.
Baseline
PASS1354ch / 14simport sys
from bisect import bisect_right
def main():
input = sys.stdin.buffer.read().split()
idx = 0
N = int(input[idx]); idx += 1
A = [int(input[idx+i]) for i in range(N)]; idx += N
# prefix[i] = total sleep time from 0 to A[i]
# Sleep intervals are [A[1],A[2]), [A[3],A[4]), ...
# i.e. [A[2k], A[2k+1]) for k=1,...,(N-1)/2
# Even-indexed (0-based) A[1],A[3],... are bed times
# Odd-indexed (0-based) A[2],A[4],... are wake times
# Build prefix sum over A array entries
prefix = [0] * N
for i in range(1, N):
prefix[i] = prefix[i-1]
if i % 2 == 0: # i is even (0-based), so this is A_{i+1} (1-based odd) = wake up
prefix[i] += A[i] - A[i-1]
# Function to get total sleep from 0 to time t
def sleep_until(t):
# Find where t falls in A
pos = bisect_right(A, t) - 1 # largest index with A[pos] <= t
ans = prefix[pos]
# If pos is odd (0-based), we're in a sleep interval [A[pos], A[pos+1])
if pos % 2 == 1:
ans += t - A[pos]
return ans
Q = int(input[idx]); idx += 1
out = []
for _ in range(Q):
l = int(input[idx]); idx += 1
r = int(input[idx]); idx += 1
out.append(str(sleep_until(r) - sleep_until(l)))
sys.stdout.write('\n'.join(out))
main()Scroll to read full code
Augmented
PASS1171ch / 27simport sys
from bisect import bisect_right
def main():
input = sys.stdin.buffer.read().split()
idx = 0
N = int(input[idx]); idx += 1
A = [int(input[idx+i]) for i in range(N)]; idx += N
# Build prefix sum of sleep time at each boundary point
# Sleep intervals are [A[1],A[2]), [A[3],A[4]), ...
# prefix[i] = total sleep from time 0 to time A[i]
prefix = [0] * N
for i in range(1, N):
if i % 2 == 0: # waking up at A[i], was asleep from A[i-1] to A[i]
prefix[i] = prefix[i-1] + (A[i] - A[i-1])
else:
prefix[i] = prefix[i-1]
def sleep_at(t):
# Total sleep from 0 to t
j = bisect_right(A, t) - 1
# j is the largest index where A[j] <= t
res = prefix[j]
# If j is odd (meaning currently in a sleep interval [A[j], A[j+1]))
if j % 2 == 1:
res += t - A[j]
return res
Q = int(input[idx]); idx += 1
out = []
for _ in range(Q):
l = int(input[idx]); idx += 1
r = int(input[idx]); idx += 1
out.append(str(sleep_at(r) - sleep_at(l)))
sys.stdout.write('\n'.join(out))
main()Scroll to read full code
Blind Evaluation
Independent evaluator. Randomized A/B labels. No knowledge of which solution used the injection.
Augmented 44, Baseline 42
| Dimension | Baseline | Augmented |
|---|---|---|
| Correctness | 10 | 10 |
| Efficiency | 9 | 9 |
| Code Structure | 7 | 8 |
| Readability | 7 | 8 |
| Robustness | 9 | 9 |
Blind Analysis
The baseline builds a prefix sum array where prefix[i] accumulates total sleep from time 0 to boundary A[i]. The loop unconditionally copies prefix[i] = prefix[i-1], then conditionally adds the sleep interval width when i is even (0-indexed), corresponding to wake-up boundaries. The augmented solution is structurally identical but uses an if/else branch rather than unconditional-copy-then-conditional-add. Every algorithmic decision is the same.
Blind Observation
The most striking observation is how nearly identical these solutions are. The algorithm, variable names, I/O pattern, output assembly, and even comment structure all converge. This is a canonical approach to this problem — prefix sums over interval boundaries with binary search for arbitrary query points — and both solutions arrive at it with minimal deviation. The only signal that distinguishes them is the prefix loop's micro-stylistic preference.
Source: LiveCodeBench Hard benchmark. Full code outputs: baseline · augmented · runner script & methodology