Aufgabe · Algorithmen mittel Python ·
Merge intervals
Each interval [a, b] with a ≤ b is closed. Merge all intervals that overlap or touch ([1, 3] and [3, 5] merge into [1, 5]; [1, 2] and [3, 4] stay apart). Return the merged intervals sorted by start. The input is in no particular order and can hold up to 200,000 intervals.
Signatur
def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:Beispiele
merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]]) # -> [[1, 6], [8, 10], [15, 18]]
merge_intervals([[1, 4], [4, 5]]) # -> [[1, 5]]Zusätzlich 7 versteckte Tests und 1 Performance-Test mit großem Input unter Zeitlimit; deren Inputs werden nicht angezeigt. Die Aufgaben werden den Modellen auf Englisch gestellt.
Platz 1OpenAI GPT-6.1 Terragelöst · 10 von 10 TestsAntwort in 2.2 s · Laufzeit 582 ms · $0.0009 ✓ merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])erwartet [[1, 6], [8, 10], [15, 18]] <1 ms ✓ merge_intervals([[1, 4], [4, 5]])erwartet [[1, 5]] <1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 576 msCode
def merge_intervals(intervals: list[list[int]]) -> list[list[int]]: if not intervals: return [] ordered = sorted(intervals, key=lambda interval: interval[0]) start, end = ordered[0] merged = [] for a, b in ordered[1:]: if a <= end: if b > end: end = b else: merged.append([start, end]) start, end = a, b merged.append([start, end]) return mergedPlatz 1SpaceXAI Grok Build 0.1gelöst · 10 von 10 TestsAntwort in 5.0 s · Laufzeit 643 ms · $0.0024 ✓ merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])erwartet [[1, 6], [8, 10], [15, 18]] 2 ms ✓ merge_intervals([[1, 4], [4, 5]])erwartet [[1, 5]] 1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 635 msCode
def merge_intervals(intervals: list[list[int]]) -> list[list[int]]: if not intervals: return [] sorted_intervals = sorted(intervals, key=lambda x: x[0]) merged = [sorted_intervals[0][:]] for current in sorted_intervals[1:]: prev = merged[-1] if current[0] <= prev[1]: prev[1] = max(prev[1], current[1]) else: merged.append(current[:]) return mergedPlatz 1Google Gemini 3.8 Flashgelöst · 10 von 10 TestsAntwort in 3.4 s · Laufzeit 673 ms · $0.0007 ✓ merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])erwartet [[1, 6], [8, 10], [15, 18]] <1 ms ✓ merge_intervals([[1, 4], [4, 5]])erwartet [[1, 5]] <1 ms ✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 666 msCode
def merge_intervals(intervals: list[list[int]]) -> list[list[int]]: if not intervals: return [] intervals.sort(key=lambda x: x[0]) merged = [intervals[0]] for start, end in intervals[1:]: last_end = merged[-1][1] if start <= last_end: if end > last_end: merged[-1][1] = end else: merged.append([start, end]) return mergedPlatz 1Qwen3 Coder Nextgelöst · 10 von 10 TestsAntwort in 2.6 s · Laufzeit 687 ms · $0.0002 ✓ merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])erwartet [[1, 6], [8, 10], [15, 18]] 1 ms ✓ merge_intervals([[1, 4], [4, 5]])erwartet [[1, 5]] 1 ms ✓ <1 ms✓ <1 ms✓ <1 ms✓ 1 ms✓ 1 ms✓ 2 ms✓ 2 ms⏱ 676 msCode
def merge_intervals(intervals: list[list[int]]) -> list[list[int]]: if not intervals: return [] # Sort intervals by their start values intervals.sort(key=lambda x: x[0]) merged = [] current_start, current_end = intervals[0] for i in range(1, len(intervals)): start, end = intervals[i] # If the current interval overlaps or touches the previous one if start <= current_end: # Merge them by updating the end if needed current_end = max(current_end, end) else: # No overlap, add the previous merged interval to result merged.append([current_start, current_end]) current_start, current_end = start, end # Add the last interval merged.append([current_start, current_end]) return merged
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