Zadanie · algorytmy średni 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.

Sygnatura

def merge_intervals(intervals: list[list[int]]) -> list[list[int]]:

Przykłady

merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])  # -> [[1, 6], [8, 10], [15, 18]]
merge_intervals([[1, 4], [4, 5]])  # -> [[1, 5]]

Oraz 7 ukrytych testów i 1 test wydajnościowy z limitem czasu; dane wejściowe nie są widoczne. Zadania są przekazywane modelom w języku angielskim.

#1OpenAI GPT-6.1 Terrarozwiązane · 10 z 10 testówodpowiedź w 2.2 s · czas wykonania 582 ms · $0.0009

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])oczekiwano [[1, 6], [8, 10], [15, 18]]	<1 ms
✓	merge_intervals([[1, 4], [4, 5]])oczekiwano [[1, 5]]	<1 ms

✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 576 ms

Kod

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 merged

#1SpaceXAI Grok Build 0.1rozwiązane · 10 z 10 testówodpowiedź w 5.0 s · czas wykonania 643 ms · $0.0024

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])oczekiwano [[1, 6], [8, 10], [15, 18]]	2 ms
✓	merge_intervals([[1, 4], [4, 5]])oczekiwano [[1, 5]]	1 ms

✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 635 ms

Kod

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 merged

#1Google Gemini 3.8 Flashrozwiązane · 10 z 10 testówodpowiedź w 3.4 s · czas wykonania 673 ms · $0.0007

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])oczekiwano [[1, 6], [8, 10], [15, 18]]	<1 ms
✓	merge_intervals([[1, 4], [4, 5]])oczekiwano [[1, 5]]	<1 ms

✓ <1 ms✓ 1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms✓ <1 ms⏱ 666 ms

Kod

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 merged

#1Qwen3 Coder Nextrozwiązane · 10 z 10 testówodpowiedź w 2.6 s · czas wykonania 687 ms · $0.0002

✓	merge_intervals([[1, 3], [2, 6], [8, 10], [15, 18]])oczekiwano [[1, 6], [8, 10], [15, 18]]	1 ms
✓	merge_intervals([[1, 4], [4, 5]])oczekiwano [[1, 5]]	1 ms

✓ <1 ms✓ <1 ms✓ <1 ms✓ 1 ms✓ 1 ms✓ 2 ms✓ 2 ms⏱ 676 ms

Kod

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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