Constraints

• 1 <= n <= 1000
• Each output row must use exactly one space between adjacent stars
• No trailing spaces are allowed

Examples

Solution

def solution(n):
    result = []
    for i in range(1, n + 1):
        result.append(' '.join(['*'] * i))
    return result

Time complexity: O(n^2). Space complexity: O(n^2).

Hints

1. Build each row independently, then join the row items with ' ' so you never create a trailing space.
2. The i-th row should contain exactly i copies of the same symbol.

Constraints

• 1 <= n <= 1000
• Numbers start at 1 and increase by 1 each time
• No trailing spaces are allowed

Examples

Solution

def solution(n):
    if n <= 0:
        return []

    result = []
    current = 1

    for row_length in range(1, n + 1):
        row = []
        for _ in range(row_length):
            row.append(str(current))
            current += 1
        result.append(" ".join(row))

    return result

Time complexity: O(n^2). Space complexity: O(n^2).

Hints

1. Keep a running number that starts at 1 and increases every time you place a value in the staircase.
2. Build each row as a list of strings, then use ' '.join(...) to format it cleanly.

Constraints

• 1 <= len(partitions) <= 10
• 0 <= total number of integers <= 200000
• Each integer fits in 32-bit signed range
• If value_range is provided, it should be a tuple (lo, hi) with lo <= hi

Examples

Solution

from collections import Counter\n\ndef solution(partitions, value_range=None):\n    def median_from_counts(counts):\n        total = sum(counts.values())\n        if total == 0:\n            return None\n\n        left_pos = (total + 1) // 2\n        right_pos = (total + 2) // 2\n        running = 0\n        left_val = None\n        right_val = None\n\n        for value in sorted(counts):\n            running += counts[value]\n            if left_val is None and running >= left_pos:\n                left_val = value\n            if right_val is None and running >= right_pos:\n                right_val = value\n                break\n\n        if left_val == right_val:\n            return left_val\n\n        avg = (left_val + right_val) / 2\n        if avg.is_integer():\n            return int(avg)\n        return avg\n\n    total_counts = Counter()\n    max_read_bytes = 0\n    raw_bytes = 0\n    sparse_bytes = 0\n\n    dense_valid = isinstance(value_range, tuple) and len(value_range) == 2 and value_range[0] <= value_range[1]\n    if dense_valid:\n        lo, hi = value_range\n        range_size = hi - lo + 1\n\n    for part in partitions:\n        local_counts = Counter(part)\n        total_counts.update(local_counts)\n\n        local_bytes = len(part) * 4\n        raw_bytes += local_bytes\n        max_read_bytes = max(max_read_bytes, local_bytes)\n        sparse_bytes += len(local_counts) * 8\n\n        if dense_valid:\n            for value in local_counts:\n                if value < lo or value > hi:\n                    dense_valid = False\n                    break\n\n    candidates = [\n        (raw_bytes, 2, 'raw'),\n        (sparse_bytes, 1, 'sparse_counts')\n    ]\n\n    if dense_valid:\n        dense_bytes = len(partitions) * range_size * 4\n        candidates.append((dense_bytes, 0, 'dense_histogram'))\n\n    sent_bytes, _, strategy = min(candidates, key=lambda item: (item[0], item[1]))\n\n    if not total_counts:\n        mode = None\n        median = None\n    else:\n        max_freq = max(total_counts.values())\n        mode = min(value for value, freq in total_counts.items() if freq == max_freq)\n        median = median_from_counts(total_counts)\n\n    read_seconds = round(max_read_bytes / 10.0, 1)\n    network_seconds = sent_bytes\n    total_seconds = round(read_seconds + network_seconds, 1)\n\n    return {\n        'mode': mode,\n        'median': median,\n        'strategy': strategy,\n        'sent_bytes': sent_bytes,\n        'read_seconds': read_seconds,\n        'network_seconds': network_seconds,\n        'total_seconds': total_seconds\n    }

Time complexity: O(N + U log U), where N is the total number of integers and U is the number of distinct values globally. Space complexity: O(U).

Hints

1. A global frequency map is enough to compute both the exact mode and the exact median, because the median only needs counts in sorted value order.
2. Compare total bytes for all valid strategies before choosing one; the cheapest communication plan is not always the same.