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question:def can_form_continuous_sequence(n: int, sequence: List[int]) -> str: Determines if the sequence can be rearranged to form a continuous increasing sequence. :param n: int - number of elements in the sequence :param sequence: List[int] - the sequence of integers :return: str - "YES" if a continuous increasing sequence can be formed, "NO" otherwise >>> can_form_continuous_sequence(5, [2, 4, 6, 3, 5]) == "YES" >>> can_form_continuous_sequence(4, [10, 8, 6, 5]) == "NO" >>> can_form_continuous_sequence(1, [1000]) == "YES" >>> can_form_continuous_sequence(3, [2, 1, 3]) == "YES" >>> can_form_continuous_sequence(3, [1, 2, 4]) == "NO" >>> can_form_continuous_sequence(5, [5, 3, 1, 4, 2]) == "YES" >>> can_form_continuous_sequence(5, [100, 200, 300, 400, 500]) == "NO" >>> can_form_continuous_sequence(4, [1, 2, 4, 5]) == "NO"

answer:def can_form_continuous_sequence(n, sequence): Determines if the sequence can be rearranged to form a continuous increasing sequence. :param n: int - number of elements in the sequence :param sequence: List[int] - the sequence of integers :return: str - "YES" if a continuous increasing sequence can be formed, "NO" otherwise sequence = sorted(sequence) for i in range(1, n): if sequence[i] != sequence[i - 1] + 1: return "NO" return "YES"

question:def can_partition(nums: List[int]) -> bool: Determine if the array can be partitioned into two subsets with equal sum. >>> can_partition([1, 5, 11, 5]) True >>> can_partition([1, 2, 3, 5]) False >>> can_partition([1]) False >>> can_partition([2]) False >>> can_partition([2, 2]) True >>> can_partition([2, 3]) False >>> can_partition([3, 3, 3, 3]) True >>> can_partition([1, 2, 3, 4, 5, 6, 7, 8, 9, 10]) False

answer:def can_partition(nums): Determine if the array can be partitioned into two subsets with equal sum. total_sum = sum(nums) # If total sum is odd, we can't partition it into two equal subsets if total_sum % 2 != 0: return False target = total_sum // 2 n = len(nums) # Initialize a dp array where dp[i] indicates whether a subset sum 'i' is possible dp = [False] * (target + 1) dp[0] = True # A sum of 0 is always possible # Process each number in nums for num in nums: for j in range(target, num - 1, -1): dp[j] = dp[j] or dp[j - num] return dp[target]

question:def max_profit(prices): Returns the max profit possible from buying and selling a stock given the prices array. If no profit can be made, returns 0. >>> max_profit([7, 1, 5, 3, 6, 4]) 5 >>> max_profit([7, 6, 4, 3, 1]) 0 >>> max_profit([2, 4, 1]) 2

answer:def max_profit(prices): Returns the max profit possible from buying and selling a stock given the prices array. If no profit can be made, returns 0. if not prices: return 0 min_price = float('inf') max_profit = 0 for price in prices: if price < min_price: min_price = price elif price - min_price > max_profit: max_profit = price - min_price return max_profit

question:def saturation_volumes(volumes): Performs saturation operations on a list of water volumes in containers until all container volumes are zero. Prints the maximum volume subtracted in each step. :param volumes: List of integers representing water volumes in containers :return: List of maximum volumes subtracted in each saturation operation pass from solution import saturation_volumes def test_saturation_volumes_sample_input(): assert saturation_volumes([4, 3, 2, 6, 1]) == [6] def test_saturation_volumes_all_zeros(): assert saturation_volumes([0, 0, 0, 0, 0]) == [] def test_saturation_volumes_single_element(): assert saturation_volumes([10]) == [10] assert saturation_volumes([0]) == [] def test_saturation_volumes_all_same(): assert saturation_volumes([5, 5, 5, 5]) == [5] def test_saturation_volumes_descending_order(): assert saturation_volumes([8, 6, 5, 4, 3, 2, 1]) == [8] def test_saturation_volumes_ascending_order(): assert saturation_volumes([1, 2, 3, 4, 5, 6]) == [6]

answer:def saturation_volumes(volumes): Performs saturation operations on a list of water volumes in containers until all container volumes are zero. Prints the maximum volume subtracted in each step. :param volumes: List of integers representing water volumes in containers :return: List of maximum volumes subtracted in each saturation operation if not volumes: return [] subtracted_volumes = [] while any(volumes): max_volume = max(volumes) if max_volume == 0: break subtracted_volumes.append(max_volume) volumes = [max(0, v - max_volume) for v in volumes] return subtracted_volumes # Example Execution (can be removed/commented out in the final submission) # print(saturation_volumes([4, 3, 2, 6, 1])) # Output should be [6]