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Maximum Number of Removal Queries That Can Be Processed I

Number: 3323

Difficulty: Hard

Paid? Yes

Companies: N/A

Problem Description

Given a 0-indexed array nums and a 0-indexed array queries, you are allowed to perform the following operation at the beginning at most once: replace nums with any subsequence of itself. Then, you process the queries one by one. For each query: • If both the first and last element of nums are less than the query value, you must stop processing further queries. • Otherwise, you choose either the first or last element (whichever is greater than or equal to the query) and remove it from nums. Return the maximum number of queries that can be processed by choosing the initial subsequence optimally and making optimal removal choices.

Key Insights

  • You are allowed a one-time operation to choose any subsequence of nums; this can be used to “prepare” nums so that more queries can be processed.
  • Once the subsequence is fixed, removal operations are allowed only at the endpoints.
  • The removal process can be viewed as splitting the k processed queries into two parts: • The first i queries are handled by removing from the front (using elements in increasing order). • The remaining k − i queries are handled by removing from the back (using elements in reverse order).
  • For a chosen split, a valid strategy exists only if the last element used from the front strictly occurs before the first element used from the back in the original array.
  • The problem becomes: for each candidate k (number of queries processed) and every split (i from 0 up to k), using a greedy pointer strategy can we find a pair of non‐overlapping matches in nums that can serve the front and back removal parts? This check guides a binary (or linear) search over k.

Space and Time Complexity

Time Complexity: O(m² * n) in the worst-case (where m = number of queries and n = length of nums) when using a naïve greedy check for each candidate k and every possible split. With precomputation/optimization, the repeated scans can be reduced. Space Complexity: O(1) or O(m²) if additional tables are used to store precomputed results.

Solution

We use a greedy matching strategy with a “split” approach. For a candidate number of queries processed (k), we try every possible split (i from 0 to k) where: • The first i queries are matched from the beginning of nums. • The remaining k − i queries are matched from the end (in reverse order) of nums. For the front part, starting with an initial pointer, we scan nums left-to-right to find, in order, an element ≥ each query in queries[0…i-1]. For the back part, we scan nums right-to-left to match queries[k-1 … i] in reverse order. If for any split the last index chosen in the front is strictly less than the first index chosen in the back, then processing k queries is feasible. We then try increasing k (e.g. via a simple linear scan of possible k values, or binary search) until the condition fails. The maximum feasible k is the answer. Key data structures: simple pointer indices in the array. The greedy approach mimics the removal process while checking the feasibility of a split.

Code Solutions

# Python solution with inline comments.
def maximumQueriesProcessed(nums, queries):
    n = len(nums)
    m = len(queries)
    
    # Helper function: check if it is possible to process k queries.
    def can_process(k):
        # Try each possible split:
        # i queries will be processed from the front, (k - i) from the back.
        for i in range(k + 1):
            # Greedily match first i queries from left of nums.
            front_idx = -1  # last matched index for front removal
            valid_front = True
            for q in range(i):
                found = False
                j = front_idx + 1
                while j < n:
                    if nums[j] >= queries[q]:
                        front_idx = j
                        found = True
                        break
                    j += 1
                if not found:
                    valid_front = False
                    break
            if not valid_front:
                continue
                
            # Greedily match (k - i) queries from the back (in reverse order).
            back_idx = n  # starting pointer just beyond the last index
            valid_back = True
            for q in range(k - 1, i - 1, -1):  # process queries from index k-1 downto i
                found = False
                j = back_idx - 1
                while j >= 0:
                    if nums[j] >= queries[q]:
                        back_idx = j
                        found = True
                        break
                    j -= 1
                if not found:
                    valid_back = False
                    break
            if not valid_back:
                continue
            
            # Ensure that the chosen front segment and back segment do not overlap.
            if front_idx < back_idx:
                return True
        return False

    max_k = 0
    # Iterate over possible values of k from 0 to m.
    for k in range(m + 1):
        if can_process(k):
            max_k = k
        else:
            break
    return max_k

# Example Usage:
nums1 = [1,2,3,4,5]
queries1 = [1,2,3,4,6]
print(maximumQueriesProcessed(nums1, queries1))  # Expected output: 4
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