Naive Approach

A brute-force solution would involve iterating through all possible subsets of the queries array and, for each subset, checking if it's possible to make the nums array a zero array. This means checking all 2ⁿ subsets of the queries. This would be very computationally expensive.

Code (Python):

from typing import List


def solve():
    def can_convert_to_zero(nums: List[int], queries: List[List[int]]) -> bool:
        temp_nums = nums[:]
        for l, r in queries:
            for i in range(l, r + 1):
                temp_nums[i] = max(0, temp_nums[i] - 1)

        return all(x == 0 for x in temp_nums)


    def max_queries_removed(nums: List[int], queries: List[List[int]]) -> int:
        n = len(queries)
        max_removed = -1

        for i in range(1 << n):
            subset_queries = []
            for j in range(n):
                if (i >> j) & 1:
                    subset_queries.append(queries[j])

            remaining_queries = [queries[k] for k in range(n) if not ((i >> k) & 1)]

            if can_convert_to_zero(nums, remaining_queries):
                max_removed = max(max_removed, bin(i).count('1'))

        return max_removed


    nums = [2, 0, 2]
    queries = [[0, 2], [0, 2], [1, 1]]
    result = max_queries_removed(nums, queries)
    print(f"Max queries removed: {result}")

    nums = [1, 1, 1, 1]
    queries = [[1, 3], [0, 2], [1, 3], [1, 2]]
    result = max_queries_removed(nums, queries)
    print(f"Max queries removed: {result}")

    nums = [1, 2, 3, 4]
    queries = [[0, 3]]
    result = max_queries_removed(nums, queries)
    print(f"Max queries removed: {result}")




if __name__ == "__main__":
    solve()

Time Complexity: O(2^Q * N * Q), where Q is the number of queries and N is the length of nums. We iterate through all 2^Q subsets, and for each subset, we iterate through nums to check if it can be converted to a zero array, using the remaining queries.

Space Complexity: O(N), for creating copies of nums.

Optimal Approach: Greedy with Query Prioritization

The core idea is to process the numbers in nums from left to right. For each number, determine the minimum number of queries needed to make that number zero. Then, prioritize the queries that cover indices where the current nums[i] is positive. The key is to sort queries based on their ending index and greedily apply the queries that end earlier.

1. Sort Queries: Sort queries based on their ending index.
2. Iterate through nums: Process the array from left to right.
3. Track Active Queries: Maintain a data structure (e.g., an array) to track how much each query has been used.
4. Decrement nums[i]: For each element, decrement it by the amounts provided by the active queries. If nums[i] is still positive, use additional queries to decrement it to zero.

Algorithm Steps:

• Initialize an array used to keep track of how many times each query has been used.
• Sort the queries array based on the ending index (right endpoint) of each query.
• Iterate through the nums array:
  • For each nums[i], calculate the net decrement applied to it by the used queries.
  • If nums[i] is still positive, iterate through the sorted queries to find those that cover index i and have remaining uses.
  • Use the minimum between nums[i] and the available decrements of each applicable query.
  • Increment a counter for the number of queries that needed to be removed.

Code (Python):

from typing import List


def solve():
    def max_queries_removed(nums: List[int], queries: List[List[int]]) -> int:
        n = len(nums)
        q = len(queries)

        # Sort queries by their end index
        queries_with_indices = [(queries[i][0], queries[i][1], i) for i in range(q)]
        queries_with_indices.sort(key=lambda x: x[1])  # Sort by the end index

        used = [0] * q  # How many times each query is used
        removed_count = 0

        for i in range(n):
            decrement = 0
            for j in range(q):
                if used[j] > 0 and queries_with_indices[j][0] <= i <= queries_with_indices[j][1]:
                    decrement += used[j]

            nums[i] -= decrement

            if nums[i] > 0:
                applicable_queries = []
                for j in range(q):
                    if queries_with_indices[j][0] <= i <= queries_with_indices[j][1] and used[j] < 1:  # Query covers index and not used
                        applicable_queries.append(j)

                if not applicable_queries:
                    return -1  # Cannot convert to zero array

                for j in applicable_queries:
                    use = min(nums[i], 1 - used[j])  # Use the query as much as possible
                    used[j] += use
                    nums[i] -= use

                    if used[j] == 1:
                        removed_count += 1

                    if nums[i] == 0:
                        break
                if nums[i] > 0:
                    return -1

        return q - removed_count

    nums = [2, 0, 2]
    queries = [[0, 2], [0, 2], [1, 1]]
    result = max_queries_removed(nums, queries)
    print(f"Max queries removed: {result}")

    nums = [1, 1, 1, 1]
    queries = [[1, 3], [0, 2], [1, 3], [1, 2]]
    result = max_queries_removed(nums, queries)
    print(f"Max queries removed: {result}")

    nums = [1, 2, 3, 4]
    queries = [[0, 3]]
    result = max_queries_removed(nums, queries)
    print(f"Max queries removed: {result}")


if __name__ == "__main__":
    solve()

Time Complexity: O(Q log Q + N * Q), where N is the length of nums and Q is the number of queries. Sorting takes O(Q log Q), and the nested loop iterates O(N * Q) in the worst case.

Space Complexity: O(Q), for storing the used array and sorting.

Edge Cases

• Empty nums or queries: Handle cases where either the input array nums or the queries array is empty.
• Impossible Case: If it is impossible to convert the nums array to a zero array even using all the queries, return -1.
• Queries with Invalid Ranges: Ensure that the queries have valid ranges within the bounds of the nums array.