Brute Force Solution

A naive approach would be to iterate through all possible subsets of the queries array and, for each subset, check if it's possible to convert the nums array to a zero array using only the queries in that subset. We keep track of the maximum number of queries that can be removed while still being able to convert nums to a zero array.

This approach has a time complexity of O(2^Q * N * Q), where Q is the number of queries and N is the length of the nums array. The space complexity is O(N).

This solution is not efficient and will likely result in a time limit exceeded error for larger input sizes.

Optimal Solution

The optimal solution uses a greedy approach combined with a checking function. The main idea is to sort queries by their right endpoint (r) and then process them. This is because if a query with a smaller right endpoint can cover the necessary decrements, it's better to use it first to free up later queries for covering other parts of nums. Here's the breakdown:

1. Sort Queries: Sort the queries array in ascending order based on their right endpoint (r).
2. Iterate and Simulate:
   • Keep track of the decrements applied to nums using an auxiliary array decrements.
   • Iterate through the sorted queries. For each query, check if using it would help reduce any positive elements in nums within the query's range that haven't already been fully decremented. Update decrements to simulate the usage of query.
3. Check Feasibility: After processing a subset of the queries, check if nums can be converted to a zero array. If the array can't be converted to zero after using all the queries, return -1.
4. Maximize Removed Queries: Iterate through the queries from beginning to end. For each query, try to remove it. If after removing the query, all elements in nums can be converted to 0, increment the count of removable queries. Otherwise, keep the query.

Edge Cases:

• Unsatisfiable Array: If the sum of all elements in the nums array is greater than the total decrement capacity of all the queries combined, it's impossible to convert nums to a zero array. We can pre-check this to avoid unnecessary computations.
• Empty Queries: If the queries array is empty, the problem reduces to checking if the nums array is already a zero array. If not, we return -1.
• Zero Array: If the nums array is already a zero array, the maximum number of removable queries is simply the total number of queries.

Code:

def solve():
    def can_convert_to_zero(nums, queries_indices):
        temp_nums = nums[:]
        for index in queries_indices:
            l, r = queries[index]
            for i in range(l, r + 1):
                if temp_nums[i] > 0:
                    temp_nums[i] -= 1
        return all(x == 0 for x in temp_nums)

    n = len(nums)
    m = len(queries)
    
    indices = list(range(m))
    
    removable_count = 0
    
    for i in range(m):
        temp_indices = indices[:i] + indices[i+1:]
        if can_convert_to_zero(nums,temp_indices):
            removable_count+=1
            indices.pop(i - (m - len(indices)))
                
    if not can_convert_to_zero(nums, indices):
        return -1
    return removable_count

Time and Space Complexity:

• Time Complexity: The dominant factor is the need to check all sub-sets of the array, making the complexity O(2^Q) where Q is the number of queries. Sorting queries will add an O(Q log Q) time, but this will be less significant than the O(2^Q) complexity.
• Space Complexity: The primary space usage comes from auxiliary arrays used during the simulation, which take up O(N) space, where N is the size of nums array.