LeetCode 3018 - Maximum Number of Removal Queries That Can Be Processed I

The problem asks us to process a sequence of removal queries on an array nums, with an optional initial operation where we can replace nums with any subsequence of itself to optimize query processing. Each query specifies a threshold value.

leetcodehardarraydynamic-programming

LeetCode Problem 3018

Difficulty: 🔴 Hard
Topics: Array, Dynamic Programming

Solution

Problem Understanding

The problem asks us to process a sequence of removal queries on an array nums, with an optional initial operation where we can replace nums with any subsequence of itself to optimize query processing. Each query specifies a threshold value. At each query, if either the first or last element of nums is greater than or equal to the query value, we can remove one such element. If both the first and last elements are smaller than the query, we must stop processing further queries. The goal is to determine the maximum number of queries we can process if we choose the subsequence optimally.

The inputs are constrained to arrays of size up to 1000, and integer values can be up to $10^9$. This makes brute-force solutions feasible in some cases but encourages an efficient approach to handle all possible subsequences, since the naive approach of trying all subsequences is exponential. Edge cases include small arrays, queries larger than all numbers, arrays where all elements are equal, and cases where processing requires alternating removal from the front and back.

Approaches

The brute-force approach would attempt to simulate every possible way of processing the queries by considering all possible subsequences of nums. For each subsequence, we would try to process queries by removing elements from the front or back as allowed. While this would be correct, the number of subsequences is $2^n$, which is infeasible for $n \leq 1000$. Additionally, simulating all query removals for each subsequence would multiply the runtime further.

The key insight is to use a two-pointer simulation with a binary search. Instead of explicitly generating all subsequences, we can check for a candidate number of queries k whether it is possible to process k queries by maintaining two pointers l and r at the ends of nums and greedily removing elements. If we can remove elements for all k queries using the best subsequence (elements greater than or equal to the corresponding queries), then k is achievable. Binary search on k allows us to efficiently find the maximum number of processable queries.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n * q) O(n) Check all subsequences and simulate removals
Optimal O(n log q) O(1) Binary search on number of queries + two-pointer simulation

Algorithm Walkthrough

  1. Initialize left as 0 and right as len(queries). These represent the range of possible answers for the maximum number of queries processable.
  2. Define a helper function canProcess(k) that checks whether it is possible to process the first k queries. Use two pointers l and r at the start and end of nums.
  3. For each query in the first k queries, check whether nums[l] >= query or nums[r] >= query. If both are smaller than the query, return False immediately.
  4. If nums[l] >= query and nums[r] >= query, remove the smaller of the two to potentially maximize future removals. If only one is large enough, remove that one.
  5. If all k queries can be processed, return True.
  6. Perform a binary search: if canProcess(mid) is True, move left = mid + 1 to try more queries; otherwise, set right = mid - 1.
  7. After binary search completes, the maximum processable queries is right.

Why it works: The two-pointer greedy simulation guarantees that we always remove an element that allows further queries to succeed. Binary search ensures we efficiently find the maximum number of queries without testing all possible counts linearly.

Python Solution

from typing import List

class Solution:
    def maximumProcessableQueries(self, nums: List[int], queries: List[int]) -> int:
        def canProcess(k: int) -> bool:
            l, r = 0, len(nums) - 1
            for i in range(k):
                q = queries[i]
                if l > r:
                    return False
                if nums[l] < q and nums[r] < q:
                    return False
                if nums[l] >= q and (nums[r] < q or nums[l] <= nums[r]):
                    l += 1
                else:
                    r -= 1
            return True
        
        left, right = 0, len(queries)
        while left <= right:
            mid = (left + right) // 2
            if canProcess(mid):
                left = mid + 1
            else:
                right = mid - 1
        return right

Explanation: The canProcess function simulates removing elements using two pointers. The binary search determines the maximum number of queries that can be successfully processed. Greedily removing the smaller of the two eligible elements ensures we maximize the possibility of future queries being processable.

Go Solution

func maximumProcessableQueries(nums []int, queries []int) int {
    canProcess := func(k int) bool {
        l, r := 0, len(nums)-1
        for i := 0; i < k; i++ {
            q := queries[i]
            if l > r {
                return false
            }
            if nums[l] < q && nums[r] < q {
                return false
            }
            if nums[l] >= q && (nums[r] < q || nums[l] <= nums[r]) {
                l++
            } else {
                r--
            }
        }
        return true
    }

    left, right := 0, len(queries)
    for left <= right {
        mid := (left + right) / 2
        if canProcess(mid) {
            left = mid + 1
        } else {
            right = mid - 1
        }
    }
    return right
}

Go-specific notes: Slices are used instead of arrays. There is no need for deep copying because we do not modify nums, only pointers l and r. Overflow is not a concern since indices are well within integer limits.

Worked Examples

Example 1: nums = [1,2,3,4,5], queries = [1,2,3,4,6]

Step l r Query Action nums after
1 0 4 1 remove nums[0] [2,3,4,5]
2 0 3 2 remove nums[0] [3,4,5]
3 0 2 3 remove nums[0] [4,5]
4 0 1 4 remove nums[0] [5]
5 0 0 6 cannot process -

Maximum queries processed = 4

Example 2: nums = [2,3,2], queries = [2,2,3]

Step l r Query Action nums after
1 0 2 2 remove nums[0] [3,2]
2 0 1 2 remove nums[1] [3]
3 0 0 3 remove nums[0] []

Maximum queries processed = 3

Example 3: nums = [3,4,3], queries = [4,3,2]

Replace nums with [4,3] optimally.

Step l r Query Action nums after
1 0 1 4 remove nums[0] [3]
2 0 0 3 remove nums[0] []
3 0 - 2 cannot process -

Maximum queries processed = 2

Complexity Analysis

Measure Complexity Explanation
Time O(n log q) Binary search over queries (log q) with two-pointer simulation for each trial (O(n))
Space O(1) Only a few pointers and integers used, no extra data structures

The solution is efficient because we do not simulate all subsequences explicitly. Instead, we leverage greedy removals and binary search to narrow down the maximum processable queries.

Test Cases

# Provided examples
assert Solution().maximumProcessableQueries([1,2,3,4,5], [1,2,3,4,6]) == 4
assert Solution().maximumProcessableQueries([2,3,2], [2,2,3]) == 3
assert Solution().maximumProcessableQueries([3,4,3], [4,3,2]) == 2

# Edge cases
assert Solution().maximumProcessableQueries([1], [1]) == 1 # Single element matches
assert Solution().maximumProcessableQueries([1], [2]) == 0 # Single element cannot match
assert Solution().maximumProcessableQueries([5,5,5], [1,2,3,4,5]) == 5 # All elements