Maximum Number of Removal Queries That Can Be Processed I

Maximum Number of Removal Queries That Can Be Processed I - Problem

The Query Processing Challenge

You're given two arrays: nums (your data array) and queries (processing requests). Your goal is to maximize the number of queries you can successfully process.

🎯 The Rules:
• One-time preprocessing: You can replace nums with any subsequence of itself (keep elements in original order)
• Query processing: Process queries in order. For each query queries[i]:
- If both first and last elements of nums are less than queries[i], processing stops
- Otherwise, remove either the first or last element (whichever is ≥ queries[i])

Challenge: Choose the optimal subsequence and removal strategy to process the maximum number of queries!

Example: nums = [2, 3, 2], queries = [3, 2, 1]
→ Keep subsequence [3, 2]
→ Process query 3: remove 3, left with [2]
→ Process query 2: remove 2, left with []
→ Query 1: no elements left, but we processed 2 queries!

Input & Output

example_1.py — Basic Case

$ Input: {"nums": [2, 3, 2], "queries": [3, 2, 1]}

› Output: 2

💡 Note: Choose subsequence [3, 2]. Process query 3 by removing 3 (count=1), then process query 2 by removing 2 (count=2). Cannot process query 1 as no elements remain.

example_2.py — All Elements Work

$ Input: {"nums": [5, 4, 3], "queries": [3, 4, 1]}

› Output: 3

💡 Note: Keep the full array [5, 4, 3]. Process query 3: remove 3 from right. Process query 4: remove 4. Process query 1: remove 5. All queries processed successfully.

example_3.py — Strategic Subsequence

$ Input: {"nums": [1, 100, 1], "queries": [90, 90]}

› Output: 1

💡 Note: Choose subsequence [100]. Can process first query 90 by removing 100, but then no elements remain for the second query. Maximum achievable is 1.

Constraints

1 ≤ nums.length ≤ 20
1 ≤ queries.length ≤ 1000
1 ≤ nums[i], queries[i] ≤ 10⁶
The subsequence must maintain the original order of elements

Visualization

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Asked in

G Google 15 a Amazon 12 f Meta 8 ⊞ Microsoft 6

This problem requires finding the optimal subsequence that maximizes query processing. The key insight is that we only need to consider contiguous intervals from the original array, since we can only remove from the ends during query processing. The interval-based DP approach with O(n²m) complexity is much more efficient than the brute force O(2ⁿm) solution while still guaranteeing the optimal result.

Common Approaches

Approach	Time	Space	Notes
✓ Dynamic Programming with Interval Selection	O(n²m)	O(1)	Use DP to find optimal contiguous intervals that can handle maximum queries
Brute Force (Try All Subsequences)	O(2^n × m)	O(n)	Generate all possible subsequences and simulate query processing for each

Dynamic Programming with Interval Selection — Algorithm Steps

For each possible starting position i in nums
For each possible ending position j >= i
Extract subarray nums[i:j+1] as our subsequence
Simulate query processing on this interval using two pointers
Track the maximum number of queries processed across all intervals

Visualization

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Step-by-Step Walkthrough

Select Interval

Choose interval [i, j] from original array

Two Pointers

Use left/right pointers to simulate queries

Greedy Choice

For each query, pick the better end to remove

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int simulateQueriesOptimal(int* arr, int arrSize, int* queries, int querySize) {
    if (arrSize == 0) return 0;
    
    int count = 0;
    int left = 0, right = arrSize - 1;
    
    for (int i = 0; i < querySize; i++) {
        if (left > right) break;
        
        if (arr[left] >= queries[i]) {
            left++;
            count++;
        } else if (arr[right] >= queries[i]) {
            right--;
            count++;
        } else {
            break;
        }
    }
    
    return count;
}

int maximumQueries(int* nums, int numsSize, int* queries, int queriesSize) {
    int maxQueries = 0;
    
    // Try all possible contiguous intervals [i, j]
    for (int i = 0; i < numsSize; i++) {
        for (int j = i; j < numsSize; j++) {
            int processed = simulateQueriesOptimal(nums + i, j - i + 1, queries, queriesSize);
            if (processed > maxQueries) {
                maxQueries = processed;
            }
        }
    }
    
    return maxQueries;
}

int main() {
    // Input parsing and solution call
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²m)

O(n²) intervals to check, each taking O(m) time to process queries

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for pointers and counters

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Dynamic Programming with Interval Selection — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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