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

✓ Brute Force (Try All Subsequences)

⏱️ Time: O(2^n × m) Space: O(n)

This approach generates every possible subsequence of nums, then simulates the entire query processing for each subsequence to find the maximum number of queries that can be processed.

Dynamic Programming with Interval Selection

⏱️ Time: O(n²m) Space: O(1)

This approach uses dynamic programming to efficiently find the best way to process queries by considering all possible intervals of the original array and simulating query processing optimally.

Brute Force (Try All Subsequences) — Algorithm Steps

Generate all possible subsequences of nums (2^n subsequences)
For each subsequence, simulate the query processing
For each query, check if we can remove first or last element
Track the maximum number of queries processed across all subsequences
Return the maximum count found

Visualization

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Step-by-Step Walkthrough

Generate Subsequence

Use bitmask to generate subsequence

Simulate Queries

Process queries using two pointers

Track Maximum

Keep track of best result found

Code -

solution.c — C

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

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

int maximumQueries(int* nums, int numsSize, int* queries, int queriesSize) {
    int maxQueries = 0;
    
    // Try all possible subsequences
    for (int mask = 1; mask < (1 << numsSize); mask++) {
        int subsequence[20];
        int subSize = 0;
        
        for (int i = 0; i < numsSize; i++) {
            if (mask & (1 << i)) {
                subsequence[subSize++] = nums[i];
            }
        }
        
        int processed = simulateQueries(subsequence, subSize, queries, queriesSize);
        if (processed > maxQueries) {
            maxQueries = processed;
        }
    }
    
    return maxQueries;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    
    while (*p && (*p == ' ' || *p == ',' || !(*p >= '0' && *p <= '9') && *p != '-')) {
        p++;
    }
    
    while (*p) {
        if (*p >= '0' && *p <= '9' || *p == '-') {
            arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
        } else {
            p++;
        }
        
        while (*p && *p != ',' && (*p < '0' || *p > '9') && *p != '-') {
            p++;
        }
        if (*p == ',') p++;
        while (*p == ' ') p++;
        
        if (*p == '\0' || *p == '\n') break;
    }
}

int main() {
    char line1[10000], line2[10000];
    fgets(line1, sizeof(line1), stdin);
    fgets(line2, sizeof(line2), stdin);
    
    // Remove newline characters
    line1[strcspn(line1, "\n")] = 0;
    line2[strcspn(line2, "\n")] = 0;
    
    int nums[20], queries[1000];
    int numsSize, queriesSize;
    
    parseArray(line1, nums, &numsSize);
    parseArray(line2, queries, &queriesSize);
    
    int result = maximumQueries(nums, numsSize, queries, queriesSize);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n × m)

2^n subsequences, each taking O(m) time to process m queries

⚠ Quadratic Growth

Space Complexity

O(n)

Space for storing current subsequence and recursion stack

✓ Linear Space

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Medium Frequency

~35 min Avg. Time

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

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Try All Subsequences) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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