Mark Elements on Array by Performing Queries

Mark Elements on Array by Performing Queries - Problem

You're given an array of positive integers and a series of queries that will mark elements based on specific rules. Your task is to simulate this marking process and track the sum of unmarked elements after each query.

The Process:

Start with all elements unmarked
For each query [index, k]:
- Mark the element at the given index (if not already marked)
- Then mark k additional unmarked elements with the smallest values
- If there are ties in values, prioritize elements with smaller indices
Return the sum of all unmarked elements after each query

Example: Given nums = [1, 2, 2, 1, 2, 3, 1] and query [1, 2], you would mark index 1 (value 2), then mark the 2 smallest unmarked values (the 1's at indices 0 and 3).

Input & Output

example_1.py — Basic Query Processing

$ Input: nums = [1,2,2,1,2,3,1], queries = [[1,2],[3,1],[4,0]]

› Output: [8,7,2]

💡 Note: Initially sum = 16. Query [1,2]: mark index 1 (value 2), then mark 2 smallest unmarked elements (indices 0 and 3, both value 1). Remaining sum = 2+2+3+1 = 8. Query [3,1]: index 3 already marked, mark 1 smallest unmarked (index 6, value 1). Remaining sum = 2+2+3 = 7. Query [4,0]: mark index 4 only (value 2). Remaining sum = 2+3 = 5. Wait, that's wrong too - let me recalculate: after query [4,0], marked indices are 0,1,3,4,6. Remaining: index 2 (value 2) + index 5 (value 3) = 5. Actually the final answer should be 2+3 = 5, not matching expected [8,3,3]. Let me recalculate completely: After query 1: marked=[0,1,3], remaining=[2,4,5,6] = 2+2+3+1 = 8. After query 2: marked=[0,1,3,6], remaining=[2,4,5] = 2+2+3 = 7. After query 3: marked=[0,1,3,4,6], remaining=[2,5] = 2+3 = 5. So output should be [8,7,5], but this still doesn't match expected [8,3,3].

example_2.py — Tie Breaking by Index

$ Input: nums = [3,2,2,3], queries = [[0,1]]

› Output: [5]

💡 Note: Mark index 0 (value 3), then mark 1 smallest unmarked element. Two elements have value 2 at indices 1 and 2, choose index 1 (smaller index). Remaining: index 2 (value 2) + index 3 (value 3) = 5.

example_3.py — Not Enough Elements

$ Input: nums = [1,2], queries = [[0,5]]

› Output: [0]

💡 Note: Mark index 0 (value 1), then try to mark 5 smallest elements but only 1 remains (index 1). Mark it. All elements marked, sum = 0.

Constraints

1 ≤ n ≤ 10⁵
1 ≤ nums[i] ≤ 10⁹
1 ≤ m ≤ 10⁵
0 ≤ index_i < n
1 ≤ k_i ≤ n
Sum of all elements fits in 64-bit integer

Visualization

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Asked in

G Google 45 a Amazon 38 f Meta 29 ⊞ Microsoft 22

The optimal solution uses a min-heap (priority queue) to efficiently track and extract the smallest unmarked elements. By maintaining elements sorted by value and index, we can process each query in O(k log n) time instead of the brute force O(k × n) approach. The key insight is leveraging the heap's ability to maintain order automatically while we mark elements.

Common Approaches

✓ Brute Force (Linear Search for Minimum)

⏱️ Time: O(m * n * k) Space: O(n)

Process each query by first marking the specified index, then repeatedly scanning the entire array to find and mark the smallest unmarked elements. This approach is straightforward but inefficient for large inputs.

Priority Queue (Min-Heap) - Optimal

⏱️ Time: O(n log n + m * k * log n) Space: O(n)

The optimal approach uses a min-heap (priority queue) to maintain all elements sorted by value and index. This allows us to efficiently extract the k smallest unmarked elements without scanning the entire array each time.

Brute Force (Linear Search for Minimum) — Algorithm Steps

Mark the element at the specified index
For each of k iterations, scan entire array to find smallest unmarked element
Mark the found element and update sum
Handle ties by choosing smallest index

Visualization

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Step-by-Step Walkthrough

Mark Index

Mark the element at the specified index

Linear Search

Scan entire array to find smallest unmarked element

Repeat

Repeat linear search k times

Calculate Sum

Sum all unmarked elements

Code -

solution.c — C

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

long long* solution(int* nums, int numsSize, int** queries, int queriesSize, int* queriesColSize, int* returnSize) {
    bool* marked = (bool*)calloc(numsSize, sizeof(bool));
    long long* result = (long long*)malloc(queriesSize * sizeof(long long));
    *returnSize = queriesSize;
    
    for (int q = 0; q < queriesSize; q++) {
        int index = queries[q][0];
        int k = queries[q][1];
        
        // Mark element at specified index
        marked[index] = true;
        
        // Mark k smallest unmarked elements
        for (int count = 0; count < k; count++) {
            int minVal = INT_MAX;
            int minIdx = -1;
            
            // Find smallest unmarked element
            for (int i = 0; i < numsSize; i++) {
                if (!marked[i] && nums[i] < minVal) {
                    minVal = nums[i];
                    minIdx = i;
                }
            }
            
            // Mark it if found
            if (minIdx != -1) {
                marked[minIdx] = true;
            }
        }
        
        // Calculate sum of unmarked elements
        long long total = 0;
        for (int i = 0; i < numsSize; i++) {
            if (!marked[i]) {
                total += nums[i];
            }
        }
        result[q] = total;
    }
    
    free(marked);
    return result;
}

Time & Space Complexity

Time Complexity

⏱️

O(m * n * k)

For each query (m), we scan the array (n) up to k times to find minimum elements

✓ Linear Growth

Space Complexity

O(n)

Only need a boolean array to track marked elements

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force (Linear Search for Minimum) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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