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Zero Array Transformation IV - Problem

🎯 Zero Array Transformation IV

You're given an integer array nums and a collection of transformation queries. Each query allows you to select any subset of indices within a specified range and decrement their values by a given amount.

Your mission: Find the minimum number of queries needed to transform the entire array into zeros! 🚀

Input:

nums: An array of positive integers
queries: Each query [li, ri, vali] lets you:

Pick any indices between li and ri (inclusive)
Subtract exactly vali from each selected index

Output: Return the minimum k such that after applying the first k queries in order, all elements become 0. If impossible, return -1.

Example: With nums = [2, 1, 3] and queries that can subtract 1 from different ranges, you need to strategically apply queries to zero out all elements efficiently!

Input & Output

basic_case.py — Python

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

› Output: 2

💡 Note: After applying first 2 queries: Query 0 reduces range [0,2] by 1 each → [1,0,2]. Query 1 reduces range [1,2] by 1 each → [1,0,1]. We can make nums[0] and nums[2] zero with remaining reductions, so k=2 works.

impossible_case.py — Python

$ Input: nums = [5, 3, 4], queries = [[0, 1, 1], [2, 2, 2]]

› Output: -1

💡 Note: Even with all queries, we can only reduce nums[0] and nums[1] by total of 1 each, and nums[2] by 2. This gives us [4,2,2] at best, which is not all zeros.

single_query.py — Python

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

› Output: 1

💡 Note: Single query can reduce entire range [0,2] by 1, making all elements zero in one step.

Constraints

1 ≤ nums.length ≤ 10⁵
0 ≤ nums[i] ≤ 10⁵
1 ≤ queries.length ≤ 10⁵
queries[i].length = 3
0 ≤ l_i ≤ r_i < nums.length
1 ≤ val_i ≤ 10⁵

Visualization

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Understanding the Visualization

Identify Search Space

We need between 0 and total_queries payments

Binary Search Logic

If k payments work, k+1 will too (monotonic property)

Greedy Simulation

For each candidate k, optimally apply first k queries

Range Reduction

Each query reduces values in its range by the specified amount

Key Takeaway

🎯 Key Insight: The monotonic property (if k works, k+1 works too) enables binary search, while greedy query application ensures optimal reduction at each step.

Asked in

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

The optimal approach uses binary search on the answer combined with greedy simulation. Since the problem has a monotonic property (if k queries work, k+1 also work), we can binary search for the minimum k. For each candidate k, we greedily simulate applying the first k queries to maximize reduction at each position.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force Simulation	O(q² × n²)	O(n)	Try k = 1, 2, 3... queries until array becomes zero
Binary Search + Greedy Simulation	O(log q × q × n)	O(n)	Binary search on answer with optimal query application

Brute Force Simulation — Algorithm Steps

For k from 1 to total queries
Simulate applying first k queries optimally
For each query, greedily select indices that need the most reduction
Check if all elements can reach zero
Return the first k that works

Visualization

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Step-by-Step Walkthrough

Try k=1

Apply first query and check if array becomes zero

Try k=2

Apply first two queries and check

Continue

Keep increasing k until success or exhaustion

Code -

solution.c — C

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

int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }

int canZero(int* nums, int n, int queries[][3], int q, int k) {
    int* arr = (int*)malloc(n * sizeof(int));
    memcpy(arr, nums, n * sizeof(int));
    
    for (int i = 0; i < k; i++) {
        int l = queries[i][0], r = queries[i][1], val = queries[i][2];
        for (int j = l; j <= min(r, n - 1); j++) {
            if (arr[j] > 0) {
                arr[j] = max(0, arr[j] - val);
            }
        }
    }
    
    for (int i = 0; i < n; i++) {
        if (arr[i] > 0) {
            free(arr);
            return 0;
        }
    }
    
    free(arr);
    return 1;
}

int minZeroArray(int* nums, int n, int queries[][3], int q) {
    for (int k = 1; k <= q; k++) {
        if (canZero(nums, n, queries, q, k)) {
            return k;
        }
    }
    return -1;
}

int main() {
    // Parse input and call solution
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(q² × n²)

q iterations, each simulating q queries on n elements with complex selection

⚠ Quadratic Growth

Space Complexity

O(n)

Copy of original array for simulation

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

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Smart Actions

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🎯 Zero Array Transformation IV

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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