Zero Array Transformation IV - Practice Coding Problems

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.

Visualization

Tap to expand

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.

Time & Space Complexity

Time Complexity

⏱️

O(log q × q × n)

log q binary search iterations, each simulating q queries on n elements

⚡ Linearithmic

Space Complexity

O(n)

Temporary arrays for simulation and difference array

⚡ Linearithmic Space

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

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

Binary Search + Greedy Simulation — Algorithm Steps

Binary search on k from 1 to total queries
For each mid value, check if first mid queries can zero array
Use greedy strategy: apply each query optimally
For each query, prioritize positions with highest remaining values
Use difference array for efficient range updates

Visualization

Tap to expand

Step-by-Step Walkthrough

Set Search Range

left=0, right=total_queries

Test Middle

mid = (left+right)/2, check if mid queries work

Narrow Range

If works: search left half, else: search right half

Find Minimum

Continue until finding the minimum valid k

Code -

solution.c — C

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

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

int canZeroWithK(int* nums, int n, int queries[][3], int q, int k) {
    int* remaining = (int*)malloc(n * sizeof(int));
    memcpy(remaining, 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 (remaining[j] > 0) {
                int reduction = min(val, remaining[j]);
                remaining[j] -= reduction;
            }
        }
    }
    
    for (int i = 0; i < n; i++) {
        if (remaining[i] > 0) {
            free(remaining);
            return 0;
        }
    }
    
    free(remaining);
    return 1;
}

int minZeroArray(int* nums, int n, int queries[][3], int q) {
    int left = 0, right = q, result = -1;
    
    while (left <= right) {
        int mid = (left + right) / 2;
        if (canZeroWithK(nums, n, queries, q, mid)) {
            result = mid;
            right = mid - 1;
        } else {
            left = mid + 1;
        }
    }
    
    return result;
}

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

Time & Space Complexity

Time Complexity

⏱️

O(log q × q × n)

log q binary search iterations, each simulating q queries on n elements

⚡ Linearithmic

Space Complexity

O(n)

Temporary arrays for simulation and difference array

⚡ Linearithmic Space

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

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🎯 Zero Array Transformation IV

Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Binary Search + Greedy Simulation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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