LeetCode #3356 — MEDIUM

Zero Array Transformation II

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given an integer array nums of length n and a 2D array queries where queries[i] = [l_i, r_i, val_i].

Each queries[i] represents the following action on nums:

Decrement the value at each index in the range [l_i, r_i] in nums by at most val_i.
The amount by which each value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence, nums becomes a Zero Array. If no such k exists, return -1.

Example 1:

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

Output: 2

Explanation:

For i = 0 (l = 0, r = 2, val = 1):
- Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
- The array will become [1, 0, 1].
For i = 1 (l = 0, r = 2, val = 1):
- Decrement values at indices [0, 1, 2] by [1, 0, 1] respectively.
- The array will become [0, 0, 0], which is a Zero Array. Therefore, the minimum value of k is 2.

Example 2:

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

Output: -1

Explanation:

For i = 0 (l = 1, r = 3, val = 2):
- Decrement values at indices [1, 2, 3] by [2, 2, 1] respectively.
- The array will become [4, 1, 0, 0].
For i = 1 (l = 0, r = 2, val = 1):
- Decrement values at indices [0, 1, 2] by [1, 1, 0] respectively.
- The array will become [3, 0, 0, 0], which is not a Zero Array.

Constraints:

1 <= nums.length <= 10⁵
0 <= nums[i] <= 5 * 10⁵
1 <= queries.length <= 10⁵
queries[i].length == 3
0 <= l_i <= r_i < nums.length
1 <= val_i <= 5

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums of length n and a 2D array queries where queries[i] = [li, ri, vali]. Each queries[i] represents the following action on nums: Decrement the value at each index in the range [li, ri] in nums by at most vali. The amount by which each value is decremented can be chosen independently for each index. A Zero Array is an array with all its elements equal to 0. Return the minimum possible non-negative value of k, such that after processing the first k queries in sequence, nums becomes a Zero Array. If no such k exists, return -1.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[2,0,2]
[[0,2,1],[0,2,1],[1,1,3]]

Example 2

[4,3,2,1]
[[1,3,2],[0,2,1]]

Core Insight

What unlocks the optimal approach

Can we apply binary search here?
Utilize a difference array to optimize the processing of queries.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3356: Zero Array Transformation II
class Solution {
    private int n;
    private int[] nums;
    private int[][] queries;

    public int minZeroArray(int[] nums, int[][] queries) {
        this.nums = nums;
        this.queries = queries;
        n = nums.length;
        int m = queries.length;
        int l = 0, r = m + 1;
        while (l < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l > m ? -1 : l;
    }

    private boolean check(int k) {
        int[] d = new int[n + 1];
        for (int i = 0; i < k; ++i) {
            int l = queries[i][0], r = queries[i][1], val = queries[i][2];
            d[l] += val;
            d[r + 1] -= val;
        }
        for (int i = 0, s = 0; i < n; ++i) {
            s += d[i];
            if (nums[i] > s) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #3356: Zero Array Transformation II
func minZeroArray(nums []int, queries [][]int) int {
	n, m := len(nums), len(queries)
	l := sort.Search(m+1, func(k int) bool {
		d := make([]int, n+1)
		for _, q := range queries[:k] {
			l, r, val := q[0], q[1], q[2]
			d[l] += val
			d[r+1] -= val
		}
		s := 0
		for i, x := range nums {
			s += d[i]
			if x > s {
				return false
			}
		}
		return true
	})
	if l > m {
		return -1
	}
	return l
}

# Accepted solution for LeetCode #3356: Zero Array Transformation II
class Solution:
    def minZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
        def check(k: int) -> bool:
            d = [0] * (len(nums) + 1)
            for l, r, val in queries[:k]:
                d[l] += val
                d[r + 1] -= val
            s = 0
            for x, y in zip(nums, d):
                s += y
                if x > s:
                    return False
            return True

        m = len(queries)
        l = bisect_left(range(m + 1), True, key=check)
        return -1 if l > m else l

// Accepted solution for LeetCode #3356: Zero Array Transformation II
impl Solution {
    pub fn min_zero_array(nums: Vec<i32>, queries: Vec<Vec<i32>>) -> i32 {
        let n = nums.len();
        let m = queries.len();
        let mut d: Vec<i64> = vec![0; n + 1];
        let (mut l, mut r) = (0_usize, m + 1);

        let check = |k: usize, d: &mut Vec<i64>| -> bool {
            d.fill(0);
            for i in 0..k {
                let (l, r, val) = (
                    queries[i][0] as usize,
                    queries[i][1] as usize,
                    queries[i][2] as i64,
                );
                d[l] += val;
                d[r + 1] -= val;
            }
            let mut s: i64 = 0;
            for i in 0..n {
                s += d[i];
                if nums[i] as i64 > s {
                    return false;
                }
            }
            true
        };

        while l < r {
            let mid = (l + r) >> 1;
            if check(mid, &mut d) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        if l > m {
            -1
        } else {
            l as i32
        }
    }
}

// Accepted solution for LeetCode #3356: Zero Array Transformation II
function minZeroArray(nums: number[], queries: number[][]): number {
    const [n, m] = [nums.length, queries.length];
    const d: number[] = Array(n + 1);
    let [l, r] = [0, m + 1];
    const check = (k: number): boolean => {
        d.fill(0);
        for (let i = 0; i < k; ++i) {
            const [l, r, val] = queries[i];
            d[l] += val;
            d[r + 1] -= val;
        }
        for (let i = 0, s = 0; i < n; ++i) {
            s += d[i];
            if (nums[i] > s) {
                return false;
            }
        }
        return true;
    };
    while (l < r) {
        const mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid;
        } else {
            l = mid + 1;
        }
    }
    return l > m ? -1 : l;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + m)

Space

O(n)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.