Zero Array Transformation IV
MediumUpdated: Aug 2, 2025
Practice on:
Problem
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:
- Select a subset of indices in the range
[li, ri]fromnums. - Decrement the value at each selected index by exactly
vali.
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.
Examples
Example 1
Input: nums = [2,0,2], queries = [[0,2,1],[0,2,1],[1,1,3]]
Output: 2
Explanation:
* **For query 0 (l = 0, r = 2, val = 1):**
* Decrement the values at indices `[0, 2]` by 1.
* The array will become `[1, 0, 1]`.
* **For query 1 (l = 0, r = 2, val = 1):**
* Decrement the values at indices `[0, 2]` by 1.
* 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:
It is impossible to make nums a Zero Array even after all the queries.
Example 3
Input: nums = [1,2,3,2,1], queries =
[[0,1,1],[1,2,1],[2,3,2],[3,4,1],[4,4,1]]
Output: 4
Explanation:
* **For query 0 (l = 0, r = 1, val = 1):**
* Decrement the values at indices `[0, 1]` by `1`.
* The array will become `[0, 1, 3, 2, 1]`.
* **For query 1 (l = 1, r = 2, val = 1):**
* Decrement the values at indices `[1, 2]` by 1.
* The array will become `[0, 0, 2, 2, 1]`.
* **For query 2 (l = 2, r = 3, val = 2):**
* Decrement the values at indices `[2, 3]` by 2.
* The array will become `[0, 0, 0, 0, 1]`.
* **For query 3 (l = 3, r = 4, val = 1):**
* Decrement the value at index 4 by 1.
* The array will become `[0, 0, 0, 0, 0]`. Therefore, the minimum value of `k` is 4.
Example 4
Input: nums = [1,2,3,2,6], queries =
[[0,1,1],[0,2,1],[1,4,2],[4,4,4],[3,4,1],[4,4,5]]
Output: 4
Constraints
1 <= nums.length <= 100 <= nums[i] <= 10001 <= queries.length <= 1000queries[i] = [li, ri, vali]0 <= li <= ri < nums.length1 <= vali <= 10
Solution
Method 1 – Brute Force Simulation
Intuition
Since the array is small (n ≤ 10), we can simulate the process for each prefix of queries. For each k, apply the first k queries in order, always greedily decrementing as much as possible in the allowed range, and check if the array becomes all zeros.
Approach
- For k from 1 to len(queries):
- Copy the original nums array.
- For each query up to k:
- For each index in the range [li, ri]:
- If nums[idx] >= vali, decrement nums[idx] by vali.
- For each index in the range [li, ri]:
- After all k queries, check if all elements are zero.
- If so, return k.
- If no such k exists, return -1.
Code
C++
class Solution {
public:
int minQueriesToZeroArray(vector<int>& nums, vector<vector<int>>& queries) {
int n = nums.size(), q = queries.size();
for (int k = 1; k <= q; ++k) {
vector<int> arr = nums;
for (int i = 0; i < k; ++i) {
int l = queries[i][0], r = queries[i][1], v = queries[i][2];
for (int j = l; j <= r; ++j) {
if (arr[j] >= v) arr[j] -= v;
}
}
bool allZero = true;
for (int x : arr) if (x != 0) allZero = false;
if (allZero) return k;
}
return -1;
}
};
Go
func minQueriesToZeroArray(nums []int, queries [][]int) int {
n, q := len(nums), len(queries)
for k := 1; k <= q; k++ {
arr := make([]int, n)
copy(arr, nums)
for i := 0; i < k; i++ {
l, r, v := queries[i][0], queries[i][1], queries[i][2]
for j := l; j <= r; j++ {
if arr[j] >= v {
arr[j] -= v
}
}
}
allZero := true
for _, x := range arr {
if x != 0 {
allZero = false
break
}
}
if allZero {
return k
}
}
return -1
}
Java
class Solution {
public int minQueriesToZeroArray(int[] nums, int[][] queries) {
int n = nums.length, q = queries.length;
for (int k = 1; k <= q; k++) {
int[] arr = nums.clone();
for (int i = 0; i < k; i++) {
int l = queries[i][0], r = queries[i][1], v = queries[i][2];
for (int j = l; j <= r; j++) {
if (arr[j] >= v) arr[j] -= v;
}
}
boolean allZero = true;
for (int x : arr) if (x != 0) allZero = false;
if (allZero) return k;
}
return -1;
}
}
Kotlin
class Solution {
fun minQueriesToZeroArray(nums: IntArray, queries: Array<IntArray>): Int {
val n = nums.size
val q = queries.size
for (k in 1..q) {
val arr = nums.copyOf()
for (i in 0 until k) {
val (l, r, v) = queries[i]
for (j in l..r) {
if (arr[j] >= v) arr[j] -= v
}
}
if (arr.all { it == 0 }) return k
}
return -1
}
}
Python
from typing import List
class Solution:
def minQueriesToZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
n, q = len(nums), len(queries)
for k in range(1, q+1):
arr = nums[:]
for i in range(k):
l, r, v = queries[i]
for j in range(l, r+1):
if arr[j] >= v:
arr[j] -= v
if all(x == 0 for x in arr):
return k
return -1
Rust
impl Solution {
pub fn min_queries_to_zero_array(nums: Vec<i32>, queries: Vec<Vec<i32>>) -> i32 {
let n = nums.len();
let q = queries.len();
for k in 1..=q {
let mut arr = nums.clone();
for i in 0..k {
let l = queries[i][0] as usize;
let r = queries[i][1] as usize;
let v = queries[i][2];
for j in l..=r {
if arr[j] >= v {
arr[j] -= v;
}
}
}
if arr.iter().all(|&x| x == 0) {
return k as i32;
}
}
-1
}
}
TypeScript
class Solution {
minQueriesToZeroArray(nums: number[], queries: number[][]): number {
const n = nums.length, q = queries.length;
for (let k = 1; k <= q; k++) {
const arr = nums.slice();
for (let i = 0; i < k; i++) {
const [l, r, v] = queries[i];
for (let j = l; j <= r; j++) {
if (arr[j] >= v) arr[j] -= v;
}
}
if (arr.every(x => x === 0)) return k;
}
return -1;
}
}
Complexity
- ⏰ Time complexity:
O(q * n^2)— For each prefix of queries (up to q), we may update up to n elements for each query, and there are up to q queries, so O(q^2 * n) in the worst case, but n is small (≤ 10). - 🧺 Space complexity:
O(n)— We use a copy of the nums array for each prefix.