LeetCode #3489 — MEDIUM

Zero Array Transformation IV

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:

Select a subset of indices in the range [l_i, r_i] from nums.
Decrement the value at each selected index by exactly val_i.

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 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 <= 10
0 <= nums[i] <= 1000
1 <= queries.length <= 1000
queries[i] = [l_i, r_i, val_i]
0 <= l_i <= r_i < nums.length
1 <= val_i <= 10

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: Select a subset of indices in the range [li, ri] from nums. 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.

Baseline thinking

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

Pattern signal: Array · Dynamic Programming

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]]

Example 3

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

Core Insight

What unlocks the optimal approach

Use dynamic programming.
For each <code>nums[i]</code>, use DP to check whether the <code>queries[.][2]</code> values (i.e., the <code>val</code> values) of the queries that affect it can form a combination with a sum equal to <code>nums[i]</code>.

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 #3489: Zero Array Transformation IV
class Solution {
  public int minZeroArray(int[] nums, int[][] queries) {
    if (Arrays.stream(nums).allMatch(num -> num == 0))
      return 0;

    final int n = nums.length;
    Set<Integer>[] subsetSums = new Set[n];

    for (int i = 0; i < n; ++i)
      subsetSums[i] = new HashSet<>(List.of(0));

    for (int k = 0; k < queries.length; ++k) {
      final int l = queries[k][0];
      final int r = queries[k][1];
      final int val = queries[k][2];
      for (int i = l; i <= r; ++i) {
        List<Integer> sums = new ArrayList<>();
        for (final int subsetSum : subsetSums[i])
          sums.add(subsetSum + val);
        subsetSums[i].addAll(sums);
      }
      if (IntStream.range(0, n).allMatch(i -> subsetSums[i].contains(nums[i])))
        return k + 1;
    }

    return -1;
  }
}

// Accepted solution for LeetCode #3489: Zero Array Transformation IV
package main

import (
	"math/big"
	"slices"
	"sort"
)

// https://space.bilibili.com/206214
func minZeroArray4(nums []int, queries [][]int) int {
	ans := (slices.Max(nums) + 9) / 10
	m := len(queries)
	cnts := make([][11]int, m+1)
	for i, x := range nums {
		if x == 0 {
			continue
		}
		for k, q := range queries {
			cnts[k+1] = cnts[k]
			if q[0] <= i && i <= q[1] {
				cnts[k+1][q[2]]++
			}
		}
		ans += sort.Search(m+1-ans, func(mx int) bool {
			mx += ans
			p := new(big.Int)
			f := big.NewInt(1)
			for v, num := range cnts[mx] {
				for pow2 := 1; num > 0; pow2 *= 2 {
					k := min(pow2, num)
					f.Or(f, p.Lsh(f, uint(v*k)))
					if f.Bit(x) > 0 {
						return true
					}
					num -= k
				}
			}
			return false
		})
		if ans > m {
			return -1
		}
	}
	return ans
}

func minZeroArray32(nums []int, queries [][]int) int {
	m := len(queries)
	cnts := make([][][11]int, m+1)
	cnts[0] = make([][11]int, len(nums))
	for k, q := range queries {
		cnts[k+1] = slices.Clone(cnts[k])
		for i := q[0]; i <= q[1]; i++ {
			cnts[k+1][i][q[2]]++
		}
	}
	ans := sort.Search(m+1, func(mx int) bool {
		p := new(big.Int)
	next:
		for i, x := range nums {
			if x == 0 {
				continue
			}
			// 多重背包（二进制优化）
			f := big.NewInt(1)
			for v, num := range cnts[mx][i] {
				for pow2 := 1; num > 0; pow2 *= 2 {
					k := min(pow2, num)
					f.Or(f, p.Lsh(f, uint(v*k)))
					if f.Bit(x) > 0 {
						continue next
					}
					num -= k
				}
			}
			return false
		}
		return true
	})
	if ans <= m {
		return ans
	}
	return -1
}

func minZeroArray3(nums []int, queries [][]int) int {
	ans := sort.Search(len(queries)+1, func(mx int) bool {
		p := new(big.Int)
	next:
		for i, x := range nums {
			if x == 0 {
				continue
			}
			cnt := [11]int{}
			for _, q := range queries[:mx] {
				if q[0] <= i && i <= q[1] {
					cnt[q[2]]++
				}
			}
			// 多重背包（二进制优化）
			f := big.NewInt(1)
			for v, num := range cnt {
				for pow2 := 1; num > 0; pow2 *= 2 {
					k := min(pow2, num)
					f.Or(f, p.Lsh(f, uint(v*k)))
					if f.Bit(x) > 0 {
						continue next
					}
					num -= k
				}
			}
			return false
		}
		return true
	})
	if ans <= len(queries) {
		return ans
	}
	return -1
}

func minZeroArray2(nums []int, queries [][]int) (ans int) {
	p := new(big.Int)
	for i, x := range nums {
		if x == 0 {
			continue
		}
		f := big.NewInt(1)
		for k, q := range queries {
			if i < q[0] || i > q[1] {
				continue
			}
			f.Or(f, p.Lsh(f, uint(q[2])))
			if f.Bit(x) > 0 {
				ans = max(ans, k+1)
				break
			}
		}
		if f.Bit(x) == 0 {
			return -1
		}
	}
	return
}

func minZeroArray(nums []int, queries [][]int) (ans int) {
	for i, x := range nums {
		if x == 0 {
			continue
		}
		f := make([]bool, x+1)
		f[0] = true
		for k, q := range queries {
			if i < q[0] || i > q[1] {
				continue
			}
			val := q[2]
			for j := x; j >= val; j-- {
				f[j] = f[j] || f[j-val]
			}
			if f[x] {
				ans = max(ans, k+1)
				break
			}
		}
		if !f[x] {
			return -1
		}
	}
	return
}

# Accepted solution for LeetCode #3489: Zero Array Transformation IV
class Solution:
  def minZeroArray(self, nums: list[int], queries: list[list[int]]) -> int:
    if all(num == 0 for num in nums):
      return 0

    n = len(nums)
    subsetSums = [{0} for _ in range(n)]

    for k, (l, r, val) in enumerate(queries):
      for i in range(l, r + 1):
        newSums = {subsetSum + val for subsetSum in subsetSums[i]}
        subsetSums[i].update(newSums)
      if all(nums[i] in subsetSums[i] for i in range(n)):
        return k + 1

    return -1

// Accepted solution for LeetCode #3489: Zero Array Transformation IV
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3489: Zero Array Transformation IV
// class Solution {
//   public int minZeroArray(int[] nums, int[][] queries) {
//     if (Arrays.stream(nums).allMatch(num -> num == 0))
//       return 0;
// 
//     final int n = nums.length;
//     Set<Integer>[] subsetSums = new Set[n];
// 
//     for (int i = 0; i < n; ++i)
//       subsetSums[i] = new HashSet<>(List.of(0));
// 
//     for (int k = 0; k < queries.length; ++k) {
//       final int l = queries[k][0];
//       final int r = queries[k][1];
//       final int val = queries[k][2];
//       for (int i = l; i <= r; ++i) {
//         List<Integer> sums = new ArrayList<>();
//         for (final int subsetSum : subsetSums[i])
//           sums.add(subsetSum + val);
//         subsetSums[i].addAll(sums);
//       }
//       if (IntStream.range(0, n).allMatch(i -> subsetSums[i].contains(nums[i])))
//         return k + 1;
//     }
// 
//     return -1;
//   }
// }

// Accepted solution for LeetCode #3489: Zero Array Transformation IV
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #3489: Zero Array Transformation IV
// class Solution {
//   public int minZeroArray(int[] nums, int[][] queries) {
//     if (Arrays.stream(nums).allMatch(num -> num == 0))
//       return 0;
// 
//     final int n = nums.length;
//     Set<Integer>[] subsetSums = new Set[n];
// 
//     for (int i = 0; i < n; ++i)
//       subsetSums[i] = new HashSet<>(List.of(0));
// 
//     for (int k = 0; k < queries.length; ++k) {
//       final int l = queries[k][0];
//       final int r = queries[k][1];
//       final int val = queries[k][2];
//       for (int i = l; i <= r; ++i) {
//         List<Integer> sums = new ArrayList<>();
//         for (final int subsetSum : subsetSums[i])
//           sums.add(subsetSum + val);
//         subsetSums[i].addAll(sums);
//       }
//       if (IntStream.range(0, n).allMatch(i -> subsetSums[i].contains(nums[i])))
//         return k + 1;
//     }
// 
//     return -1;
//   }
// }

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 × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.

Zero Array Transformation IV

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

State misses one required dimension