LeetCode 3489 - Zero Array Transformation IV

The problem asks us to determine the minimum number of sequential queries needed to transform an integer array nums into a Zero Array, where every element is zero.

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LeetCode Problem 3489

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming

Solution

Problem Understanding

The problem asks us to determine the minimum number of sequential queries needed to transform an integer array nums into a Zero Array, where every element is zero. Each query specifies a subarray [li, ri] and a value vali, and the query allows us to decrement any subset of indices in that subarray by exactly vali. The challenge is to find the smallest k such that after applying the first k queries in order, it is possible to choose subsets in each query that make the array all zeros. If it is impossible to achieve a Zero Array with all queries, the result should be -1.

The input consists of nums, a small array with length up to 10, and queries, a larger array with up to 1000 entries. This means the problem has a combinatorial element because for each query, we can select any combination of indices in the specified range, which could be up to $2^{10} = 1024$ possibilities per query. However, the small size of nums allows us to use approaches that explore combinations efficiently.

Important points include handling non-negative decrements and respecting the query order. Edge cases include arrays that are already zero, queries that cannot reduce elements to zero exactly, and sequences where a single query could fully zero out some elements.

Approaches

Brute Force

The brute-force approach would simulate all possible subsets for each query in order. For each prefix of queries, we would attempt every combination of indices in the query range to decrement and recursively check if we can reach a Zero Array. This guarantees correctness but is extremely slow because the number of combinations grows exponentially: for each query, there are $2^{r-l+1}$ subsets. With up to 1000 queries, this is infeasible even for small nums.

Key Insight

The key insight is that since nums has at most 10 elements, we can represent the array as a tuple of counts and use a breadth-first search (BFS) or dynamic programming with bitmasking to track all reachable states. For each query, we only need to consider states that decrement each element by vali if the element is greater than or equal to vali. This transforms the problem into a search over a manageable number of states: the number of distinct non-negative arrays bounded by the values in nums.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^(n * k)) O(n) Explore all subsets for each query. Exponential in nums size and number of queries.
BFS / State Search O(k * ∏(nums[i]+1)) O(∏(nums[i]+1)) Explore reachable states with BFS using tuples; feasible since n <= 10 and nums[i] <= 1000.

Algorithm Walkthrough

  1. Represent the current state of nums as a tuple, since tuples are hashable and can be stored in a visited set.

  2. Initialize a BFS queue with the initial state and a counter for the number of queries applied.

  3. For each state in the queue, iterate through the first k queries:

  4. For the current query [l, r, val], generate all subsets of indices in [l, r] that can be decremented without going below zero.

  5. For each valid subset, create a new state by decrementing the selected indices by val.

  6. If the new state is all zeros, return k as the minimum number of queries.

  7. Otherwise, enqueue the new state if it has not been visited yet.

  8. If BFS finishes without reaching the zero array, return -1.

Why it works: The algorithm explores all possible reachable states of nums in order of the number of queries applied. By storing visited states, it avoids redundant work, and the BFS guarantees that the first time we encounter a Zero Array corresponds to the minimum number of queries required.

Python Solution

from typing import List
from collections import deque

class Solution:
    def minZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
        n = len(nums)
        initial_state = tuple(nums)
        visited = set([initial_state])
        queue = deque([(initial_state, 0)])
        
        while queue:
            state, q_idx = queue.popleft()
            if all(x == 0 for x in state):
                return q_idx
            if q_idx >= len(queries):
                continue

            l, r, val = queries[q_idx]
            # Generate all possible subsets in [l, r]
            for mask in range(1 << (r - l + 1)):
                new_state = list(state)
                valid = True
                for i in range(r - l + 1):
                    if (mask >> i) & 1:
                        if new_state[l + i] < val:
                            valid = False
                            break
                        new_state[l + i] -= val
                if valid:
                    new_tuple = tuple(new_state)
                    if new_tuple not in visited:
                        visited.add(new_tuple)
                        queue.append((new_tuple, q_idx + 1))
            # Also consider skipping this query entirely
            queue.append((state, q_idx + 1))
        
        return -1

This implementation uses BFS to explore all reachable states. Each query can be fully applied to subsets of indices or skipped. The visited set ensures no state is revisited, avoiding infinite loops. The BFS guarantees that the first time we encounter a Zero Array corresponds to the minimal k.

Go Solution

package main

func minZeroArray(nums []int, queries [][]int) int {
    type state struct {
        arr []int
        qIdx int
    }

    n := len(nums)
    initial := make([]int, n)
    copy(initial, nums)
    
    visited := map[string]bool{}
    queue := []state{{arr: initial, qIdx: 0}}
    key := func(a []int) string {
        s := ""
        for _, v := range a {
            s += string(rune(v)) + ","
        }
        return s
    }
    visited[key(initial)] = true

    for len(queue) > 0 {
        cur := queue[0]
        queue = queue[1:]
        
        zero := true
        for _, v := range cur.arr {
            if v != 0 {
                zero = false
                break
            }
        }
        if zero {
            return cur.qIdx
        }
        if cur.qIdx >= len(queries) {
            continue
        }
        
        l, r, val := queries[cur.qIdx][0], queries[cur.qIdx][1], queries[cur.qIdx][2]
        size := r - l + 1
        
        // Generate all subsets
        for mask := 0; mask < (1 << size); mask++ {
            valid := true
            newArr := make([]int, n)
            copy(newArr, cur.arr)
            for i := 0; i < size; i++ {
                if (mask>>i)&1 == 1 {
                    if newArr[l+i] < val {
                        valid = false
                        break
                    }
                    newArr[l+i] -= val
                }
            }
            if valid {
                k := key(newArr)
                if !visited[k] {
                    visited[k] = true
                    queue = append(queue, state{arr: newArr, qIdx: cur.qIdx + 1})
                }
            }
        }
        // Skip this query
        queue = append(queue, state{arr: cur.arr, qIdx: cur.qIdx + 1})
    }

    return -1
}

In Go, we represent the array state as a slice and use a map to track visited states using a string key. The algorithm otherwise mirrors the Python BFS implementation.

Worked Examples

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

Step Query Applied Array State Notes
0 - [2,0,2] initial
1 [0,2,1] [1,0,1] decrement first query
2 [0,2,1] [0,0,0] second query achieves zero array

Result: k = 2

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

All BFS states explored, cannot reach zero array. Result: -1.

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

Final zero array achieved at k = 4.

Complexity Analysis

Measure Complexity Explanation
Time O(len(queries) * 2^n * ∏(nums[i]+1)) BFS explores all reachable states; each query can create 2^n subsets. Feasible because n <= 10.
Space O(∏(nums[i]+1)) Stores all visited states in memory using a tuple or string key.

Since nums is small, this exponential growth is