LeetCode 3356 - Zero Array Transformation II

The problem asks us to determine the minimum number of sequential queries required to transform an array nums into a Zero Array, where all elements are zero. Each query specifies a subarray [li, ri] and a value vali.

leetcodemediumarraytwo-pointersbinary-searchprefix-sum

LeetCode Problem 3356

Difficulty: 🟡 Medium
Topics: Array, Two Pointers, Binary Search, Prefix Sum

Solution

Problem Understanding

The problem asks us to determine the minimum number of sequential queries required to transform an array nums into a Zero Array, where all elements are zero. Each query specifies a subarray [li, ri] and a value vali. We are allowed to decrement each element within that subarray by any value up to vali independently. This means we can choose a different decrement amount for each index in the range, as long as we do not exceed vali. The task is to find the smallest k such that applying the first k queries in order allows us to make all elements zero. If it is impossible with all queries, we return -1.

The input nums represents the initial state of the array, and queries represents a sequence of operations. The output is the minimum number of queries that can achieve a zero array or -1 if no combination works.

Constraints are significant:

  • 1 <= nums.length <= 10^5 and 1 <= queries.length <= 10^5, which implies a naive approach that applies every query individually to each index would be too slow.
  • 0 <= nums[i] <= 5 * 10^5 and 1 <= vali <= 5, which means that decrement values are small but array elements can be large, which hints that an efficient range-summing technique could work.
  • Each query only modifies a contiguous range, which suggests that prefix sums or a difference array can optimize the application of multiple queries.

Important edge cases include arrays where all elements are initially zero, arrays where no combination of queries can reach zero, and queries where vali is less than some elements, requiring careful accumulation of decrements.

Approaches

Brute-Force Approach

A straightforward approach would simulate each query in sequence. For each query, iterate through the range [li, ri] and decrement each element by the maximum allowed vali or the remaining value needed to reach zero. After each query, check if the array has become a Zero Array. This works correctly but is too slow because in the worst case it requires O(n * m) operations, where n is the array length and m is the number of queries. Given the constraints of 10^5 for both, this results in up to 10^10 operations, which is infeasible.

Optimal Approach

The key observation is that each query contributes at most vali decrement to a contiguous subarray. We do not need to simulate individual decrements; we only need to know the total maximum decrement possible at each index after the first k queries. This can be efficiently computed using a difference array. The difference array allows applying a range increment/decrement in O(1) per query, and then a prefix sum gives the total accumulated decrement at each index. With this approach, we can perform a binary search over k, checking whether the first k queries can reduce all elements to zero. This reduces the complexity from O(n * m) to O(n log m).

Approach Time Complexity Space Complexity Notes
Brute Force O(n * m) O(n) Apply each query element-wise, check zero array
Optimal O(n log m) O(n) Use difference array + prefix sum + binary search on number of queries

Algorithm Walkthrough

  1. Initialize a binary search range from 1 to len(queries) to find the minimum k such that the first k queries make nums a zero array.
  2. Define a helper function can_zero(k) that checks if applying the first k queries can reduce all elements of nums to zero. This function uses a difference array.
  3. Build a difference array diff of length n + 1 initialized to zero. For each query i in the first k queries, increment diff[li] by vali and decrement diff[ri + 1] by vali. This encodes the maximum decrement each index can get.
  4. Compute the prefix sum over diff to get the total decrement possible for each index.
  5. Check feasibility: for each index j in nums, if nums[j] > decrement[j], it is impossible to reduce it to zero, and can_zero(k) returns false. Otherwise, return true.
  6. Perform binary search: if can_zero(mid) is true, search the left half; otherwise, search the right half. Track the minimum k found.
  7. Return the result: the minimum k if found, otherwise -1.

Why it works

The difference array allows us to efficiently accumulate the maximum decrement each index can receive without applying every decrement individually. Binary search ensures that we find the smallest prefix length of queries that can zero the array. Each index is checked against its maximum possible decrement to ensure feasibility.

Python Solution

from typing import List

class Solution:
    def minZeroArray(self, nums: List[int], queries: List[List[int]]) -> int:
        n = len(nums)
        m = len(queries)
        
        def can_zero(k: int) -> bool:
            diff = [0] * (n + 1)
            for i in range(k):
                l, r, val = queries[i]
                diff[l] += val
                if r + 1 < n:
                    diff[r + 1] -= val
            # Prefix sum to get total decrement at each index
            total = 0
            for i in range(n):
                total += diff[i]
                if nums[i] > total:
                    return False
            return True
        
        left, right = 1, m
        ans = -1
        while left <= right:
            mid = (left + right) // 2
            if can_zero(mid):
                ans = mid
                right = mid - 1
            else:
                left = mid + 1
        return ans

Explanation

We use a helper can_zero(k) to check if the first k queries suffice. The difference array allows range updates in O(1). We then perform a prefix sum to compute the total decrement at each index and compare with nums[i]. Binary search over k finds the minimal number of queries needed.

Go Solution

func minZeroArray(nums []int, queries [][]int) int {
    n := len(nums)
    m := len(queries)
    
    canZero := func(k int) bool {
        diff := make([]int, n+1)
        for i := 0; i < k; i++ {
            l, r, val := queries[i][0], queries[i][1], queries[i][2]
            diff[l] += val
            if r+1 < n {
                diff[r+1] -= val
            }
        }
        total := 0
        for i := 0; i < n; i++ {
            total += diff[i]
            if nums[i] > total {
                return false
            }
        }
        return true
    }
    
    left, right := 1, m
    ans := -1
    for left <= right {
        mid := (left + right) / 2
        if canZero(mid) {
            ans = mid
            right = mid - 1
        } else {
            left = mid + 1
        }
    }
    return ans
}

Explanation

The Go version mirrors the Python implementation. Slice indexing and allocation are handled with make. The difference array is n+1 in length. Binary search and prefix sum are implemented in the same manner, with attention to zero-based indexing.

Worked Examples

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

Step diff array (after queries) prefix sum nums check
k=1 [1,0,0, -1] [1,1,1] [2,0,2] → cannot zero
k=2 [2,0,0, -2] [2,2,2] [2,0,2] ≤ [2,2,2] → can zero

Minimum k = 2.

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

Step diff array prefix sum nums check
k=2 [1,2,1,-2] [1,3,4,2] [4,3,2,1] > prefix sum → cannot zero

Return -1.

Complexity Analysis

Measure Complexity Explanation
Time O(n log m) Binary search over queries (log m) with O(n) prefix sum check per step
Space O(n) Difference array of size n+1

The binary search reduces the number of queries checked, and the difference array ensures we do not perform O(n) operations per query individually.

Test Cases

# Provided examples
assert Solution().minZeroArray([