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Zero Array Transformation IV

Number: 3795

Difficulty: Medium

Paid? No

Companies: Google

Problem Description

You are given an integer array nums and a list of queries. Each query is of the form [l, r, val] and lets you choose any subset of indices in the range [l, r] and decrement the value at those indices by exactly val. Your goal is to determine the minimum non-negative k such that after processing the first k queries (in sequence) it is possible to make nums a Zero Array (i.e. all elements 0) by choosing an optimal subset for each query. If it is impossible even after all queries, return -1.

Key Insights

  • Each query allows a binary decision for each index in its range: either subtract val or leave that index untouched.
  • For each element, the total decrement must equal its initial value; hence, its required value must be representable as a sum (using each available query at most once) of the query values available at that index.
  • Since operations on different indices are independent (the same query can be applied to many indices), the problem reduces to checking for each element whether its target value has a subset-sum representation using the values from the queries that cover it.
  • We can determine the minimum k by progressively considering a prefix of the queries (or by binary searching over k) and verifying the subset-sum condition for all indices.

Space and Time Complexity

Time Complexity: O(n * Q * T) per check in the worst-case, where n is the length of nums (n ≤ 10), Q is the number of queries (up to 1000) and T is the maximum target value (up to 1000). Binary search over Q gives an overall complexity of O(log Q * n * Q * T).
Space Complexity: O(T) for the subset sum DP per element.

Solution

The solution leverages binary search over the number of queries used (k). For each candidate prefix (first k queries), we iterate over each index in nums. For a given index, we construct the list of query values available (i.e. queries that cover the index). Since a query decrements by a fixed value and can be applied at most once to that index per query, the problem of making that element zero is now equivalent to a subset sum problem: determining if the initial value of the element can be constructed by summing a subset of the available query values.
We solve the subset sum using dynamic programming by maintaining a bitmask of reachable sums. If every element’s target value is reachable with the allowed queries for that index, then the array can be reduced to zero. Using binary search over k, we find the minimum k that satisfies this condition.
The same logic is implemented in each of the provided code solutions.

Code Solutions

# Python solution for Zero Array Transformation IV

def can_zero(nums, queries_prefix):
    # For each index, check if its value can be reached by a subset of allowed query values
    n = len(nums)
    for index in range(n):
        # Collect query values that affect this index
        allowed = [val for (l, r, val) in queries_prefix if l <= index <= r]
        target = nums[index]
        # dp will be an integer acting as a bitmask:
        # if the bit at position i is 1, then sum i is reachable.
        dp = 1  # dp[0] is true initially
        for val in allowed:
            # Shift dp by val and OR it with dp so far
            dp |= dp << val
            # Limit the dp bitmask to only care up to target sum
            dp &= (1 << (target + 1)) - 1
        # Check if target sum is reachable
        if not (dp >> target) & 1:
            return False
    return True

def min_k_to_zero(nums, queries):
    # Binary search the minimum k for which nums is zeroizable
    left, right = 0, len(queries)
    answer = -1
    while left <= right:
        mid = (left + right) // 2
        if can_zero(nums, queries[:mid]):
            answer = mid
            right = mid - 1  # Try to find a smaller k
        else:
            left = mid + 1
    return answer

# Example test case
nums = [2, 0, 2]
queries = [[0, 2, 1], [0, 2, 1], [1, 1, 3]]
print(min_k_to_zero(nums, queries))  # Expected output: 2
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