We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Zero Array Transformation III

Difficulty: Medium

Problem Description

You are given an integer array nums of length n and a 2D array queries where queries[i] = [l_i, r_i]. Each queries[i] represents the following action on nums:

  • Decrement the value at each index in the range [l_i, r_i] in nums by at most 1.
  • The amount by which the value is decremented can be chosen independently for each index.

A Zero Array is an array with all its elements equal to 0.

Return the maximum number of elements that can be removed from queries, such that nums can still be converted to a zero array using the remaining queries. If it is not possible to convert nums to a zero array, return -1.

Key Insights

  • Each query allows for a specific range of indices in nums to be decremented.
  • The sum of decrements needed to turn nums into a zero array must be met by the remaining queries after removing some.
  • An efficient approach involves calculating the contribution of each query and determining how many can be removed while still satisfying the total decrement requirement.

Space and Time Complexity

Time Complexity: O(n + m log m), where n is the number of elements in nums and m is the number of queries. This includes sorting the queries and processing each in relation to the cumulative decrement. Space Complexity: O(n + m), to store the necessary data structures for tracking the contributions from queries and the cumulative decrements.

Solution

The solution involves calculating the total decrements required to convert nums to a zero array, and then using a greedy approach to determine which queries can be safely removed. We can sort the queries based on their impact and iteratively check if the required decrements can still be satisfied.

  1. Calculate the total sum of nums.
  2. For each query, compute the range it affects and keep track of the contributions.
  3. Sort the queries based on their range size or their impact on the total decrement.
  4. Iterate through the sorted queries, maintaining a count of the decrements contributed by the remaining queries until the total decrement requirement is met.
  5. Return the maximum number of removable queries.

Code Solutions

def maxRemovableQueries(nums, queries):
    total = sum(nums)  # Total decrements needed to reach a zero array
    n = len(nums)
    m = len(queries)

    # Create an array to track the impact of each query
    impact = [0] * (n + 1)
    
    for l, r in queries:
        impact[l] += 1
        if r + 1 < n:
            impact[r + 1] -= 1

    # Calculate the actual decrement impact per index
    decrement = [0] * n
    current = 0
    for i in range(n):
        current += impact[i]
        decrement[i] = current

    # Sort queries based on their range length
    queries.sort(key=lambda x: (x[1] - x[0] + 1))

    # Attempt to remove queries from the end
    removable_count = 0
    for l, r in reversed(queries):
        # Calculate the current decrement impact
        total_decrement = sum(decrement[i] for i in range(l, r + 1))
        if total - total_decrement <= 0:  # If we can achieve zero array
            return removable_count  # Return the count of removed queries
        removable_count += 1

    return -1  # If we can't convert nums to a zero array
← Back to All Questions