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Minimum Time to Make Array Sum At Most x

Difficulty: Hard

Problem Description

You are given two 0-indexed integer arrays nums1 and nums2 of equal length. Every second, for all indices 0 <= i < nums1.length, value of nums1[i] is incremented by nums2[i]. After this is done, you can do the following operation:

  • Choose an index 0 <= i < nums1.length and make nums1[i] = 0.

You are also given an integer x. Return the minimum time in which you can make the sum of all elements of nums1 to be less than or equal to x, or -1 if this is not possible.

Key Insights

  • The problem involves simulating the growth of nums1 over time based on the increments specified in nums2.
  • The goal is to determine the minimum seconds required to ensure the sum of nums1 is less than or equal to x.
  • If the total growth of nums1 surpasses x and cannot be reduced sufficiently with the operations allowed, the answer should be -1.
  • The operations must be optimized to achieve the target sum while minimizing the time taken.

Space and Time Complexity

Time Complexity: O(n log n) - due to sorting operations and possible evaluations of each increment time step. Space Complexity: O(n) - for storing results during calculations.

Solution

To solve the problem, we can simulate the increment of nums1 at each second while tracking the changes. We will maintain a priority queue (or a sorted list) of the maximum values that can be set to zero, allowing us to effectively manage which element to zero out to minimize the sum of nums1.

  1. Initialize a variable to track the current time in seconds.
  2. Calculate the initial sum of nums1.
  3. Increment nums1 based on nums2 each second and update the sum.
  4. After each increment, check if the sum is less than or equal to x.
  5. If not, determine which element to set to zero based on the highest impact (largest current value).
  6. Repeat until the sum is adjusted to be less than or equal to x, or until it's clear that it's impossible.

Code Solutions

def minTimeToMakeArraySumAtMostX(nums1, nums2, x):
    import heapq
    
    n = len(nums1)
    current_sum = sum(nums1)
    if current_sum <= x:
        return 0
    
    increments = []
    for i in range(n):
        increments.append((nums2[i], nums1[i]))

    increments.sort(reverse=True)  # Sort by increments in descending order
    time = 0
    
    while current_sum > x:
        time += 1
        for i in range(n):
            nums1[i] += nums2[i]  # Increment nums1 by nums2
            
        current_sum = sum(nums1)
        
        if current_sum <= x:
            return time
        
        # Choose the maximum element to set to zero
        max_index = max(range(n), key=lambda i: nums1[i])
        current_sum -= nums1[max_index]
        nums1[max_index] = 0  # Set the chosen element to zero

        # Check if we can still achieve the target
        if current_sum > x and all(v == 0 for v in nums1):
            return -1
      
    return time
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