How do you solve Minimum Time to Make Array Sum At Most x on LeetCode?

Minimum Time to Make Array Sum At Most x (LeetCode #2809) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Setting an index to zero should be done in the order of the increment rates to minimize the overall time.

What is the time complexity of Minimum Time to Make Array Sum At Most x?

The optimal solution for Minimum Time to Make Array Sum At Most x runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the pairs of nums1 and nums2.

What is the brute force solution for Minimum Time to Make Array Sum At Most x?

The brute force approach for Minimum Time to Make Array Sum At Most x is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves simulating every second and checking all possible combinations of indices to set to zero. This is straightforward but inefficient as it explores all possibilities. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Time to Make Array Sum At Most x?

Minimum Time to Make Array Sum At Most x uses the following concepts: Array, Dynamic Programming, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Minimum Time to Make Array Sum At Most x?

Always think about the constraints and whether a brute force solution is feasible. Look for patterns in the problem that can lead to optimization, such as sorting or greedy approaches. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Minimum Time to Make Array Sum At Most x?

Not considering the order of operations when setting indices to zero. Failing to check if the sum can ever be reduced to x before proceeding with operations.

#2809

Minimum Time to Make Array Sum At Most x

Hard

ArrayDynamic ProgrammingSortingGreedySorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(n)

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Intuition

Time O(n log n)Space O(n)

The optimal solution leverages sorting and a greedy approach. By prioritizing which indices to set to zero based on their increment rates, we can minimize the time needed to achieve the target sum.

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Algorithm

3 steps

1Step 1: Pair each element of nums1 with its corresponding element in nums2 and sort these pairs based on the values in nums2 in descending order.
2Step 2: Initialize a variable to track the current sum of nums1 and the time taken.
3Step 3: Iterate through the sorted pairs, incrementing the time and adjusting the current sum until it is less than or equal to x.

solution.py12 lines

1def minTime(nums1, nums2, x):
2    n = len(nums1)
3    current_sum = sum(nums1)
4    pairs = sorted(zip(nums1, nums2), key=lambda p: p[1])
5    time = 0
6    for i in range(n):
7        if current_sum <= x:
8            return time
9        current_sum += pairs[i][1]
10        current_sum -= pairs[i][0]
11        time += 1
12    return time if current_sum <= x else -1

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Complexity note: The time complexity is O(n log n) due to the sorting step, while the space complexity is O(n) for storing the pairs of nums1 and nums2.

1Setting an index to zero should be done in the order of the increment rates to minimize the overall time.
2The sum of nums1 can only be reduced by setting elements to zero, so careful selection is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.