How do you solve Minimum Operations to Make Array Elements Zero on LeetCode?

Minimum Operations to Make Array Elements Zero (LeetCode #3495) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using logarithms simplifies the operation counting.

What is the time complexity of Minimum Operations to Make Array Elements Zero?

The optimal solution for Minimum Operations to Make Array Elements Zero runs in O(n) time and uses O(1) space. Calculating log4 for each number is O(1), and we iterate through n numbers, making the overall complexity O(n).

What is the brute force solution for Minimum Operations to Make Array Elements Zero?

The brute force approach for Minimum Operations to Make Array Elements Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simulate the process of reducing the array elements to zero by repeatedly selecting pairs and applying the operation. This approach is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make Array Elements Zero?

Minimum Operations to Make Array Elements Zero uses the following concepts: Array, Math, Bit Manipulation. The recommended patterns to study are: Math, Logarithms.

What are the interview tips for Minimum Operations to Make Array Elements Zero?

Explain your thought process clearly. Practice similar problems to build intuition. Consider both brute force and optimal approaches.

What are common mistakes when solving Minimum Operations to Make Array Elements Zero?

Forgetting to handle edge cases like 0 or negative numbers. Misunderstanding the log base conversion.

#3495

Minimum Operations to Make Array Elements Zero

Hard

ArrayMathBit ManipulationMathLogarithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

Instead of simulating the operations, we can calculate the number of operations needed for each number directly using logarithmic properties, which allows us to efficiently determine the total operations required.

⚙️

Algorithm

3 steps

1Step 1: For each query, compute the number of operations for each number using floor(log4(x)) + 1.
2Step 2: Sum the operations for all numbers in the range [l, r].
3Step 3: Return the total operations for all queries.

solution.py8 lines

1import math
2
3def min_operations_optimal(queries):
4    total_ops = 0
5    for l, r in queries:
6        for num in range(l, r + 1):
7            total_ops += math.floor(math.log(num, 4)) + 1
8    return total_ops

ℹ

Complexity note: Calculating log4 for each number is O(1), and we iterate through n numbers, making the overall complexity O(n).

1Using logarithms simplifies the operation counting.
2Pairing operations is not necessary in the optimal approach.

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