What is the time complexity of Minimum Numbers of Function Calls to Make Target Array?

The optimal solution for Minimum Numbers of Function Calls to Make Target Array runs in O(n log m) time and uses O(1) space. This complexity arises because for each number, we may need to perform log m operations (where m is the maximum number in nums) to reduce it to zero.

What is the brute force solution for Minimum Numbers of Function Calls to Make Target Array?

The brute force approach for Minimum Numbers of Function Calls to Make Target Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves incrementing each element of the array until we reach the target values in nums. This method is straightforward but inefficient, as it may require many operations for larger numbers. The optimal Optimal Solution approach improves this to O(n log m).

What algorithm or data structure is used to solve Minimum Numbers of Function Calls to Make Target Array?

Minimum Numbers of Function Calls to Make Target Array uses the following concepts: Array, Greedy, Bit Manipulation. The recommended patterns to study are: Greedy, Bit Manipulation.

What are the interview tips for Minimum Numbers of Function Calls to Make Target Array?

Always clarify the operations allowed before diving into the solution. Think about edge cases, such as when nums contains zeros or large numbers. Practice explaining your thought process as you work through the problem.

What are common mistakes when solving Minimum Numbers of Function Calls to Make Target Array?

Not considering the effect of doubling or halving on the number of calls. Forgetting to check if the number is even or odd before deciding the operation.

#1558

Minimum Numbers of Function Calls to Make Target Array

Medium

ArrayGreedyBit ManipulationGreedyBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log m)Space O(1)

The optimal solution works backwards from nums to arr, using division and counting operations. By halving the numbers when possible, we minimize the number of function calls needed to reach the target.

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Algorithm

4 steps

1Step 1: Initialize a counter for function calls.
2Step 2: For each number in nums, while it is greater than 0, check if it is even or odd.
3Step 3: If even, divide by 2 (this counts as one function call). If odd, subtract 1 (this counts as one function call).
4Step 4: Continue until all numbers in nums are reduced to 0.

solution.py15 lines

1# Full working Python code
2
3def minFunctionCalls(nums):
4    calls = 0
5    for num in nums:
6        while num > 0:
7            if num % 2 == 0:
8                num //= 2
9            else:
10                num -= 1
11            calls += 1
12    return calls
13
14# Example usage
15print(minFunctionCalls([1, 5]))  # Output: 5

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Complexity note: This complexity arises because for each number, we may need to perform log m operations (where m is the maximum number in nums) to reduce it to zero.

1Working backwards from the target can simplify the problem.
2Dividing by 2 when possible reduces the number of operations significantly.

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