What is the time complexity of Minimum Operations to Make Binary Array Elements Equal to One I?

The optimal solution for Minimum Operations to Make Binary Array Elements Equal to One I runs in O(n) time and uses O(1) space. This complexity is optimal as we only traverse the array once, making it linear in terms of time.

What is the brute force solution for Minimum Operations to Make Binary Array Elements Equal to One I?

The brute force approach for Minimum Operations to Make Binary Array Elements Equal to One I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible triplet of consecutive elements in the array and flipping them if needed. This is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make Binary Array Elements Equal to One I?

Minimum Operations to Make Binary Array Elements Equal to One I uses the following concepts: Array, Bit Manipulation, Queue, Sliding Window, Prefix Sum. The recommended patterns to study are: Greedy, Sliding Window.

What are the interview tips for Minimum Operations to Make Binary Array Elements Equal to One I?

Always think about edge cases and how they might affect your solution. Explain your thought process clearly while coding, as it helps the interviewer understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Minimum Operations to Make Binary Array Elements Equal to One I?

Not considering the edge cases where the array length is less than 3. Failing to track the state of flips correctly, which can lead to incorrect results.

#3191

Minimum Operations to Make Binary Array Elements Equal to One I

Medium

ArrayBit ManipulationQueueSliding WindowPrefix SumGreedySliding Window

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a greedy strategy to flip elements as we encounter them. By processing the array in a single pass, we can efficiently determine the minimum number of flips needed.

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Algorithm

4 steps

1Step 1: Initialize a counter for operations and a variable to track the current state of flips.
2Step 2: Iterate through the array from index 0 to n-1.
3Step 3: If the current element is 0 and the current state of flips is even (indicating it appears as 0), flip the next three elements and increment the operation counter.
4Step 4: Adjust the current state of flips accordingly.

solution.py14 lines

1# Full working Python code
2
3def min_operations(nums):
4    operations = 0
5    flips = 0
6    n = len(nums)
7    for i in range(n):
8        if (nums[i] + flips) % 2 == 0:
9            if i + 2 < n:
10                flips += 1
11                operations += 1
12            else:
13                return -1
14    return operations

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Complexity note: This complexity is optimal as we only traverse the array once, making it linear in terms of time.

1Flipping three consecutive elements can change the parity of the elements, which is crucial for determining the number of operations.
2The greedy approach allows us to minimize operations by making decisions based on the current state of the array.

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