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

The optimal solution for Minimum Operations to Make Binary Array Elements Equal to One II runs in O(n) time and uses O(1) space. The time complexity is O(n) because we traverse the array once, and the space complexity is O(1) since we only use a few extra variables.

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

The brute force approach for Minimum Operations to Make Binary Array Elements Equal to One II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible index to flip and counting the number of operations needed to turn the entire array into 1s. This method is straightforward but inefficient due to repeated calculations. 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 II?

Minimum Operations to Make Binary Array Elements Equal to One II uses the following concepts: Array, Dynamic Programming, Greedy. The recommended patterns to study are: Greedy, Array.

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

Always consider edge cases, such as arrays already filled with 1s. Explain your thought process clearly; understanding the problem is more important than jumping to code. Practice recognizing patterns in problems, as many can be solved using similar strategies.

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

Not considering the cumulative effect of flips, leading to unnecessary operations. Overcomplicating the problem by trying to simulate each flip instead of tracking the state.

#3192

Minimum Operations to Make Binary Array Elements Equal to One II

Medium

ArrayDynamic ProgrammingGreedyGreedyArray

LeetCode ↗

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 involves iterating through the array while keeping track of the current state of flips. This way, we can determine when a flip is necessary based on the cumulative effect of previous flips, allowing us to minimize the number of operations.

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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 left to right.
3Step 3: If the current element, considering the flips, is 0, increment the operations counter and toggle the current state of flips.
4Step 4: Continue until the end of the array and return the operations counter.

solution.py10 lines

1# Full working Python code
2
3def minOperations(nums):
4    operations = 0
5    flip = 0
6    for num in nums:
7        if (num + flip) % 2 == 0:  # If current effective value is 0
8            operations += 1
9            flip ^= 1  # Toggle flip state
10    return operations

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Complexity note: The time complexity is O(n) because we traverse the array once, and the space complexity is O(1) since we only use a few extra variables.

1Flipping affects all subsequent elements, so we need to track the cumulative effect of flips.
2The optimal solution leverages the parity of the current element and the flip state to minimize operations.

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