Problem Description

You have a binary array nums containing only 0s and 1s. Your goal is to make all elements in the array equal to 1 using a specific operation.

The operation you can perform is:

• Select any 3 consecutive elements in the array and flip all of them
• Flipping means changing 0 to 1 and 1 to 0
• You can perform this operation any number of times (including zero)

For example, if you have [0, 1, 1] and apply the operation, it becomes [1, 0, 0].

The problem asks you to find the minimum number of operations needed to turn all elements into 1. If it's impossible to achieve this goal, return -1.

The solution works by traversing the array from left to right. When encountering a 0 at position i, it must be flipped to become 1. Since each operation flips exactly 3 consecutive elements, flipping at position i affects positions i, i+1, and i+2. The key insight is that if we encounter a 0, we must flip it along with the next two elements. If there aren't enough elements remaining (less than 3 elements from current position to end), it's impossible to complete the transformation, so we return -1.

The code uses XOR operation (^= 1) to flip bits efficiently - when XORing with 1, a 0 becomes 1 and a 1 becomes 0. Each time a flip operation is performed, the operation counter increases by 1.

Intuition

The key observation is that we need to process the array from left to right in a greedy manner. Why? Consider the leftmost 0 in the array - this position must eventually become 1. Since our operation flips exactly 3 consecutive elements, the only way to change this leftmost 0 to 1 is by including it in a flip operation.

Once we decide to flip at position i (because nums[i] = 0), we affect positions i, i+1, and i+2. After this flip, position i becomes 1 and won't need attention again. This is crucial - we never need to revisit or flip an already processed position because doing so would undo our work.

This greedy approach works because:

1. Each 0 we encounter must be flipped exactly once to become 1
2. Flipping a position multiple times is wasteful (even number of flips returns to original state, odd number equals one flip)
3. By processing left to right, we ensure each position is handled optimally without interference from future operations

The impossibility case arises naturally: if we find a 0 in the last two positions, we can't perform a valid 3-element flip operation, making it impossible to turn all elements to 1.

This leads to the simple algorithm: scan left to right, whenever we see a 0, flip it along with the next two elements. The XOR operation (^= 1) elegantly handles the flipping since 0 ^ 1 = 1 and 1 ^ 1 = 0.

Learn more about Queue, Prefix Sum and Sliding Window patterns.

Solution Approach

The solution implements a sequential traversal with simulation approach. We iterate through the array once from left to right and perform flip operations as needed.

Here's the step-by-step implementation:

1. Initialize counter: Start with ans = 0 to track the number of operations performed.

2. Traverse the array: Use enumeration to get both index i and value x for each element.

3. Check for zeros: When we encounter x == 0, we know this position needs to be flipped.

4. Boundary check: Before flipping, verify if i + 2 >= len(nums). If true, we don't have enough elements to perform a 3-element flip, so return -1.

5. Perform the flip: Instead of flipping the current element (which we know will become 1), we only flip the next two elements:

   • nums[i + 1] ^= 1 - flips the element at position i + 1
   • nums[i + 2] ^= 1 - flips the element at position i + 2

   Note: We don't flip nums[i] because we're already processing it and know it will become 1 after this operation.

6. Count the operation: Increment ans += 1 for each flip operation performed.

7. Return result: After traversing the entire array, return ans as the minimum number of operations.

The XOR operation (^= 1) is used for flipping because it efficiently toggles bits: 0 ^ 1 = 1 and 1 ^ 1 = 0.

The time complexity is O(n) where n is the length of the array, as we traverse it once. The space complexity is O(1) as we only use a constant amount of extra space for the counter variable.

Example Walkthrough

Let's walk through the solution with the array nums = [0, 1, 0, 1, 1].

Initial state: [0, 1, 0, 1, 1], operations = 0

Step 1: We start at index 0 and find nums[0] = 0.

• Since it's 0, we need to flip it along with the next two elements
• Check if we have enough elements: 0 + 2 = 2 < 5 ✓
• Flip elements at positions 0, 1, and 2:
  • Position 0: 0 → 1
  • Position 1: 1 → 0
  • Position 2: 0 → 1
• Array becomes: [1, 0, 1, 1, 1]
• Operations = 1

Step 2: Move to index 1 and find nums[1] = 0.

• Since it's 0, we need to flip it along with the next two elements
• Check if we have enough elements: 1 + 2 = 3 < 5 ✓
• Flip elements at positions 1, 2, and 3:
  • Position 1: 0 → 1
  • Position 2: 1 → 0
  • Position 3: 1 → 0
• Array becomes: [1, 1, 0, 0, 1]
• Operations = 2

Step 3: Move to index 2 and find nums[2] = 0.

• Since it's 0, we need to flip it along with the next two elements
• Check if we have enough elements: 2 + 2 = 4 < 5 ✓
• Flip elements at positions 2, 3, and 4:
  • Position 2: 0 → 1
  • Position 3: 0 → 1
  • Position 4: 1 → 0
• Array becomes: [1, 1, 1, 1, 0]
• Operations = 3

Step 4: Move to index 3 and find nums[3] = 1.

• Since it's already 1, skip to the next position

Step 5: Move to index 4 and find nums[4] = 0.

• Since it's 0, we need to flip it
• Check if we have enough elements: 4 + 2 = 6 ≥ 5 ✗
• We can't perform a valid 3-element flip operation
• Return -1 (impossible)

In this example, we cannot convert all elements to 1 because the last element is 0 and we don't have enough elements to perform a flip operation.

Let's consider a successful example: nums = [0, 0, 0]

Initial state: [0, 0, 0], operations = 0

Step 1: At index 0, find nums[0] = 0.

• Flip positions 0, 1, 2: [0, 0, 0] → [1, 1, 1]
• Operations = 1

Result: All elements are 1, return 1 operation.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. The algorithm iterates through the array once using a single for loop. In each iteration, it performs constant time operations: checking if an element equals 0, performing XOR operations on at most two subsequent elements, and incrementing a counter. Even though there are modifications to elements ahead in the array, each element is visited only once by the main loop iterator.

The space complexity is O(1). The algorithm only uses a constant amount of extra space for variables ans, i, and x. The modifications are done in-place on the input array nums, and no additional data structures that grow with the input size are created.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Modifying the Array While Iterating

The most significant pitfall in this solution is that it modifies the input array in-place during iteration. This can lead to:

• Side effects: The original array is permanently modified, which might not be expected by the caller
• Debugging difficulties: Hard to trace the state changes during execution
• Reusability issues: Cannot run the function twice on the same array without restoring it first

Solution: Create a copy of the array at the beginning:

def minOperations(self, nums: List[int]) -> int:
    nums = nums.copy()  # Work with a copy to preserve original
    operation_count = 0
    # ... rest of the code remains the same

2. Not Understanding Why We Don't Flip nums[i]

A common misconception is thinking we need to explicitly flip nums[i] when we encounter a 0. The code doesn't flip nums[i] because:

• We're already at position i and know it's 0
• After the operation, it will become 1 (which is what we want)
• We only need to update the next two elements for future iterations

Pitfall Example: Someone might incorrectly write:

if value == 0:
    if index + 2 >= len(nums):
        return -1
    nums[index] ^= 1  # WRONG: This would flip it back to 0!
    nums[index + 1] ^= 1
    nums[index + 2] ^= 1

3. Incorrect Boundary Check

Using index + 2 > len(nums) instead of index + 2 >= len(nums) would cause an index out of bounds error:

# Wrong boundary check
if index + 2 > len(nums):  # This allows index + 2 == len(nums)
    return -1
nums[index + 2] ^= 1  # Would crash when index + 2 == len(nums)

Solution: Always use >= for the boundary check to ensure we have at least 3 elements starting from the current position.

4. Assuming Greedy Always Works Without Proof

While the greedy approach (always flip when encountering a 0) works for this problem, not understanding why it's optimal can lead to applying similar logic incorrectly in related problems.

Key insight: Once we encounter a 0 at position i, there's only one way to make it 1 - by flipping the window starting at i. Delaying this flip doesn't provide any advantage because:

• The 0 must eventually be flipped
• Each position can only be the start of one flip operation
• The order of non-overlapping flips doesn't matter