Solution Approach

The solution implements a single-pass algorithm that efficiently tracks consecutive 1s in the array. Let's walk through the implementation step by step:

Variable Initialization:

ans = cnt = 0

We initialize two variables to 0:

• ans: stores the maximum number of consecutive 1s found so far
• cnt: tracks the current consecutive 1s count

Main Loop:

for x in nums:

We iterate through each element x in the array sequentially.

Counting Logic:

if x:  cnt += 1  ans = max(ans, cnt) else:  cnt = 0

For each element, we check:

• If x is 1 (truthy in Python), we:
  • Increment cnt by 1 to extend our current consecutive count
  • Update ans using max(ans, cnt) to keep track of the longest sequence seen so far
• If x is 0, we:
  • Reset cnt to 0 since the consecutive sequence is broken

Return Result:

return ans

After processing all elements, ans contains the maximum consecutive 1s count.

Example Walkthrough: For nums = [1, 1, 0, 1, 1, 1]:

• Index 0: x=1, cnt=1, ans=1
• Index 1: x=1, cnt=2, ans=2
• Index 2: x=0, cnt=0, ans=2 (remains unchanged)
• Index 3: x=1, cnt=1, ans=2 (remains unchanged)
• Index 4: x=1, cnt=2, ans=2 (remains unchanged)
• Index 5: x=1, cnt=3, ans=3 (new maximum)
• Return: 3

The algorithm's efficiency comes from processing each element exactly once with constant-time operations, achieving O(n) time complexity with O(1) space complexity.

Example Walkthrough

Let's trace through the algorithm with nums = [1, 0, 1, 1, 0, 1]:

Initial State:

• ans = 0 (maximum consecutive 1s found)
• cnt = 0 (current consecutive 1s count)

Step-by-step execution:

Index 0: x = 1

• Since x is 1, we increment cnt: cnt = 0 + 1 = 1
• Update ans: ans = max(0, 1) = 1
• State: cnt = 1, ans = 1

Index 1: x = 0

• Since x is 0, we reset cnt: cnt = 0
• ans remains unchanged at 1
• State: cnt = 0, ans = 1

Index 2: x = 1

• Since x is 1, we increment cnt: cnt = 0 + 1 = 1
• Update ans: ans = max(1, 1) = 1
• State: cnt = 1, ans = 1

Index 3: x = 1

• Since x is 1, we increment cnt: cnt = 1 + 1 = 2
• Update ans: ans = max(1, 2) = 2 (new maximum!)
• State: cnt = 2, ans = 2

Index 4: x = 0

• Since x is 0, we reset cnt: cnt = 0
• ans remains unchanged at 2
• State: cnt = 0, ans = 2

Index 5: x = 1

• Since x is 1, we increment cnt: cnt = 0 + 1 = 1
• Update ans: ans = max(2, 1) = 2
• State: cnt = 1, ans = 2

Final Result: Return ans = 2

The maximum consecutive 1s in the array is 2, found at indices 2-3.

Time and Space Complexity

The time complexity is O(n), where n is the length of the array nums. This is because the algorithm iterates through the array exactly once, performing constant-time operations (comparison, addition, and max operation) for each element.

The space complexity is O(1). The algorithm only uses two additional variables (ans and cnt) to track the maximum consecutive ones and the current consecutive count, regardless of the input size. No additional data structures that scale with the input are created.

Common Pitfalls

1. Forgetting to Update Maximum Before Resetting

A common mistake is resetting the counter when encountering a 0 without first checking if the current streak should update the maximum. While the provided solution correctly updates ans immediately when incrementing cnt, some implementations might try to optimize by only updating at the end of a streak:

Incorrect Approach:

def findMaxConsecutiveOnes(self, nums: List[int]) -> int:  max_consecutive = 0  current_consecutive = 0   for num in nums:  if num == 1:  current_consecutive += 1  else:  max_consecutive = max(max_consecutive, current_consecutive) # Only updating here  current_consecutive = 0   return max_consecutive # Missing final check!

Issue: This fails when the array ends with consecutive 1s (e.g., [0, 1, 1, 1] would return 0 instead of 3).

Solution: Always check after the loop ends or update the maximum immediately when incrementing:

# Option 1: Add final check return max(max_consecutive, current_consecutive)  # Option 2: Update immediately (as in original solution) if num == 1:  current_consecutive += 1  max_consecutive = max(max_consecutive, current_consecutive)

2. Using Wrong Variable Names Leading to Confusion

When using short variable names like cnt and ans, it's easy to mix them up, especially in longer functions:

Error-Prone:

ans = cnt = 0 for x in nums:  if x:  ans += 1 # Accidentally incrementing wrong variable  cnt = max(ans, cnt) # Variables swapped

Solution: Use descriptive variable names:

max_count = current_count = 0 for num in nums:  if num == 1:  current_count += 1  max_count = max(max_count, current_count)

3. Incorrect Handling of Edge Cases

Not considering edge cases can lead to runtime errors:

Potential Issues:

• Empty array: nums = []
• Single element: nums = [1] or nums = [0]
• All zeros: nums = [0, 0, 0]
• All ones: nums = [1, 1, 1]

Solution: The algorithm naturally handles these cases, but always verify:

def findMaxConsecutiveOnes(self, nums: List[int]) -> int:  if not nums: # Explicit empty check if needed  return 0   max_consecutive = current_consecutive = 0  # Rest of implementation...

4. Using Unnecessary Space with Additional Data Structures

Some might be tempted to store all consecutive sequences first:

Space-Inefficient:

def findMaxConsecutiveOnes(self, nums: List[int]) -> int:  sequences = [] # Unnecessary list  current = 0   for num in nums:  if num == 1:  current += 1  else:  if current > 0:  sequences.append(current)  current = 0   if current > 0:  sequences.append(current)   return max(sequences) if sequences else 0

Solution: Maintain only the maximum value during iteration (as in the original solution) to achieve O(1) space complexity.