# Leetcode: Max Consecutive Ones

Max Consecutive Ones

Given a binary array, find the maximum number of consecutive 1s in this array.

```Example 1:
Input: [1,1,0,1,1,1]
Output: 3
Explanation: The first two digits or the last three digits are consecutive 1s.
The maximum number of consecutive 1s is 3.

```

Note:

• The input array will only contain 0 and 1.
• The length of input array is a positive integer and will not exceed 10,000

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```class Solution(object):
## Basic Ideas:  counter
##               If found one 0, reset the counter to 0. Otherwise counter + 1
##
## Complexity: Time O(n), Space O(1)
def findMaxConsecutiveOnes(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
max_count, counter = 0, 0
for num in nums:
if num == 1:
counter = counter+1
max_count = max(max_count, counter)
else:
counter = 0
return max_count

## Ideas: Two pointers
##              i points to the start of consecutive of 1s
##              j points to the next element of the end of consecutive of 1s.
##
##  Assumption: [1, 0, 1], we should return 1
## Complexity: Time O(n) Space O(1)
def findMaxConsecutiveOnes_v1(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
max_count = 0
length  = len(nums)
i = 0
while i < length:
if nums[i] == 0:
i += 1
continue
# nums[i] == 1
j = i + 1
while j < length and nums[j] == 1:
j += 1
# nums[j] == 0 and nums[j-1] == 1
max_count = max(max_count, j-i)
# move to next
i = j + 1
return max_count

# s = Solution()
# print s.findMaxConsecutiveOnes() # 0
# print s.findMaxConsecutiveOnes([1, 0, 1]) # 1
# print s.findMaxConsecutiveOnes() # 1
```

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