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LeetCode: Max Consecutive Ones

Posted on January 20, 2018July 26, 2020 by braindenny

Max Consecutive Ones



Similar Problems:

  • LeetCode: Max Consecutive Ones
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  • LeetCode: Minimum Swaps to Group All 1’s Together
  • CheatSheet: Leetcode For Code Interview
  • CheatSheet: Common Code Problems & Follow-ups
  • Tag: #array

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):
    ## https://code.dennyzhang.com/max-consecutive-ones
    ## 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]) # 0
# print s.findMaxConsecutiveOnes([1, 0, 1]) # 1
# print s.findMaxConsecutiveOnes([1]) # 1
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