485 Max Consecutive Ones

1. Description

Given a binary array nums, return the maximum number of consecutive 1's in the array.

 

constraints :

• 1 <= nums.length <= \(10^{5}\)
• nums[i] is either 0 or 1.

2. Algorithms

1. Start
2. Declare three variables count, i, and max_num. Initialize them with zero.
3. Repeat the steps until i == len(nums).
   1. If nums[i] equals to zero, update max_num with max value between count and max_num. After that, initialize count to zero.
   2. Increment count and zero by 1.
4. Return the max between final count and max_num.
5. End

3. Codes

class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        # i : Index for iterating through nums. 
        # count : Count of consecutive 1's in the array. 
        # max_num : Storing max count of consecutive 1's 
        i, count, max_num = 0, 0, 0
        
        while i < len(nums) : 
            if nums[i] == 0 : 
                max_num = max(count, max_num) 
                count = -1 
            count += 1
            i += 1
        
        return max(count, max_num)

4. Conclusion