Leetcode 485 - Max Consecutive Ones solution

Solution one:

The idea here is to maintain two variables at the same time. current_max stores the maximum occurrences of 1 and maximum just stores the max value of current_max variable.

class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        maximum = 0
        current_max = 0

        for i in nums:
            if i == 1:
                current_max += 1
                maximum = max(maximum, current_max)
            else:
                current_max = 0

        return maximum

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int high = 0;
        int cur = 0;
        
        for(int i : nums){
            if(i == 1)
                cur += 1;
            else
                cur = 0;
                
            high = Math.max(high, cur);
        }
        
        return high;
    }
}

110111 current_max = 1

110111 current_max = 2

110111 current_max = 0

110111 current_max = 1

110111 current_max = 2

110111 current_max = 3

Solution Two:

Faster than the previous approach because max fucntion is not running all the times.

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int high = 0;
        int cur = 0;
        
        for(int i : nums){
            if(i == 1){
                cur += 1;                
            } else{
                high = Math.max(high, cur);
                cur = 0;
            }
        }
        // if last element is one
        high = Math.max(high, cur);

        return high;
    }
}

Solution Three:

For O(n) space we can reassign values to the input array.

class Solution:
    def findMaxConsecutiveOnes(self, nums):
        for i in range(1, len(nums)):
            if nums[i]:
                nums[i] += nums[i - 1]
        return max(nums)

Transform the input eg: [0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1] to [0, 1, 2, 3, 0, 1, 2, 0, 0, 1, 2, 3, 4] and return the max element.