Approach: Running counter

def findMaxConsecutiveOnes(nums):
    best = cur = 0
    for x in nums:
        cur = cur + 1 if x == 1 else 0
        best = max(best, cur)
    return best

Complexity: O(n) time, O(1) space.

Full Solution — all languages

class Solution:
    def findMaxConsecutiveOnes(self, nums):
        best = cur = 0
        for x in nums:
            cur = cur + 1 if x == 1 else 0
            best = max(best, cur)
        return best

import java.util.*;
class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int best = 0, cur = 0;
        for (int x : nums) {
            cur = x == 1 ? cur + 1 : 0;
            best = Math.max(best, cur);
        }
        return best;
    }
}

#include <bits/stdc++.h> using namespace std;
class Solution {
    public: int findMaxConsecutiveOnes(vector<int>& nums) {
        int best = 0, cur = 0;
        for (int x : nums) {
            cur = x == 1 ? cur + 1 : 0;
            best = max(best, cur);
        }
        return best;
    }
};

/**
 * @param {number[]} nums
 * @return {number}
 */ var findMaxConsecutiveOnes = function(nums) {
    let best = 0, cur = 0;
    for (const x of nums) {
        cur = x === 1 ? cur + 1 : 0;
        if (cur > best) best = cur;
    }
    return best;
};

Max Consecutive Ones

Example 1

Example 2

Constraints