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485. Max Consecutive Ones

Description

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

 

Example 1:

Input: nums = [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.

Example 2:

Input: nums = [1,0,1,1,0,1]
Output: 2

 

Constraints:

  • 1 <= nums.length <= 105
  • nums[i] is either 0 or 1.

Solutions

Solution 1: Single Pass

Thinking

Longest run of ones. Checking every subarray is \(O(n^2)\); one scan is enough.

A \(1\) grows the current run and updates the answer; a \(0\) resets the counter.

Zeros separate runs, so the counter never joins two blocks.

We can iterate through the array, using a variable \(\textit{cnt}\) to record the current number of consecutive 1s, and another variable \(\textit{ans}\) to record the maximum number of consecutive 1s.

When we encounter a 1, we increment \(\textit{cnt}\) by one, and then update \(\textit{ans}\) to be the maximum of \(\textit{cnt}\) and \(\textit{ans}\) itself, i.e., \(\textit{ans} = \max(\textit{ans}, \textit{cnt})\). Otherwise, we reset \(\textit{cnt}\) to 0.

After the iteration ends, we return the value of \(\textit{ans}\).

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\).

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class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        ans = cnt = 0
        for x in nums:
            if x:
                cnt += 1
                ans = max(ans, cnt)
            else:
                cnt = 0
        return ans

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

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class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int ans = 0, cnt = 0;
        for (int x : nums) {
            if (x) {
                ans = max(ans, ++cnt);
            } else {
                cnt = 0;
            }
        }
        return ans;
    }
};

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func findMaxConsecutiveOnes(nums []int) (ans int) {
    cnt := 0
    for _, x := range nums {
        if x == 1 {
            cnt++
            ans = max(ans, cnt)
        } else {
            cnt = 0
        }
    }
    return
}

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function findMaxConsecutiveOnes(nums: number[]): number {
    let [ans, cnt] = [0, 0];
    for (const x of nums) {
        if (x) {
            ans = Math.max(ans, ++cnt);
        } else {
            cnt = 0;
        }
    }
    return ans;
}

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impl Solution {
    pub fn find_max_consecutive_ones(nums: Vec<i32>) -> i32 {
        let mut ans = 0;
        let mut cnt = 0;

        for &x in nums.iter() {
            if x == 1 {
                cnt += 1;
                ans = ans.max(cnt);
            } else {
                cnt = 0;
            }
        }

        ans
    }
}

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/**
 * @param {number[]} nums
 * @return {number}
 */
var findMaxConsecutiveOnes = function (nums) {
    let [ans, cnt] = [0, 0];
    for (const x of nums) {
        if (x) {
            ans = Math.max(ans, ++cnt);
        } else {
            cnt = 0;
        }
    }
    return ans;
};

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class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function findMaxConsecutiveOnes($nums) {
        $ans = $cnt = 0;

        foreach ($nums as $x) {
            if ($x == 1) {
                $cnt += 1;
                $ans = max($ans, $cnt);
            } else {
                $cnt = 0;
            }
        }

        return $ans;
    }
}

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