Maximum Consecutive Ones

Given an array arr[] = { 1, 1, 0, 1, 1, 1, 0, 1, 1 }, We need to find the length of the longest sequence of consecutive 1s.
For example, in this case, the maximum consecutive ones are 3.

Optimal Approach:

• Initialize cnt = 0 and maxx = 0.
• Traverse the array:
  • If the current element is 1, increment cnt and update maxx if maxx < cnt.
  • If the current element is 0, reset cnt to 0.
• After traversal, maxx will keep the length of the maximum sequence of consecutive 1s.

int findMaxConsecutiveOnes(vector<int> &nums)
{
    int maxx = 0;
    int cnt = 0;
    for(int i = 0; i < nums.size(); i++)
    {
        if(nums[i] == 1)
        {
            cnt++;
            maxx = max(maxx, cnt); 
        }
        else cnt = 0;
    }
    return maxx;
}

Time Complexity
O(n): We traverse the array once.

Space Complexity
O(1): No extra space is used apart from a few integer variables (maxx, cnt), making it a constant space solution.

🎯 Practice

🔗 Maximum Consecutive Ones