Max Consecutive Ones

Source: Dev.to

Problem Statement

Given a binary array nums, return the maximum number of consecutive 1’s present in the array. Example Input: nums = [1,1,0,1,1,1]

Output: 3

Explanation: The longest sequence of consecutive 1’s is [1,1,1] Length = 3

For every index, if the element is 1, start moving forward and count how many consecutive 1’s exist. Keep updating the maximum length found so far. Since for every position we may scan ahead again, many elements get visited multiple times. O(N²)

O(1)

class Solution { public int findMaxConsecutiveOnes(int[] nums) {

    int maxLen = 0;

    for (int i = 0; i < nums.length; i++) {

        int count = 0;

        for (int j = i; j < nums.length; j++) {

            if (nums[j] == 1) {
                count++;
                maxLen = Math.max(maxLen, count);
            } else {
                break;
            }
        }
    }

    return maxLen;
}

}

Notice that we only need the length of the current streak of 1’s. Whenever we encounter: 1 → extend the current streak. 0 → streak breaks, reset count. Thus, a single traversal is enough. No extra data structure is required. Initialize:

curCount = 0 maxCount = 0 Traverse the array. If current element is 1:

Increment curCount

Update maxCount

If current element is 0:

Reset curCount = 0

Return maxCount. A consecutive sequence continues only while we keep seeing 1’s. Whenever a 0 appears, the sequence breaks and a new count must start after it. So maintaining only the current streak length is sufficient. class Solution { public int findMaxConsecutiveOnes(int[] nums) {

    int maxCount = 0;
    int curCount = 0;

    for (int num : nums) {

        if (num == 1) {
            curCount++;
            maxCount = Math.max(maxCount, curCount);
        } else {
            curCount = 0;
        }
    }

    return maxCount;
}

}

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

Element Current Count Max Count

1 1 1

1 2 2

0 0 2

1 1 2

1 2 2

1 3 3

Final Answer: 3

This pattern appears whenever the problem asks for: Longest consecutive occurrence Longest streak Continuous segment Maximum run length Common approach: Maintain Current Streak Update Global Maximum Reset on Break Condition

Examples: Max Consecutive Ones Longest Increasing Continuous Segment Longest Repeating Character Block Continuous Attendance/Activity Problems Since only the current streak of consecutive 1’s matters, we maintain a running count, reset it whenever a 0 appears, and continuously update the maximum streak, achieving O(N) time and O(1) space.