Skip to main content

Max Consecutive Ones

Problem Statement:​

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

✅ Solution: Linear Scan​

int findMaxConsecutiveOnes(vector<int>& binaryArray) {
    int maxConsecutiveOnes = 0;      // Stores the maximum count of consecutive 1s
    int currentCount = 0;            // Tracks the current streak of 1s

    for (int i = 0; i < binaryArray.size(); i++) {
        if (binaryArray[i] == 1) {
            currentCount++;          // Extend the streak when we find a 1
        } else {
            // Update the max if current streak ends
            maxConsecutiveOnes = max(maxConsecutiveOnes, currentCount);
            currentCount = 0;        // Reset the streak
        }
    }

    // Final check in case the array ends with a streak of 1s
    maxConsecutiveOnes = max(maxConsecutiveOnes, currentCount);

    return maxConsecutiveOnes;
}

📝 How It Works​

  • You scan the array once, keeping a running count of consecutive 1s.
  • Each time you see a 1, you increment currentCount.
  • If you see a 0, you:
    • Compare currentCount with maxConsecutiveOnes and update it if needed.
    • Reset currentCount to 0.
  • After the loop ends, you again update maxConsecutiveOnes to handle the case where the array ends with 1s.

🧩 Key Logic​

There’s no recurrence here — it’s a simple one-pass comparison-based logic:

maxConsecutiveOnes = max(maxConsecutiveOnes, currentCount)

⏱️ Time & Space Complexity​

MetricValue
⏱ TimeO(n)
🗂 SpaceO(1)
  • Single pass through the array.
  • Only two integer variables used.

⚠️ Edge Cases​

  • All elements are 1 → final max is updated at the end.
  • All elements are 0 → maxConsecutiveOnes remains 0.
  • Single element → works correctly for both 0 and 1.

💡 Other Approaches​

ApproachTimeSpaceNotes
Linear Scan (this)O(n)O(1)✅ Most efficient
Segment TreeO(log n)O(n)❌ Overkill
Sliding Window (Fixed Size)Not applicable here–❌ Only works for fixed-length windows

🔁 Related Problems​

💬

Discussion & Doubts