Max consecutive ones

Problem Overview

The problem "Max Consecutive Ones" is a common question that involves the manipulation and traversal of arrays. The task is to find the maximum number of consecutive 1s in a binary array, which is an array consisting only of 0s and 1s. This problem is often used to test an understanding of array traversal and basic counting techniques.

Understanding the Problem

Given a binary array, the goal is to traverse the array to find the longest contiguous segment consisting solely of 1s. For instance, if the input array is [1, 1, 0, 1, 1, 1], the maximum stretch of consecutive 1s is three.

Understanding the problem is crucial as it forms the foundation for developing an efficient algorithm. The primary requirement is to dynamically track sequences of 1s and determine the longest sequence encountered during the iteration over the array.

Key DSA Concepts and Theory

Arrays

Arrays are a fundamental data structure in computer science and programming. They are a collection of elements identified by index or key. In our problem, the binary array is a simple form of data representation that we need to parse and analyze.

Iteration

The process of traversing the array is a form of iteration. Iteration is essential in this scenario as it allows us to evaluate each element in sequence and apply conditional logic to achieve the desired outcome.

Counters

A counter is a basic programming construct often used in problems that require counting elements in a data structure. In this problem, a counter is used to keep track of the number of consecutive 1s.

Solution Approach

The solution involves a single pass through the array, utilizing two variables: one to count the current number of consecutive 1s and another to store the maximum count encountered so far.

Steps:

1. Initialize Counters: Set maxCount and currentCount to zero.
2. Traverse the Array: Iterate over each element in the array.
   • If the element is 1, increment currentCount.
   • If the element is 0, compare currentCount with maxCount and update maxCount if necessary. Then reset currentCount to zero.
3. Final Comparison: After the loop, a final comparison is needed to ensure the maximum count is captured if the array ends with 1s.

C++ Code

#include <vector>
#include <algorithm>

int findMaxConsecutiveOnes(std::vector<int>& nums) {
    int maxCount = 0, currentCount = 0;
    
    for (int num : nums) {
        if (num == 1) {
            currentCount++;
            maxCount = std::max(maxCount, currentCount);
        } else {
            currentCount = 0;
        }
    }
    
    return maxCount;
}

Java Code

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int maxCount = 0, currentCount = 0;

        for (int num : nums) {
            if (num == 1) {
                currentCount++;
                maxCount = Math.max(maxCount, currentCount);
            } else {
                currentCount = 0;
            }
        }

        return maxCount;
    }
}

Time and Space Complexity Analysis

The above solution has a time complexity of O(n), where n is the length of the array. This is because we make a single pass over the array, evaluating each element once.

The space complexity is O(1) since the solution uses only a fixed amount of additional space regardless of the input size. It efficiently works in place.

Common Mistakes to Avoid

• Resetting Max Count: Ensure that the maximum count is only updated after comparisons, particularly when re-encountering a 0.
• Edge Cases: Remember to handle edge cases, such as when the array contains no 1s, is empty, or consists entirely of 1s.

Similar Problems on LeetCode

• Problem 485: Max Consecutive Ones II
• Problem 485: Max Consecutive Ones III
• Problem 1052: Grumpy Bookstore Owner

Additional Resources and References

To deepen your understanding and practice the concepts involved in this problem, refer to:

• Data Structures - Arrays
• LeetCode Arrays Tutorial
• Iterative Techniques

These resources provide insightful information about arrays, iterations, and efficiently applying similar patterns to solving algorithmic problems.