Maximum consecutive one’s (or zeros) in a binary array

Given a binary array arr[] consisting of only 0s and 1s, find the length of the longest contiguous sequence of either 1s or 0s in the array.

Examples :

Input: arr[] = [0, 1, 0, 1, 1, 1, 1]
Output: 4
Explanation: The maximum number of consecutive 1’s in the array is 4 from index 3-6.

Input: arr[] = [0, 0, 1, 0, 1, 0]
Output: 2
Explanation: The maximum number of consecutive 0’s in the array is 2 from index 0-1.

Input: arr[] = [0, 0, 0, 0]
Output: 4
Explanation: The maximum number of consecutive 0’s in the array is 4.

[Approach - 1] Using Simple Traversal - O(n) Time and O(1) Space

The idea is to traverse the array once and count how many times the same value repeats consecutively. The thought process is to use a counter that increases when the current element matches the previous one. If a change is detected, we update the maximum streak and reset the count.

Steps-By-Step Approach:

• Initialize two variables: maxCount = 0 to store result and count = 1 for current streak length.
• Iterate through the array starting from index 1 to compare each element with the previous one.
• If arr[i] equals arr[i - 1], it means the streak continues, so increment count by 1.
• If they differ, update maxCount with the current count and reset count = 1 for a new streak.
• Continue this comparison until the loop finishes checking all elements in the array.
• After the loop, take the maximum of maxCount and count to cover the last streak.
• Return maxCount which holds the maximum number of consecutive 1s or 0s in the array.

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[Approach - 2] Using Bit Manipulation - O(n) Time and O(1) Space

The idea is to use XOR (^) to check if consecutive elements are the same. As XOR of two numbers is 0 if both numbers are same. So, If prev ^ num == 0, the sequence continues; else we update the maximum streak and reset the count.

Steps to implement the above idea:

• Initialize three variables: maxCount = 0, count = 0, and prev = -1 to track current and previous bits.
• Traverse the array, use XOR to compare current num with prev, if equal, increment count for the current streak.
• If prev and num differ, update maxCount with the current count, then reset count to 1.
• Update prev to current num in every iteration to track the previous element for the next comparison.
• After loop ends, compare and return the maximum of maxCount and count to cover last streak.

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