Open In App

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

Given binary array, find count of maximum number of consecutive 1’s present in the array.

Examples :

```Input  : arr[] = {1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1}
Output : 4

Input  : arr[] = {0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1}
Output : 1```

A simple solution is consider every subarray and count 1’s in every subarray. Finally return size of largest subarray with all 1’s. An efficient solution is traverse array from left to right. If we see a 1, we increment count and compare it with maximum so far. If we see a 0, we reset count as 0.

Implementation:

## CPP

 `// C++ program to count maximum consecutive``// 1's in a binary array.``#include``using` `namespace` `std;` `// Returns count of maximum consecutive 1's``// in binary array arr[0..n-1]``int` `getMaxLength(``bool` `arr[], ``int` `n)``{``    ``int` `count = 0; ``//initialize count``    ``int` `result = 0; ``//initialize max` `    ``for` `(``int` `i = 0; i < n; i++)``    ``{``        ``// Reset count when 0 is found``        ``if` `(arr[i] == 0)``            ``count = 0;` `        ``// If 1 is found, increment count``        ``// and update result if count becomes``        ``// more.``        ``else``        ``{``            ``count++;``//increase count``            ``result = max(result, count);``        ``}``    ``}` `    ``return` `result;``}` `// Driver code``int` `main()``{``    ``bool` `arr[] = {1, 1, 0, 0, 1, 0, 1, 0,``                  ``1, 1, 1, 1};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr[0]);``    ``cout << getMaxLength(arr, n) << endl;``    ``return` `0;``}`

## Java

 `// Java program to count maximum consecutive``// 1's in a binary array.``class` `GFG {``    ` `    ``// Returns count of maximum consecutive 1's``    ``// in binary array arr[0..n-1]``    ``static` `int` `getMaxLength(``boolean` `arr[], ``int` `n)``    ``{``        ` `        ``int` `count = ``0``; ``//initialize count``        ``int` `result = ``0``; ``//initialize max``    ` `        ``for` `(``int` `i = ``0``; i < n; i++)``        ``{``            ` `            ``// Reset count when 0 is found``            ``if` `(arr[i] == ``false``)``                ``count = ``0``;``    ` `            ``// If 1 is found, increment count``            ``// and update result if count becomes``            ``// more.``            ``else``            ``{``                ``count++;``//increase count``                ``result = Math.max(result, count);``            ``}``        ``}``    ` `        ``return` `result;``    ``}``    ` `    ``// Driver method``    ``public` `static` `void` `main(String[] args)``    ``{``        ``boolean` `arr[] = {``true``, ``true``, ``false``, ``false``,``                         ``true``, ``false``, ``true``, ``false``,``                           ``true``, ``true``, ``true``, ``true``};``                           ` `        ``int` `n = arr.length;``        ` `        ``System.out.println(getMaxLength(arr, n));``    ``}``}` `// This code is contributed by Anant Agarwal.`

## Python3

 `# Python 3 program to count``# maximum consecutive 1's``# in a binary array.` `# Returns count of maximum``# consecutive 1's in binary``# array arr[0..n-1]``def` `getMaxLength(arr, n):` `    ``# initialize count``    ``count ``=` `0``    ` `    ``# initialize max``    ``result ``=` `0` `    ``for` `i ``in` `range``(``0``, n):``    ` `        ``# Reset count when 0 is found``        ``if` `(arr[i] ``=``=` `0``):``            ``count ``=` `0` `        ``# If 1 is found, increment count``        ``# and update result if count``        ``# becomes more.``        ``else``:``            ` `            ``# increase count``            ``count``+``=` `1``            ``result ``=` `max``(result, count)``        ` `    ``return` `result` `# Driver code``arr ``=` `[``1``, ``1``, ``0``, ``0``, ``1``, ``0``, ``1``,``             ``0``, ``1``, ``1``, ``1``, ``1``]``n ``=` `len``(arr)` `print``(getMaxLength(arr, n))` `# This code is contributed by Smitha Dinesh Semwal`

## C#

 `// C# program to count maximum``// consecutive 1's in a binary array.``using` `System;` `class` `GFG {``    ` `    ``// Returns count of maximum consecutive``    ``// 1's in binary array arr[0..n-1]``    ``static` `int` `getMaxLength(``bool` `[]arr, ``int` `n)``    ``{``        ` `        ``int` `count = 0; ``//initialize count``        ``int` `result = 0; ``//initialize max``    ` `        ``for` `(``int` `i = 0; i < n; i++)``        ``{``            ` `            ``// Reset count when 0 is found``            ``if` `(arr[i] == ``false``)``                ``count = 0;``    ` `            ``// If 1 is found, increment count``            ``// and update result if count``            ``// becomes more.``            ``else``            ``{``                ``count++; ``//increase count``                ``result = Math.Max(result, count);``            ``}``        ``}``    ` `        ``return` `result;``    ``}``    ` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``bool` `[]arr = {``true``, ``true``, ``false``, ``false``,``                      ``true``, ``false``, ``true``, ``false``,``                      ``true``, ``true``, ``true``, ``true``};``                            ` `        ``int` `n = arr.Length;``        ` `        ``Console.Write(getMaxLength(arr, n));``    ``}``}` `// This code is contributed by Nitin Mittal.`

## PHP

 ``

## Javascript

 ``

Output

`4`

Time Complexity : O(n)
Auxiliary Space : O(1)

Approach 2:
Another approach to solve this problem is to use the bitwise AND operation to count the number of consecutive 1’s in the binary representation of the given number. We iterate over the bits of the number from the right and use a mask to check the value of each bit. If the bit is 1, we increment a counter. If the bit is 0, we reset the counter to zero. We update the maxCount if the counter is greater than maxCount.

Here’s the implementation of the above approach in C++:

## C++

 `#include ``#include ` `using` `namespace` `std;` `int` `maxConsecutiveOnes(vector<``int``>& nums) {``    ``int` `max_count = 0, current_count = 0, mask = 0;``    ``for` `(``int` `i = 0; i < nums.size(); i++) {``        ``if` `(nums[i] == 1) {``            ``mask = (mask << 1) | 1;``        ``} ``else` `{``            ``mask = mask << 1;``        ``}``        ``if` `((nums[i] & mask) != 0) {``            ``current_count++;``        ``} ``else` `{``            ``max_count = max(max_count, current_count);``            ``current_count = 0;``            ``mask = 0;``        ``}``    ``}``    ``max_count = max(max_count, current_count);``    ``return` `max_count;``}` `int` `main() {``    ``vector<``int``> nums = {1, 1, 0, 0, 1, 0, 1, 0,``                  ``1, 1, 1, 1};``    ``int` `max_ones = maxConsecutiveOnes(nums);``    ``cout << ``"Maximum consecutive ones: "` `<< max_ones << endl;``    ``return` `0;``}`

Output

```Maximum consecutive ones: 4
```

Time Complexity: O(logn), where n is the decimal representation of the given number.
Space Complexity: O(1).

Exercise:
Maximum consecutive zeros in a binary array.

This article is contributed by Smarak Chopdar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.