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

• Difficulty Level : Easy
• Last Updated : 24 Dec, 2021

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 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.

## 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

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## Javascript

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Output:

`4`

Time Complexity : O(n)
Auxiliary Space : O(1)
Exercise:
Maximum consecutive zeros in a binary array.
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