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

Given a binary circular array of size N, the task is to find the count maximum number of consecutive 1’s present in the circular array.

Examples:

Input: arr[] = {1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1}
Output: 6
The last 4 and first 2 positions have 6 consecutive ones.

Input: a[] = {0, 1, 0, 1, 0, 1, 0, 1, 0, 1}
Output: 1

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The following steps can be followed to solve the above problem:

1. Instead of creating an array of size 2*N to implement the circular array, we can use the modulus operator to traverse the array circularly.
2. Iterate from 0 to 2*N and find the consecutive number of 1’s as:
• Traverse array from left to right.
• Everytime while traversing, calculate the current index as (i%N) in order to traverse the array circularly when i>N.
• 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.
3. Break out of the loop when a[i]==0 and i>=n to reduce the time complexity in some cases.

Below is the implementation of the above approach:

 `// C++ program to count maximum consecutive ` `// 1's in a binary circular array ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function to return the count of maximum  ` `// consecutive 1's in a binary circular array ` `int` `getMaxLength(``bool` `arr[], ``int` `n) ` `{ ` `    ``// intitialize count ` `    ``int` `count = 0; ` ` `  `    ``// initialize max ` `    ``int` `result = 0; ` ` `  `    ``for` `(``int` `i = 0; i < 2 * n; i++) { ` ` `  `        ``// Reset count when 0 is found ` `        ``// use mod operator for circular array ` `        ``if` `(arr[i % n] == 0) { ` `            ``count = 0; ` ` `  `            ``// Condition to reduce time complexity ` `            ``if` `(i >= n) ` `                ``break``; ` `        ``} ` ` `  `        ``// If 1 is found, increment count ` `        ``// and update result if count becomes ` `        ``// more. ` `        ``else` `{ ` `            ``// increase count ` `            ``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); ` `    ``cout << getMaxLength(arr, n) << endl; ` `    ``return` `0; ` `} `

 `// Java program to count maximum consecutive  ` `// 1's in a binary circular array  ` `class` `GfG ` `{  ` ` `  `// Function to return the count of maximum  ` `// consecutive 1's in a binary circular array  ` `static` `int` `getMaxLength(``int` `arr[], ``int` `n)  ` `{  ` `    ``// intitialize count  ` `    ``int` `count = ``0``;  ` ` `  `    ``// initialize max  ` `    ``int` `result = ``0``;  ` ` `  `    ``for` `(``int` `i = ``0``; i < ``2` `* n; i++) ` `    ``{  ` ` `  `        ``// Reset count when 0 is found  ` `        ``// use mod operator for circular array  ` `        ``if``(arr[i % n] == ``0``)  ` `        ``{  ` `            ``count = ``0``;  ` ` `  `            ``// Condition to reduce time complexity  ` `            ``if` `(i >= n)  ` `                ``break``;  ` `        ``}  ` ` `  `        ``// If 1 is found, increment count  ` `        ``// and update result if count becomes  ` `        ``// more.  ` `        ``else` `        ``{  ` `            ``// increase count  ` `            ``count++;  ` `            ``result = Math.max(result, count);  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `result;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `main(String[] args)  ` `{  ` `    ``int` `arr[] = ``new` `int``[] { ``1``, ``1``, ``0``, ``0``, ``1``, ``0``,  ` `                            ``1``, ``0``, ``1``, ``1``, ``1``, ``1` `};  ` `    ``int` `n = arr.length;  ` `    ``System.out.println(getMaxLength(arr, n));  ` `}  ` `}  ` ` `  `// This code is contributed by Prerna Saini `

 `# Python 3 program to count maximum consecutive ` `# 1's in a binary circular array ` ` `  `# Function to return the count of  ` `# maximum consecutive 1's in a  ` `# binary circular array ` `def` `getMaxLength(arr, n): ` `     `  `    ``# intitialize count ` `    ``count ``=` `0` ` `  `    ``# initialize max ` `    ``result ``=` `0` ` `  `    ``for` `i ``in` `range``(``2` `*` `n): ` `         `  `        ``# Reset count when 0 is found ` `        ``# use mod operator for circular array ` `        ``if` `(arr[i ``%` `n] ``=``=` `0``): ` `            ``count ``=` `0` ` `  `            ``# Condition to reduce time complexity ` `            ``if` `(i >``=` `n): ` `                ``break` ` `  `        ``# 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 ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``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 ` `# Surendra_Gangwar `

 `// C# program to count maximum consecutive  ` `// 1's in a binary circular array  ` `using` `System; ` ` `  `class` `GFG ` `{  ` ` `  `// Function to return the count of maximum  ` `// consecutive 1's in a binary circular array  ` `static` `int` `getMaxLength(``int``[] arr, ``int` `n)  ` `{  ` `    ``// intitialize count  ` `    ``int` `count = 0;  ` ` `  `    ``// initialize max  ` `    ``int` `result = 0;  ` ` `  `    ``for` `(``int` `i = 0; i < 2 * n; i++) ` `    ``{  ` ` `  `        ``// Reset count when 0 is found  ` `        ``// use mod operator for circular array  ` `        ``if``(arr[i % n] == 0)  ` `        ``{  ` `            ``count = 0;  ` ` `  `            ``// Condition to reduce time complexity  ` `            ``if` `(i >= n)  ` `                ``break``;  ` `        ``}  ` ` `  `        ``// If 1 is found, increment count  ` `        ``// and update result if count becomes  ` `        ``// more.  ` `        ``else` `        ``{  ` `            ``// increase count  ` `            ``count++;  ` `            ``result = Math.Max(result, count);  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `result;  ` `}  ` ` `  `// Driver code  ` `public` `static` `void` `Main()  ` `{  ` `    ``int``[] arr = ``new` `int``[] { 1, 1, 0, 0, 1, 0,  ` `                            ``1, 0, 1, 1, 1, 1 };  ` `    ``int` `n = arr.Length;  ` `    ``Console.WriteLine(getMaxLength(arr, n));  ` `}  ` `}  ` ` `  `// This code is contributed by Code_Mech. `

 `= ``\$n``)  ` `                ``break``;  ` `        ``}  ` ` `  `        ``// If 1 is found, increment count  ` `        ``// and update result if count becomes  ` `        ``// more.  ` `        ``else` `        ``{  ` `            ``// increase count  ` `            ``\$count``++;  ` `            ``\$result` `= max(``\$result``, ``\$count``);  ` `        ``}  ` `    ``}  ` ` `  `    ``return` `\$result``;  ` `}  ` ` `  `// Driver code  ` `\$arr` `= ``array``(1, 1, 0, 0, 1, 0,  ` `             ``1, 0, 1, 1, 1, 1 );  ` `\$n` `= sizeof(``\$arr``);  ` `echo``(getMaxLength(``\$arr``, ``\$n``));  ` ` `  `// This code is contributed by Code_Mech ` `?> `

Output:
```6
```

Striver(underscore)79 at Codechef and codeforces D

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