Maximum consecutive one’s (or zeros) in a binary circular array
Last Updated :
11 Jul, 2025
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
A Naive Solution is create an array of size 2*N where N is size of input array. We copy the input array twice in this array. Now we need to find the longest consecutive 1s in this binary array in one traversal.
Efficient Solution : The following steps can be followed to solve the above problem:
- 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.
- 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.
- 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++
// C++ program to count maximum consecutive
// 1's in a binary circular array
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of maximum
// consecutive 1's in a binary circular array
int getMaxLength(bool arr[], int n)
{
// Starting index
int start = 0;
// To store the maximum length of the
// prefix of the given array with all 1s
int preCnt = 0;
while (start < n && arr[start] == 1) {
preCnt++;
start++;
}
// Ending index
int end = n - 1;
// To store the maximum length of the
// suffix of the given array with all 1s
int suffCnt = 0;
while (end >= 0 && arr[end] == 1) {
suffCnt++;
end--;
}
// The array contains all 1s
if (start > end)
return n;
// Find the maximum length subarray
// with all 1s from the remaining not
// yet traversed subarray
int midCnt = 0;
// To store the result for middle 1s
int result = 0;
for (int i = start; i <= end; i++) {
if (arr[i] == 1) {
midCnt++;
result = max(result, midCnt);
}
else {
midCnt = 0;
}
}
// (preCnt + suffCnt) is the subarray when
// the given array is assumed to be circular
return max(result, preCnt + suffCnt);
}
// 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);
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)
{
// Starting index
int start = 0;
// To store the maximum length of the
// prefix of the given array with all 1s
int preCnt = 0;
while (start < n && arr[start] == 1) {
preCnt++;
start++;
}
// Ending index
int end = n - 1;
// To store the maximum length of the
// suffix of the given array with all 1s
int suffCnt = 0;
while (end >= 0 && arr[end] == 1) {
suffCnt++;
end--;
}
// The array contains all 1s
if (start > end)
return n;
// Find the maximum length subarray
// with all 1s from the remaining not
// yet traversed subarray
int midCnt = 0;
// To store the result for middle 1s
int result = 0;
for (int i = start; i <= end; i++) {
if (arr[i] == 1) {
midCnt++;
result = Math.max(result, midCnt);
}
else {
midCnt = 0;
}
}
// (preCnt + suffCnt) is the subarray when
// the given array is assumed to be circular
return Math.max(result, preCnt + suffCnt);
}
// 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):
# Starting index
start = 0
# To store the maximum length of the
# prefix of the given array with all 1s
preCnt = 0
while(start < n and arr[start] == 1):
preCnt = preCnt + 1
start = start + 1
# Ending index
end = n - 1
# To store the maximum length of the
# suffix of the given array with all 1s
suffCnt = 0
while(end >= 0 and arr[end] == 1):
suffCnt = suffCnt + 1
end = end - 1
# The array contains all 1s
if(start > end):
return n
# Find the maximum length subarray
# with all 1s from the remaining not
# yet traversed subarray
midCnt = 0
i = start
# To store the result for middle 1s
result = 0
while(i <= end):
if(arr[i] == 1):
midCnt = midCnt + 1
result = max(result, midCnt)
else:
midCnt = 0
i = i + 1
# (preCnt + suffCnt) is the subarray when
# the given array is assumed to be circular
return max(result, preCnt + suffCnt)
# 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)
{
// Starting index
int start = 0;
// To store the maximum length of the
// prefix of the given array with all 1s
int preCnt = 0;
while (start < n && arr[start] == 1) {
preCnt++;
start++;
}
// Ending index
int end = n - 1;
// To store the maximum length of the
// suffix of the given array with all 1s
int suffCnt = 0;
while (end >= 0 && arr[end] == 1) {
suffCnt++;
end--;
}
// The array contains all 1s
if (start > end)
return n;
// Find the maximum length subarray
// with all 1s from the remaining not
// yet traversed subarray
int midCnt = 0;
// To store the result for middle 1s
int result = 0;
for (int i = start; i <= end; i++) {
if (arr[i] == 1) {
midCnt++;
result = Math.Max(result, midCnt);
}
else {
midCnt = 0;
}
}
// (preCnt + suffCnt) is the subarray when
// the given array is assumed to be circular
return Math.Max(result, preCnt + suffCnt);
}
// Driver code
public static void Main()
{
int[] arr = new int[] { 1, 1, 0, 0, 1, 0,
1, 0, 1, 1, 1, 0 };
int n = arr.Length;
Console.WriteLine(getMaxLength(arr, n));
}
}
// This code is contributed by Code_Mech.
<script>
// javascript 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
function getMaxLength(arr , n)
{
// Starting index
var start = 0;
// To store the maximum length of the
// prefix of the given array with all 1s
var preCnt = 0;
while (start < n && arr[start] == 1) {
preCnt++;
start++;
}
// Ending index
var end = n - 1;
// To store the maximum length of the
// suffix of the given array with all 1s
var suffCnt = 0;
while (end >= 0 && arr[end] == 1) {
suffCnt++;
end--;
}
// The array contains all 1s
if (start > end)
return n;
// Find the maximum length subarray
// with all 1s from the remaining not
// yet traversed subarray
var midCnt = 0;
// To store the result for middle 1s
var result = 0;
for (i = start; i <= end; i++) {
if (arr[i] == 1) {
midCnt++;
result = Math.max(result, midCnt);
} else {
midCnt = 0;
}
}
// (preCnt + suffCnt) is the subarray when
// the given array is assumed to be circular
return Math.max(result, preCnt + suffCnt);
}
// Driver code
var arr = [ 1, 1, 0, 0, 1, 0, 1, 0, 1, 1, 1, 1 ];
var n = arr.length;
document.write(getMaxLength(arr, n));
// This code is contributed by umadevi9616
</script>
Time Complexity: O(N), as we are using a loop to traverse N times. Where N is the number of elements in the array.
Auxiliary Space: O(1), as we are not using any extra space.
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