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<bits/stdc++.h>
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

<?php
// PHP 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]
function getMaxLength($arr, $n)
{
    // initialize count
    $count = 0;

    // initialize max
    $result = 0;

    for ($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
        {
            // 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) / sizeof($arr[0]);
echo getMaxLength($arr, $n) ;

// This code is contributed by nitin mittal.
?>

Javascript

<script>

// JavaScript 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]
function getMaxLength(arr, n) {
    // initialize count
    let count = 0;

    // initialize max
    let result = 0;

    for (let 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 {
            // increase count
            count++;
            result = Math.max(result, count);
        }
    }

    return result;
}

// Driver code
let arr = new Array(1, 1, 0, 0, 1, 0,
    1, 0, 1, 1, 1, 1);
let n = arr.length;
document.write(getMaxLength(arr, n));

// This code is contributed by gfgking

</script>

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 <iostream>
#include <vector>

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.

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