Longest subarray of non-empty cells after removal of at most a single empty cell

Given a binary array arr[], the task is to find the longest subarray of non-empty cells after the removal of at most 1 empty cell.

The array indices filled with 0 are known as empty cell whereas the indices filled with 1 are known as non-empty cells.

Examples:

Input: arr[] = {1, 1, 0, 1} 
Output: 3 
Explanation:
Removal of 0 modifies the array to {1, 1, 1}, thus maximizing the length of the subarray to 3.
Input: arr[] = {1, 1, 1, 1, 1} 
Output: 5 

Approach:
The idea is to store the frequencies of 1 in the prefixes and suffixes of every index to calculate longest consecutive sequences of 1’s on both the directions from a particular index. Follow the steps below to solve the problem: 

• Initialize two arrays l[] and r[] which stores the length of longest consecutive 1s in the array arr[] from left and right side of the array respectively.
• Iterate over the input array over indices (0, N) and increase count by 1 for every arr[i] = 1. Otherwise, store the value of count till the (i – 1)^th index in l[i] reset count to zero. 
  
• Similarly, repeat the above steps by traversing over indices [N – 1, 0] store the count from right in r[].
• For every i^th index index which contains 0, calculate the length of non-empty subarray possible by removal of that 0, which is equal to l[i] + r[i].
• Compute the maximum of all such lengths and print the result.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find the maximum length
// of a subarray of 1s after removing
// at most one 0
int longestSubarray(int a[], int n)
{
    // Stores the count of consecutive
    // 1's from left
    int l[n];

    // Stores the count of consecutive
    // 1's from right
    int r[n];

    // Traverse left to right
    for (int i = 0, count = 0;
        i < n; i++) {

        // If cell is non-empty
        if (a[i] == 1)

            // Increase count
            count++;

        // If cell is empty
        else {

            // Store the count of
            // consecutive 1's
            // till (i - 1)-th index
            l[i] = count;
            count = 0;
        }
    }

    // Traverse from right to left
    for (int i = n - 1, count = 0;
        i >= 0; i--) {

        if (a[i] == 1)
            count++;

        else {

            // Store the count of
            // consecutive 1s
            // till (i + 1)-th index
            r[i] = count;
            count = 0;
        }
    }

    // Stores the length of
    // longest subarray
    int ans = -1;
    for (int i = 0; i < n; ++i) {

        if (a[i] == 0)

            // Store the maximum
            ans = max(ans, l[i] + r[i]);
    }

    // If array a contains only 1s
    // return n else return ans
    return ans < 0 ? n : ans;
}

// Driver Code
int main()
{
    int arr[] = { 0, 1, 1, 1, 0, 1,
                0, 1, 1 };

    int n = sizeof(arr) / sizeof(arr[0]);

    cout << longestSubarray(arr, n);

    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function to find the maximum length
// of a subarray of 1s after removing
// at most one 0
public static int longestSubarray(int[] a,
                                int n)
{

    // Stores the count of consecutive
    // 1's from left
    int[] l = new int[n];

    // Stores the count of consecutive
    // 1's from right
    int[] r = new int[n];

    // Traverse left to right
    for(int i = 0, count = 0;
            i < n; i++)
    {

    // If cell is non-empty
    if (a[i] == 1)

        // Increase count
        count++;

    // If cell is empty
    else
    {

        // Store the count of
        // consecutive 1's
        // till (i - 1)-th index
        l[i] = count;
        count = 0;
    }
    }

    // Traverse from right to left
    for(int i = n - 1, count = 0;
            i >= 0; i--)
    {
    if (a[i] == 1)
        count++;

    else
    {

        // Store the count of
        // consecutive 1s
        // till (i + 1)-th index
        r[i] = count;
        count = 0;
        }
    }

    // Stores the length of
    // longest subarray
    int ans = -1;
    for(int i = 0; i < n; ++i)
    {
    if (a[i] == 0)

        // Store the maximum
        ans = Math.max(ans, l[i] + r[i]);
    }

    // If array a contains only 1s
    // return n else return ans
    return ans < 0 ? n : ans;
}

// Driver code
public static void main(String[] args)
{
    int[] arr = { 0, 1, 1, 1, 0,
                1, 0, 1, 1 };
    int n = arr.length;

    System.out.println(longestSubarray(arr, n));
}
}

// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach

# Function to find the maximum length
# of a subarray of 1s after removing
# at most one 0
def longestSubarray(a, n):

    # Stores the count of consecutive
    # 1's from left
    l = [0] * (n)

    # Stores the count of consecutive
    # 1's from right
    r = [0] * (n)

    count = 0

    # Traverse left to right
    for i in range(n):

        # If cell is non-empty
        if (a[i] == 1):

            # Increase count
            count += 1

        # If cell is empty
        else:

            # Store the count of
            # consecutive 1's
            # till (i - 1)-th index
            l[i] = count
            count = 0

    count = 0
    # Traverse from right to left
    for i in range(n - 1, -1, -1):
        if (a[i] == 1):
            count += 1

        else:

            # Store the count of
            # consecutive 1s
            # till (i + 1)-th index
            r[i] = count
            count = 0

    # Stores the length of
    # longest subarray
    ans = -1
    for i in range(n):
        if (a[i] == 0):

            # Store the maximum
            ans = max(ans, l[i] + r[i])

    # If array a contains only 1s
    # return n else return ans
    return ans < 0 and n or ans

# Driver code
arr = [ 0, 1, 1, 1, 0, 1, 0, 1, 1 ]

n = len(arr)

print(longestSubarray(arr, n))

# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;

class GFG{

// Function to find the maximum length
// of a subarray of 1s after removing
// at most one 0
public static int longestSubarray(int[] a,
                                  int n)
{

    // Stores the count of consecutive
    // 1's from left
    int[] l = new int[n];

    // Stores the count of consecutive
    // 1's from right
    int[] r = new int[n];

    // Traverse left to right
    for(int i = 0, count = 0; i < n; i++)
    {

        // If cell is non-empty
        if (a[i] == 1)

            // Increase count
            count++;

        // If cell is empty
        else
        {

            // Store the count of
            // consecutive 1's
            // till (i - 1)-th index
            l[i] = count;
            count = 0;
        }
    }

    // Traverse from right to left
    for(int i = n - 1, count = 0;
            i >= 0; i--)
    {
    if (a[i] == 1)
        count++;

    else
    {

        // Store the count of
        // consecutive 1s
        // till (i + 1)-th index
        r[i] = count;
        count = 0;
        }
    }

    // Stores the length of
    // longest subarray
    int ans = -1;
    for(int i = 0; i < n; ++i)
    {
        if (a[i] == 0)

            // Store the maximum
            ans = Math.Max(ans, l[i] + r[i]);
    }

    // If array a contains only 1s
    // return n else return ans
    return ans < 0 ? n : ans;
}


// Driver code
public static void Main()
{
    int[] arr = { 0, 1, 1, 1, 0,
                  1, 0, 1, 1 };
    int n = arr.Length;

    Console.Write(longestSubarray(arr, n));
}
}

// This code is contributed by sanjoy_62

Javascript

<script>
// javascript program for the above approach
// Function to find the maximum length
    // of a subarray of 1s after removing
    // at most one 0
    function longestSubarray(a , n)
    {

        // Stores the count of consecutive
        // 1's from left
        var l = Array(n).fill(0);

        // Stores the count of consecutive
        // 1's from right
        var r = Array(n).fill(0);

        // Traverse left to right
        for (i = 0, count = 0; i < n; i++)
        {

            // If cell is non-empty
            if (a[i] == 1)

                // Increase count
                count++;

            // If cell is empty
            else {

                // Store the count of
                // consecutive 1's
                // till (i - 1)-th index
                l[i] = count;
                count = 0;
            }
        }

        // Traverse from right to left
        for (i = n - 1, count = 0; i >= 0; i--) {
            if (a[i] == 1)
                count++;

            else {

                // Store the count of
                // consecutive 1s
                // till (i + 1)-th index
                r[i] = count;
                count = 0;
            }
        }

        // Stores the length of
        // longest subarray
        var ans = -1;
        for (i = 0; i < n; ++i) {
            if (a[i] == 0)

                // Store the maximum
                ans = Math.max(ans, l[i] + r[i]);
        }

        // If array a contains only 1s
        // return n else return ans
        return ans < 0 ? n : ans;
    }

    // Driver code
        var arr = [ 0, 1, 1, 1, 0, 1, 0, 1, 1 ];
        var n = arr.length;
        document.write(longestSubarray(arr, n));

// This code is contributed by Rajput-Ji
</script>

Output:

4

Time Complexity: O(N) where n is number of elements in given array. As, we are using a loop to traverse N times so it will cost us O(N) time 
Auxiliary Space: O(N), as we are using extra space.

Related Topic: Subarrays, Subsequences, and Subsets in Array