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

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  • Difficulty Level : Medium
  • Last Updated : 11 Jul, 2022
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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:
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:
 

 

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 ith 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


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