Maximize length of Subarray of 1's after removal of a pair of consecutive Array elements
Last Updated :
15 Jul, 2025
Given a binary array arr[] consisting of N elements, the task is to find the maximum possible length of a subarray of only 1’s, after deleting a single pair of consecutive array elements. If no such subarray exists, print -1.
Examples:
Input: arr[] = {1, 1, 1, 0, 0, 1}
Output: 4
Explanation:
Removal of the pair {0, 0} modifies the array to {1, 1, 1, 1}, thus maximizing the length of the longest possible subarray consisting only of 1's.
Input: arr[] = {1, 1, 1, 0, 0, 0, 1}
Output: 3
Explanation:
Removal of any consecutive pair from the subarray {0, 0, 0, 1} maintains the longest possible subarray of 1's, i.e. {1, 1, 1}.
Naive Approach:
The simplest approach to solve the problem is to generate all possible pairs of consecutive elements from the array and for each pair, calculate the maximum possible length of a subarray of 1's. Finally, print the maximum possible length of such a subarray obtained.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach:
Follow the steps given below to solve the problem:
- Initialize an auxiliary 2D vector V.
- Keep track of all the contiguous subarrays consisting of only 1's.
- Store the length, starting index, and ending index of the subarrays.
- Now, calculate the number of 0's between any two subarrays of 1's.
- Based on the count of 0's obtained, update the maximum length of subarrays possible:
- If exactly two 0's are present between two subarrays, update the maximum possible length by comparing the combined length of both subarrays.
- If exactly one 0 is present between two subarrays, update the maximum possible length by comparing the combined length of both subarrays - 1.
- Finally, print the maximum possible length obtained
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the maximum
// subarray length of ones
int maxLen(int A[], int N)
{
// Stores the length, starting
// index and ending index of the
// subarrays
vector<vector<int> > v;
for (int i = 0; i < N; i++) {
if (A[i] == 1) {
// S : starting index
// of the sub-array
int s = i, len;
// Traverse only continuous 1s
while (A[i] == 1 && i < N) {
i++;
}
// Calculate length of the
// sub-array
len = i - s;
// v[i][0] : Length of subarray
// v[i][1] : Starting Index of subarray
// v[i][2] : Ending Index of subarray
v.push_back({ len, s, i - 1 });
}
}
// If no such sub-array exists
if (v.size() == 0) {
return -1;
}
int ans = 0;
// Traversing through the subarrays
for (int i = 0; i < v.size() - 1; i++) {
// Update maximum length
ans = max(ans, v[i][0]);
// v[i+1][1] : Starting index of
// the next Sub-Array
// v[i][2] : Ending Index of the
// current Sub-Array
// v[i+1][1] - v[i][2] - 1 : Count of
// zeros between the two sub-arrays
if (v[i + 1][1] - v[i][2] - 1 == 2) {
// Update length of both subarrays
// to the maximum
ans = max(ans, v[i][0] + v[i + 1][0]);
}
if (v[i + 1][1] - v[i][2] - 1 == 1) {
// Update length of both subarrays - 1
// to the maximum
ans = max(ans, v[i][0] + v[i + 1][0] - 1);
}
}
// Check if the last subarray has
// the maximum length
ans = max(v[v.size() - 1][0], ans);
return ans;
}
// Driver Code
int main()
{
int arr[] = { 1, 0, 1, 0, 0, 1 };
int N = sizeof(arr) / sizeof(arr[0]);
cout << maxLen(arr, N) << endl;
return 0;
}
// Java program to implement
// the above approach
import java.io.*;
import java.util.*;
class GFG {
// Function to find the maximum
// subarray length of ones
static int maxLen(int A[], int N)
{
// Stores the length, starting
// index and ending index of the
// subarrays
List<List<Integer> > v = new ArrayList<>();
for (int i = 0; i < N; i++) {
if (A[i] == 1) {
// S : starting index
// of the sub-array
int s = i, len;
// Traverse only continuous 1s
while (i < N && A[i] == 1) {
i++;
}
// Calculate length of the
// sub-array
len = i - s;
// v[i][0] : Length of subarray
// v[i][1] : Starting Index of subarray
// v[i][2] : Ending Index of subarray
v.add(Arrays.asList(len, s, i - 1));
}
}
// If no such sub-array exists
if (v.size() == 0) {
return -1;
}
int ans = 0;
// Traversing through the subarrays
for (int i = 0; i < v.size() - 1; i++) {
// Update maximum length
ans = Math.max(ans, v.get(i).get(0));
// v[i+1][1] : Starting index of
// the next Sub-Array
// v[i][2] : Ending Index of the
// current Sub-Array
// v[i+1][1] - v[i][2] - 1 : Count of
// zeros between the two sub-arrays
if (v.get(i + 1).get(1) - v.get(i).get(2) - 1
== 2) {
// Update length of both subarrays
// to the maximum
ans = Math.max(ans,
v.get(i).get(0)
+ v.get(i + 1).get(0));
}
if (v.get(i + 1).get(1) - v.get(i).get(2) - 1
== 1) {
// Update length of both subarrays - 1
// to the maximum
ans = Math.max(
ans, v.get(i).get(0)
+ v.get(i + 1).get(0) - 1);
}
}
// Check if the last subarray has
// the maximum length
ans = Math.max(v.get(v.size() - 1).get(0), ans);
return ans;
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 1, 0, 1, 0, 0, 1 };
int N = arr.length;
System.out.println(maxLen(arr, N));
}
}
// This code is contributed by offbeat
# Python3 program to implement
# the above approach
# Function to find the maximum
# subarray length of ones
def maxLen(A, N):
# Stores the length, starting
# index and ending index of the
# subarrays
v = []
i = 0
while i < N:
if (A[i] == 1):
# S : starting index
# of the sub-array
s = i
# Traverse only continuous 1s
while (i < N and A[i] == 1):
i += 1
# Calculate length of the
# sub-array
le = i - s
# v[i][0] : Length of subarray
# v[i][1] : Starting Index of subarray
# v[i][2] : Ending Index of subarray
v.append([le, s, i - 1])
i += 1
# If no such sub-array exists
if (len(v) == 0):
return -1
ans = 0
# Traversing through the subarrays
for i in range(len(v) - 1):
# Update maximum length
ans = max(ans, v[i][0])
# v[i+1][1] : Starting index of
# the next Sub-Array
# v[i][2] : Ending Index of the
# current Sub-Array
# v[i+1][1] - v[i][2] - 1 : Count of
# zeros between the two sub-arrays
if (v[i + 1][1] - v[i][2] - 1 == 2):
# Update length of both subarrays
# to the maximum
ans = max(ans, v[i][0] + v[i + 1][0])
if (v[i + 1][1] - v[i][2] - 1 == 1):
# Update length of both subarrays - 1
# to the maximum
ans = max(ans, v[i][0] + v[i + 1][0] - 1)
# Check if the last subarray has
# the maximum length
ans = max(v[len(v) - 1][0], ans)
return ans
# Driver Code
if __name__ == "__main__":
arr = [1, 0, 1, 0, 0, 1]
N = len(arr)
print(maxLen(arr, N))
# This code is contributed by chitranayal
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the maximum
// subarray length of ones
static int maxLen(int []A, int N)
{
// Stores the length, starting
// index and ending index of the
// subarrays
List<List<int>> v = new List<List<int>>();
for (int i = 0; i < N; i++)
{
if (A[i] == 1)
{
// S : starting index
// of the sub-array
int s = i, len;
// Traverse only continuous 1s
while (i < N && A[i] == 1)
{
i++;
}
// Calculate length of the
// sub-array
len = i - s;
// v[i,0] : Length of subarray
// v[i,1] : Starting Index of subarray
// v[i,2] : Ending Index of subarray
List<int> l = new List<int>{len, s, i - 1};
v.Add(l);
}
}
// If no such sub-array exists
if (v.Count == 0)
{
return -1;
}
int ans = 0;
// Traversing through the subarrays
for (int i = 0; i < v.Count - 1; i++)
{
// Update maximum length
ans = Math.Max(ans, v[i][0]);
// v[i+1,1] : Starting index of
// the next Sub-Array
// v[i,2] : Ending Index of the
// current Sub-Array
// v[i+1,1] - v[i,2] - 1 : Count of
// zeros between the two sub-arrays
if (v[i + 1][1] - v[i][2] - 1
== 2) {
// Update length of both subarrays
// to the maximum
ans = Math.Max(ans,
v[i][0]
+ v[i + 1][0]);
}
if (v[i + 1][1] - v[i][2] - 1
== 1)
{
// Update length of both subarrays - 1
// to the maximum
ans = Math.Max(
ans, v[i][0]
+ v[i + 1][0] - 1);
}
}
// Check if the last subarray has
// the maximum length
ans = Math.Max(v[v.Count - 1][0], ans);
return ans;
}
// Driver Code
public static void Main(String []args)
{
int []arr = { 1, 0, 1, 0, 0, 1 };
int N = arr.Length;
Console.WriteLine(maxLen(arr, N));
}
}
// This code is contributed by Princi Singh
<script>
// Javascript program to implement
// the above approach
// Function to find the maximum
// subarray length of ones
function maxLen(A, N)
{
// Stores the length, starting
// index and ending index of the
// subarrays
var v = [];
for (var i = 0; i < N; i++) {
if (A[i] == 1) {
// S : starting index
// of the sub-array
var s = i, len;
// Traverse only continuous 1s
while (A[i] == 1 && i < N) {
i++;
}
// Calculate length of the
// sub-array
len = i - s;
// v[i][0] : Length of subarray
// v[i][1] : Starting Index of subarray
// v[i][2] : Ending Index of subarray
v.push([len, s, i - 1]);
}
}
// If no such sub-array exists
if (v.length == 0) {
return -1;
}
var ans = 0;
// Traversing through the subarrays
for (var i = 0; i < v.length - 1; i++) {
// Update maximum length
ans = Math.max(ans, v[i][0]);
// v[i+1][1] : Starting index of
// the next Sub-Array
// v[i][2] : Ending Index of the
// current Sub-Array
// v[i+1][1] - v[i][2] - 1 : Count of
// zeros between the two sub-arrays
if (v[i + 1][1] - v[i][2] - 1 == 2) {
// Update length of both subarrays
// to the maximum
ans = Math.max(ans, v[i][0] + v[i + 1][0]);
}
if (v[i + 1][1] - v[i][2] - 1 == 1) {
// Update length of both subarrays - 1
// to the maximum
ans = Math.max(ans, v[i][0] + v[i + 1][0] - 1);
}
}
// Check if the last subarray has
// the maximum length
ans = Math.max(v[v.length - 1][0], ans);
return ans;
}
// Driver Code
var arr = [1, 0, 1, 0, 0, 1];
var N = arr.length;
document.write( maxLen(arr, N));
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
Method 2:
Prefix and Suffix Arrays
Count the number of 1's between the occurrences of 0's. If we find a 0 then we need to find the maximum length of 1's if we skip those 2 consecutive elements. We can use the concepts of Prefix and Suffix arrays to solve the problem.
Find the length of consecutive 1s from the start of the array and store the count in the prefix array. Find the length of consecutive 1s from the ending of the array and store the count in the ending array.
We traverse both the arrays and find the maximum no. of 1s.
Algorithm:
- Create two arrays prefix and suffix of length n.
- Initialise prefix[0]=0,prefix[1]=0 and suffix[n-1]=0,suffix[n-2]=0. This tells us that there are no 1s before the first 2 elements and after the last 2 elements.
- Run the loop from 2 to n-1:
- If arr[i-2]==1
- else if arr[i-2]==0:
- Run the loop from n-3 to 0:
- If arr[i+2]==1
- else if arr[i-2]==0:
- Initialize answer=INT_MIN
- for i=0 to n-2: //Count the number of 1's by skipping the current and the next element.
- answer=max(answer,prefix[i+1]+suffix[i]
- print answer
Implementation:
C++
// C++ program to find the maximum count of 1s
#include <bits/stdc++.h>
using namespace std;
void maxLengthOf1s(vector<int> arr, int n)
{
vector<int> prefix(n, 0);
for (int i = 2; i < n; i++)
{
// If arr[i-2]==1 then we increment the
// count of occurrences of 1's
if (arr[i - 2]
== 1)
prefix[i] = prefix[i - 1] + 1;
// else we initialise the count with 0
else
prefix[i] = 0;
}
vector<int> suffix(n, 0);
for (int i = n - 3; i >= 0; i--)
{
// If arr[i+2]==1 then we increment the
// count of occurrences of 1's
if (arr[i + 2] == 1)
suffix[i] = suffix[i + 1] + 1;
// else we initialise the count with 0
else
suffix[i] = 0;
}
int ans = 0;
for (int i = 0; i < n - 1; i++)
{
// We get the maximum count by
// skipping the current and the
// next element.
ans = max(ans, prefix[i + 1] + suffix[i]);
}
cout << ans << "\n";
}
// Driver Code
int main()
{
int n = 6;
vector<int> arr = { 1, 1, 1, 0, 1, 1 };
maxLengthOf1s(arr, n);
return 0;
}
// Java program to find the maximum count of 1s
class GFG{
public static void maxLengthOf1s(int arr[], int n)
{
int prefix[] = new int[n];
for(int i = 2; i < n; i++)
{
// If arr[i-2]==1 then we increment
// the count of occurrences of 1's
if (arr[i - 2] == 1)
prefix[i] = prefix[i - 1] + 1;
// Else we initialise the count with 0
else
prefix[i] = 0;
}
int suffix[] = new int[n];
for(int i = n - 3; i >= 0; i--)
{
// If arr[i+2]==1 then we increment
// the count of occurrences of 1's
if (arr[i + 2] == 1)
suffix[i] = suffix[i + 1] + 1;
// Else we initialise the count with 0
else
suffix[i] = 0;
}
int ans = 0;
for(int i = 0; i < n - 1; i++)
{
// We get the maximum count by
// skipping the current and the
// next element.
ans = Math.max(ans, prefix[i + 1] +
suffix[i]);
}
System.out.println(ans);
}
// Driver code
public static void main(String[] args)
{
int n = 6;
int arr[] = { 1, 1, 1, 0, 1, 1 };
maxLengthOf1s(arr, n);
}
}
// This code is contributed by divyeshrabadiya07
# Python program to find the maximum count of 1s
def maxLengthOf1s(arr, n):
prefix = [0 for i in range(n)]
for i in range(2, n):
# If arr[i-2]==1 then we increment
# the count of occurrences of 1's
if(arr[i - 2] == 1):
prefix[i] = prefix[i - 1] + 1
# Else we initialise the count with 0
else:
prefix[i] = 0
suffix = [0 for i in range(n)]
for i in range(n - 3, -1, -1):
# If arr[i+2]==1 then we increment
# the count of occurrences of 1's
if(arr[i + 2] == 1):
suffix[i] = suffix[i + 1] + 1
# Else we initialise the count with 0
else:
suffix[i] = 0
ans = 0
for i in range(n - 1):
# We get the maximum count by
# skipping the current and the
# next element.
ans = max(ans, prefix[i + 1] + suffix[i])
print(ans)
# Driver code
n = 6
arr = [1, 1, 1, 0, 1, 1]
maxLengthOf1s(arr, n)
# This code is contributed by avanitrachhadiya2155
// C# program to find the maximum count of 1s
using System;
class GFG{
static void maxLengthOf1s(int[] arr, int n)
{
int[] prefix = new int[n];
for(int i = 2; i < n; i++)
{
// If arr[i-2]==1 then we increment
// the count of occurrences of 1's
if (arr[i - 2] == 1)
prefix[i] = prefix[i - 1] + 1;
// Else we initialise the count with 0
else
prefix[i] = 0;
}
int[] suffix = new int[n];
for(int i = n - 3; i >= 0; i--)
{
// If arr[i+2]==1 then we increment
// the count of occurrences of 1's
if (arr[i + 2] == 1)
suffix[i] = suffix[i + 1] + 1;
// Else we initialise the count with 0
else
suffix[i] = 0;
}
int ans = 0;
for(int i = 0; i < n - 1; i++)
{
// We get the maximum count by
// skipping the current and the
// next element.
ans = Math.Max(ans, prefix[i + 1] + suffix[i]);
}
Console.WriteLine(ans);
}
// Driver code
static void Main()
{
int n = 6;
int[] arr = { 1, 1, 1, 0, 1, 1 };
maxLengthOf1s(arr, n);
}
}
// This code is contributed by divyesh072019
<script>
// Javascript program to find
// the maximum count of 1s
function maxLengthOf1s(arr, n)
{
let prefix = new Array(n);
prefix.fill(0);
for(let i = 2; i < n; i++)
{
// If arr[i-2]==1 then we increment
// the count of occurrences of 1's
if (arr[i - 2] == 1)
prefix[i] = prefix[i - 1] + 1;
// Else we initialise the count with 0
else
prefix[i] = 0;
}
let suffix = new Array(n);
suffix.fill(0);
for(let i = n - 3; i >= 0; i--)
{
// If arr[i+2]==1 then we increment
// the count of occurrences of 1's
if (arr[i + 2] == 1)
suffix[i] = suffix[i + 1] + 1;
// Else we initialise the count with 0
else
suffix[i] = 0;
}
let ans = 0;
for(let i = 0; i < n - 1; i++)
{
// We get the maximum count by
// skipping the current and the
// next element.
ans = Math.max(ans, prefix[i + 1] + suffix[i]);
}
document.write(ans);
}
let n = 6;
let arr = [ 1, 1, 1, 0, 1, 1 ];
maxLengthOf1s(arr, n);
</script>
Time Complexity :O(n)
Auxiliary Space :O(n)
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