# Maximize length of Subarray of 1’s after removal of a pair of consecutive Array elements

Given a Binary array arr[] consisting of N elements, the task is to find the maximum possible length of 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:
Explanation:
Removal of the pair {0, 0} modifies the array to {1, 1, 1, 1} thus maximizing the length of longest possible subarray consisting only of 1’s.

Input: arr[] = {1, 1, 1, 0, 0, 0, 1}
Output:
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 subarray of 1‘s . Finally print the maximum possible length of such 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 these count of 0’s obtained, update the maximum length of subarrays possible:
• If exactly two 0‘s are present between two subarrays, update maximum possible length by comparing to the combined length of both subarrays.
• If exactly one 0 is present between two subarrays, update maximum possible length by comparing to the combined length of both subarrays – 1.
• Finally, print the maixmum possible length obtained

Below is the implementation of the above approach:

## C++

 `// C++ program to implement  ` `// the above approach  ` `#include   ` `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 > 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 continous 1s  ` `            ``while` `(A[i] == 1 && i < N) {  ` `                ``i++;  ` `            ``}  ` ` `  `            ``// Calculate length of the  ` `            ``// sub-array  ` `            ``len = i - s;  ` ` `  `            ``// v[i] : Length of subarray  ` `            ``// v[i] : Starting Index of subarray  ` `            ``// v[i] : 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]);  ` ` `  `        ``// v[i+1] : Starting index of  ` `        ``// the next Sub-Array  ` ` `  `        ``// v[i] : Ending Index of the  ` `        ``// current Sub-Array  ` ` `  `        ``// v[i+1] - v[i] - 1 : Count of  ` `        ``// zeros between the two sub-arrays  ` `        ``if` `(v[i + 1] - v[i] - 1 == 2) {  ` ` `  `            ``// Update length of both subarrays  ` `            ``// to the maximum  ` `            ``ans = max(ans, v[i] + v[i + 1]);  ` `        ``}  ` `        ``if` `(v[i + 1] - v[i] - 1 == 1) {  ` ` `  `            ``// Update length of both subarrays - 1  ` `            ``// to the maximum  ` `            ``ans = max(ans, v[i] + v[i + 1] - 1);  ` `        ``}  ` `    ``}  ` ` `  `    ``// Check if the last subarray has  ` `    ``// the maximum length  ` `    ``ans = max(v[v.size() - 1], ans);  ` ` `  `    ``return` `ans;  ` `}  ` ` `  `// Driver Code  ` `int` `main()  ` `{  ` `    ``int` `arr[] = { 1, 0, 1, 0, 0, 1 };  ` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);  ` `    ``cout << maxLen(arr, N) << endl;  ` `    ``return` `0;  ` `}  `

## Java

 `// Java program for 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> 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 continous 1s  ` `            ``while` `(i < N && A[i] == ``1``) ` `            ``{  ` `                ``i++;  ` `            ``}  ` ` `  `            ``// Calculate length of the  ` `            ``// sub-array  ` `            ``len = i - s;  ` ` `  `            ``// v[i] : Length of subarray  ` `            ``// v[i] : Starting Index of subarray  ` `            ``// v[i] : 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] : Starting index of  ` `        ``// the next Sub-Array  ` ` `  `        ``// v[i] : Ending Index of the  ` `        ``// current Sub-Array  ` ` `  `        ``// v[i+1] - v[i] - 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

 `# 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 continous 1s  ` `            ``while` `(i < N ``and` `A[i] ``=``=` `1``): ` `                ``i ``+``=` `1` `             `  `            ``# Calculate length of the  ` `            ``# sub-array  ` `            ``le ``=` `i ``-` `s ` ` `  `            ``# v[i] : Length of subarray  ` `            ``# v[i] : Starting Index of subarray  ` `            ``# v[i] : 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] : Starting index of  ` `        ``# the next Sub-Array  ` ` `  `        ``# v[i] : Ending Index of the  ` `        ``# current Sub-Array  ` ` `  `        ``# v[i+1] - v[i] - 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 `

Output:

```2
```

Time Complexity: O(N)
Auxiliary Space: O(N)

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Improved By : offbeat, chitranayal