Maximum length of subarray consisting of same type of element on both halves of sub-array

Given an array arr[] of N integers, the task is to find the maximum length of sub-array consisting of the same type of element on both halves of the sub-array. Also, the elements on both halves differ from each other.

Examples:

Input: arr[] = {2, 3, 4, 4, 5, 5, 6, 7, 8, 10}
Output: 4
Explanation:
{2, 3}, {3, 4}, {4, 4, 5, 5}, {5, 6}, etc, are the valid sub-arrays where both halves have only one type of element. 
{4, 4, 5, 5} is the sub-array having maximum length.
Hence, the output is 4. 

Input: arr[] = {1, 7, 7, 10, 10, 7, 7, 7, 8, 8, 8, 9}
Output: 6
Explanation:
{1, 7}, {7, 7, 10, 10}, {7, 7, 7, 8, 8, 8}, {8, 9}, etc, are the valid sub-arrays where both halves have only one type of element. 
{7, 7, 7, 8, 8, 8} is the sub-array having maximum length.
Hence, the output is 6. 

Naive Approach: The naive idea is to generate all possible subarray and check any subarray with maximum length can be divided into two halves such that all the elements in both the halves are the same.

Time Complexity: O(N³)
Auxiliary Space: O(1)

Efficient Approach: To solve this problem the idea is to use the concept of Prefix Sum. Follow the steps below to solve the problem: 

1. Traverse the array from the start in the forward direction and store the continuous occurrence of an integer for each index in an array forward[].
2. Similarly, traverse the array from the end in the reverse direction and store the continuous occurrence of an integer for each index in an array backward[].
3. Store the maximum of min(forward[i], backward[i+1])*2, for all the index where arr[i]!=arr[i+1].
4. Print the value obtained in the above step.

Below is the implementation of the above approach:

C++

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

// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
void maxLengthSubArray(int A[], int N)
{

    // To store continuous occurrence
    // of the element
    int forward[N], backward[N];

    // To store continuous
    // forward occurrence
    for (int i = 0; i < N; i++) {

        if (i == 0
            || A[i] != A[i - 1]) {
            forward[i] = 1;
        }
        else
            forward[i] = forward[i - 1] + 1;
    }

    // To store continuous
    // backward occurrence
    for (int i = N - 1; i >= 0; i--) {

        if (i == N - 1
            || A[i] != A[i + 1]) {
            backward[i] = 1;
        }
        else
            backward[i] = backward[i + 1] + 1;
    }

    // To store the maximum length
    int ans = 0;

    // Find maximum length
    for (int i = 0; i < N - 1; i++) {

        if (A[i] != A[i + 1])
            ans = max(ans,
                    min(forward[i],
                        backward[i + 1])
                        * 2);
    }

    // Print the result
    cout << ans;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 1, 2, 3, 4, 4,
                4, 6, 6, 6, 9 };

    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    maxLengthSubArray(arr, N);
    return 0;
}

Java

// Java program for the above approach
class GFG{

// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
static void maxLengthSubArray(int A[], int N)
{

    // To store continuous occurrence
    // of the element
    int forward[] = new int[N];
    int backward[] = new int[N];

    // To store continuous
    // forkward occurrence
    for(int i = 0; i < N; i++)
    {
        if (i == 0 || A[i] != A[i - 1])
        {
            forward[i] = 1;
        }
        else
            forward[i] = forward[i - 1] + 1;
    }

    // To store continuous
    // backward occurrence
    for(int i = N - 1; i >= 0; i--)
    {
        if (i == N - 1 || A[i] != A[i + 1])
        {
            backward[i] = 1;
        }
        else
            backward[i] = backward[i + 1] + 1;
    }

    // To store the maximum length
    int ans = 0;

    // Find maximum length
    for(int i = 0; i < N - 1; i++)
    {
        if (A[i] != A[i + 1])
            ans = Math.max(ans,
                           Math.min(forward[i],
                                    backward[i + 1]) * 2);
    }

    // Print the result
    System.out.println(ans);
}

// Driver Code
public static void main(String[] args)
{

    // Given array
    int arr[] = { 1, 2, 3, 4, 4,
                  4, 6, 6, 6, 9 };

    // Size of the array
    int N = arr.length;

    // Function call
    maxLengthSubArray(arr, N);
}
}

// This code is contributed by rutvik_56

Python3

# Python3 program for the above approach

# Function that finds the maximum
# length of the sub-array that
# contains equal element on both
# halves of sub-array
def maxLengthSubArray(A, N):

    # To store continuous occurrence
    # of the element
    forward = [0] * N
    backward = [0] * N

    # To store continuous
    # forward occurrence
    for i in range(N):
            if i == 0 or A[i] != A[i - 1]:
                forward[i] = 1
            else:
                forward[i] = forward[i - 1] + 1

    # To store continuous
    # backward occurrence
    for i in range(N - 1, -1, -1):
        if i == N - 1 or A[i] != A[i + 1]:
            backward[i] = 1
        else:
            backward[i] = backward[i + 1] + 1

    # To store the maximum length
    ans = 0

    # Find maximum length
    for i in range(N - 1):
        if (A[i] != A[i + 1]):
            ans = max(ans,
                    min(forward[i],
                        backward[i + 1]) * 2);

    # Print the result
    print(ans)

# Driver Code

# Given array
arr = [ 1, 2, 3, 4, 4, 4, 6, 6, 6, 9 ]

# Size of the array
N = len(arr)

# Function call
maxLengthSubArray(arr, N)

# This code is contributed by yatinagg

C#

// C# program for the above approach
using System;
class GFG{

// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
static void maxLengthSubArray(int []A, int N)
{

    // To store continuous occurrence
    // of the element
    int []forward = new int[N];
    int []backward = new int[N];

    // To store continuous
    // forkward occurrence
    for(int i = 0; i < N; i++)
    {
        if (i == 0 || A[i] != A[i - 1])
        {
            forward[i] = 1;
        }
        else
            forward[i] = forward[i - 1] + 1;
    }

    // To store continuous
    // backward occurrence
    for(int i = N - 1; i >= 0; i--)
    {
        if (i == N - 1 || A[i] != A[i + 1])
        {
            backward[i] = 1;
        }
        else
            backward[i] = backward[i + 1] + 1;
    }

    // To store the maximum length
    int ans = 0;

    // Find maximum length
    for(int i = 0; i < N - 1; i++)
    {
        if (A[i] != A[i + 1])
            ans = Math.Max(ans,
                           Math.Min(forward[i],
                                    backward[i + 1]) * 2);
    }

    // Print the result
    Console.WriteLine(ans);
}

// Driver Code
public static void Main(String[] args)
{

    // Given array
    int []arr = { 1, 2, 3, 4, 4,
                  4, 6, 6, 6, 9 };

    // Size of the array
    int N = arr.Length;

    // Function call
    maxLengthSubArray(arr, N);
}
}

// This code is contributed by Princi Singh

Javascript

<script>

// Javascript program for the above approach

// Function that finds the maximum
// length of the sub-array that
// contains equal element on both
// halves of sub-array
function maxLengthSubArray(A, N)
{

    // To store continuous occurrence
    // of the element
    let forward = Array.from({length: N}, (_, i) => 0);
    let backward = Array.from({length: N}, (_, i) => 0);

    // To store continuous
    // forkward occurrence
    for(let i = 0; i < N; i++)
    {
        if (i == 0 || A[i] != A[i - 1])
        {
            forward[i] = 1;
        }
        else
            forward[i] = forward[i - 1] + 1;
    }

    // To store continuous
    // backward occurrence
    for(let i = N - 1; i >= 0; i--)
    {
        if (i == N - 1 || A[i] != A[i + 1])
        {
            backward[i] = 1;
        }
        else
            backward[i] = backward[i + 1] + 1;
    }

    // To store the maximum length
    let ans = 0;

    // Find maximum length
    for(let i = 0; i < N - 1; i++)
    {
        if (A[i] != A[i + 1])
            ans = Math.max(ans,
                           Math.min(forward[i],
                                    backward[i + 1]) * 2);
    }

    // Print the result
    document.write(ans);
}


// Driver Code

    // Given array
    let arr = [ 1, 2, 3, 4, 4,
                  4, 6, 6, 6, 9 ];

    // Size of the array
    let N = arr.length;

    // Function call
    maxLengthSubArray(arr, N);

</script>

Output:

6

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