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++

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

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Python3

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

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Output: 

6

 

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

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Improved By : yatinagg