Maximum length sub-array which satisfies the given conditions

Given a binary array arr[], the task is to find the length of the longest sub-array of the given array such that if the sub-array is divided into two equal-sized sub-arrays then both the sub-arrays either contain all 0s or all 1s. For example, the two sub-arrays must be of the form {0, 0, 0, 0} and {1, 1, 1, 1} or {1, 1, 1} and {0, 0, 0} and not {0, 0, 0} and {0, 0, 0}

Examples:

Input: arr[] = {1, 1, 1, 0, 0, 1, 1} 
Output: 4 
{1, 1, 0, 0} and {0, 0, 1, 1} are the maximum length valid sub-arrays.

Input: arr[] = {1, 1, 0, 0, 0, 1, 1, 1, 1} 
Output: 6 
{0, 0, 0, 1, 1, 1} is the only valid sub-array with maximum length. 

Approach: For every two consecutive elements of the array say arr[i] and arr[j] where j = i + 1, treat them as the middle two elements of the required sub-array. In order for this sub-array to be a valid sub-array arr[i] must not be equal to arr[j]. If it can be a valid sub-array then its size is 2. Now, try to extend this sub-array to a bigger size by decrementing i and incrementing j at the same time and all the elements before index i and after index j must be equal to arr[i] and arr[j] respectively. Print the size of the longest such sub-array found so far.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;

// Function to return the maximum length
// of the required sub-array
int maxLength(int arr[], int n)
{
    int maxLen = 0;

    // For the first consecutive
    // pair of elements
    int i = 0;
    int j = i + 1;

    // While a consecutive pair
    // can be selected
    while (j < n) {

        // If current pair forms a
        // valid sub-array
        if (arr[i] != arr[j]) {

            // 2 is the length of the
            // current sub-array
            maxLen = max(maxLen, 2);

            // To extend the sub-array both ways
            int l = i - 1;
            int r = j + 1;

            // While elements at indices l and r
            // are part of a valid sub-array
            while (l >= 0 && r < n && arr[l] == arr[i]
                   && arr[r] == arr[j]) {
                l--;
                r++;
            }

            // Update the maximum length so far
            maxLen = max(maxLen, 2 * (r - j));
        }

        // Select the next consecutive pair
        i++;
        j = i + 1;
    }

    // Return the maximum length
    return maxLen;
}

// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << maxLength(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function to return the maximum length
    // of the required sub-array
    static int maxLength(int arr[], int n)
    {
        int maxLen = 0;

        // For the first consecutive
        // pair of elements
        int i = 0;
        int j = i + 1;

        // While a consecutive pair
        // can be selected
        while (j < n) {

            // If current pair forms a
            // valid sub-array
            if (arr[i] != arr[j]) {

                // 2 is the length of the
                // current sub-array
                maxLen = Math.max(maxLen, 2);

                // To extend the sub-array both ways
                int l = i - 1;
                int r = j + 1;

                // While elements at indices l and r
                // are part of a valid sub-array
                while (l >= 0 && r < n && arr[l] == arr[i] && arr[r] == arr[j]) {
                    l--;
                    r++;
                }

                // Update the maximum length so far
                maxLen = Math.max(maxLen, 2 * (r - j));
            }

            // Select the next consecutive pair
            i++;
            j = i + 1;
        }

        // Return the maximum length
        return maxLen;
    }

    // Driver code
    public static void main(String[] args)
    {
        int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
        int n = arr.length;

        System.out.println(maxLength(arr, n));
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of the approach

# Function to return the maximum length
# of the required sub-array
def maxLength(arr, n):
    maxLen = 0

    # For the first consecutive
    # pair of elements
    i = 0
    j = i + 1

    # While a consecutive pair
    # can be selected
    while (j < n):

        # If current pair forms a
        # valid sub-array
        if (arr[i] != arr[j]):

            # 2 is the length of the
            # current sub-array
            maxLen = max(maxLen, 2)

            # To extend the sub-array both ways
            l = i - 1
            r = j + 1

            # While elements at indices l and r
            # are part of a valid sub-array
            while (l >= 0 and r < n and arr[l] == arr[i]
                and arr[r] == arr[j]):
                l-= 1
                r+= 1

            # Update the maximum length so far
            maxLen = max(maxLen, 2 * (r - j))

        # Select the next consecutive pair
        i+= 1
        j = i + 1

    # Return the maximum length
    return maxLen

# Driver code

arr =[1, 1, 1, 0, 0, 1, 1]
n = len(arr)

print(maxLength(arr, n))

# This code is contributed by mohit kumar 29

C#

// C# implementation of the approach
using System;

class GFG {

    // Function to return the maximum length
    // of the required sub-array
    static int maxLength(int[] arr, int n)
    {
        int maxLen = 0;

        // For the first consecutive
        // pair of elements
        int i = 0;
        int j = i + 1;

        // While a consecutive pair
        // can be selected
        while (j < n) {

            // If current pair forms a
            // valid sub-array
            if (arr[i] != arr[j]) {

                // 2 is the length of the
                // current sub-array
                maxLen = Math.Max(maxLen, 2);

                // To extend the sub-array both ways
                int l = i - 1;
                int r = j + 1;

                // While elements at indices l and r
                // are part of a valid sub-array
                while (l >= 0 && r < n && arr[l] == arr[i] && arr[r] == arr[j]) {
                    l--;
                    r++;
                }

                // Update the maximum length so far
                maxLen = Math.Max(maxLen, 2 * (r - j));
            }

            // Select the next consecutive pair
            i++;
            j = i + 1;
        }

        // Return the maximum length
        return maxLen;
    }

    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 1, 1, 1, 0, 0, 1, 1 };
        int n = arr.Length;

        Console.WriteLine(maxLength(arr, n));
    }
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the maximum length
// of the required sub-array
function maxLength(arr, n)
{
    let maxLen = 0;

    // For the first consecutive
    // pair of elements
    let i = 0;
    let j = i + 1;

    // While a consecutive pair
    // can be selected
    while (j < n) {

        // If current pair forms a
        // valid sub-array
        if (arr[i] != arr[j]) {

            // 2 is the length of the
            // current sub-array
            maxLen = Math.max(maxLen, 2);

            // To extend the sub-array both ways
            let l = i - 1;
            let r = j + 1;

            // While elements at indices l and r
            // are part of a valid sub-array
            while (l >= 0 && r < n && arr[l] == arr[i]
                   && arr[r] == arr[j]) {
                l--;
                r++;
            }

            // Update the maximum length so far
            maxLen = Math.max(maxLen, 2 * (r - j));
        }

        // Select the next consecutive pair
        i++;
        j = i + 1;
    }

    // Return the maximum length
    return maxLen;
}

// Driver code
    let arr = [ 1, 1, 1, 0, 0, 1, 1 ];
    let n = arr.length;

    document.write(maxLength(arr, n));

</script>

Output

4

Time Complexity: O(n²), where n is the length of the given array.
Auxiliary Space: O(1)

Alternate approach: We can maintain the max length of previous similar elements and try to form subarray with the next different contiguous element and maximize the subarray length.

Below is the implementation of the above approach:  

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the maximum length
// of the required sub-array
int maxLength(int a[], int n)
{

    // To store the maximum length
    // for a valid subarray
    int maxLen = 0;

    // To store the count of contiguous
    // similar elements for previous
    // group and the current group
    int prev_cnt = 0, curr_cnt = 1;
    for (int i = 1; i < n; i++) {

        // If current element is equal to
        // the previous element then it is
        // a part of the same group
        if (a[i] == a[i - 1])
            curr_cnt++;

        // Else update the previous group
        // and start counting elements
        // for the new group
        else {
            prev_cnt = curr_cnt;
            curr_cnt = 1;
        }

        // Update the maximum possible length for a group
        maxLen = max(maxLen, min(prev_cnt, curr_cnt));
    }

    // Return the maximum length of the valid subarray
    return (2 * maxLen);
}

// Driver code
int main()
{
    int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << maxLength(arr, n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the maximum length
// of the required sub-array
static int maxLength(int a[], int n)
{

    // To store the maximum length
    // for a valid subarray
    int maxLen = 0;

    // To store the count of contiguous
    // similar elements for previous
    // group and the current group
    int prev_cnt = 0, curr_cnt = 1;
    for (int i = 1; i < n; i++)
    {

        // If current element is equal to
        // the previous element then it is
        // a part of the same group
        if (a[i] == a[i - 1])
            curr_cnt++;

        // Else update the previous group
        // and start counting elements
        // for the new group
        else
        {
            prev_cnt = curr_cnt;
            curr_cnt = 1;
        }

        // Update the maximum possible length for a group
        maxLen = Math.max(maxLen,
                 Math.min(prev_cnt, curr_cnt));
    }

    // Return the maximum length
    // of the valid subarray
    return (2 * maxLen);
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 1, 1, 0, 0, 1, 1 };
    int n = arr.length;

    System.out.println(maxLength(arr, n));
}
}

// This code is contributed by Rajput-Ji

Python3

# Python3 implementation of the approach
# Function to return the maximum length
# of the required sub-array
def maxLength(a, n):

    # To store the maximum length
    # for a valid subarray
    maxLen = 0;

    # To store the count of contiguous
    # similar elements for previous
    # group and the current group
    prev_cnt = 0; curr_cnt = 1;
    for i in range(1, n):

        # If current element is equal to
        # the previous element then it is
        # a part of the same group
        if (a[i] == a[i - 1]):
            curr_cnt += 1;

        # Else update the previous group
        # and start counting elements
        # for the new group
        else:
            prev_cnt = curr_cnt;
            curr_cnt = 1;

        # Update the maximum possible
        # length for a group
        maxLen = max(maxLen, min(prev_cnt,
                                 curr_cnt));

    # Return the maximum length
    # of the valid subarray
    return (2 * maxLen);

# Driver code
arr = [ 1, 1, 1, 0, 0, 1, 1 ];
n = len(arr);

print(maxLength(arr, n));

# This code is contributed by Rajput-Ji

C#

// C# implementation of the approach
using System;

class GFG
{

// Function to return the maximum length
// of the required sub-array
static int maxLength(int[] a, int n)
{

    // To store the maximum length
    // for a valid subarray
    int maxLen = 0;

    // To store the count of contiguous
    // similar elements for previous
    // group and the current group
    int prev_cnt = 0, curr_cnt = 1;
    for (int i = 1; i < n; i++)
    {

        // If current element is equal to
        // the previous element then it is
        // a part of the same group
        if (a[i] == a[i - 1])
            curr_cnt++;

        // Else update the previous group
        // and start counting elements
        // for the new group
        else
        {
            prev_cnt = curr_cnt;
            curr_cnt = 1;
        }

        // Update the maximum possible length for a group
        maxLen = Math.Max(maxLen,
                 Math.Min(prev_cnt, curr_cnt));
    }

    // Return the maximum length
    // of the valid subarray
    return (2 * maxLen);
}

// Driver code
public static void Main()
{
    int[] arr = { 1, 1, 1, 0, 0, 1, 1 };
    int n = arr.Length;

    Console.WriteLine(maxLength(arr, n));
}
}

// This code is contributed by Code_Mech.

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the maximum length
// of the required sub-array
function maxLength(a, n)
{

    // To store the maximum length
    // for a valid subarray
    let maxLen = 0;

    // To store the count of contiguous
    // similar elements for previous
    // group and the current group
    let prev_cnt = 0, curr_cnt = 1;
    for (let i = 1; i < n; i++) {

        // If current element is equal to
        // the previous element then it is
        // a part of the same group
        if (a[i] == a[i - 1])
            curr_cnt++;

        // Else update the previous group
        // and start counting elements
        // for the new group
        else {
            prev_cnt = curr_cnt;
            curr_cnt = 1;
        }

        // Update the maximum possible length for a group
        maxLen = Math.max(maxLen, Math.min(prev_cnt, curr_cnt));
    }

    // Return the maximum length of the valid subarray
    return (2 * maxLen);
}

// Driver code
    let arr = [ 1, 1, 1, 0, 0, 1, 1 ];
    let n = arr.length;

    document.write(maxLength(arr, n));

</script>

Output

4

Time Complexity: O(n), where n is the length of the given array.
Auxiliary Space: O(1)