满足给定条件的最大长度子数组

给定二进制数组arr [] ，任务是找到给定数组的最长子数组的长度，这样，如果将子数组分为两个相等大小的子数组，则两个子数组都包含全0或全1 。例如，两个子数组的格式必须为{0，0，0，0}和{1，1，1，1}或{1，1，1}和{0，0，0}，而不能为{0，0，0}和{0，0，0}

例子：

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.

方法：对于数组的每两个连续元素，说arr [i]和arr [j] ，其中j = i + 1 ，将它们视为所需子数组的中间两个元素。为了使该子数组成为有效的子数组， arr [i]不得等于arr [j] 。如果它可以是有效的子数组，则其大小为2 。现在，尝试通过同时递减i和j来递增此子数组的大小，并且索引i之前和索引j之后的所有元素必须分别等于arr [i]和arr [j] 。打印到目前为止找到的最长的此类子数组的大小。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

输出：

4

替代方法：我们可以维持先前相似元素的最大长度，并尝试与下一个不同的连续元素形成子数组，并最大化子数组的长度。

以下是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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 previus 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 previus 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(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 previus 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 previus 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.

输出：

4