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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 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);        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

Output:

4

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

Below is the imlementation of the above approach:

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);        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

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.

Output:

4

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