# Maximum length sub-array which satisfies the given conditions

• Difficulty Level : Easy
• Last Updated : 18 Jun, 2021

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:
{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:
{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 using namespace std; // Function to return the maximum length// of the required sub-arrayint 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 codeint 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 approachclass 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-arraydef 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 approachusing 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



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 maximize the subarray length.

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-arrayint 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 codeint 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 approachclass GFG{     // Function to return the maximum length// of the required sub-arraystatic 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 codepublic 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-arraydef 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 codearr = [ 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 approachusing System; class GFG{     // Function to return the maximum length// of the required sub-arraystatic 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 codepublic 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



Output:

4

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