# Maximum length of subarray such that sum of the subarray is even

• Difficulty Level : Medium
• Last Updated : 04 Mar, 2021

Given an array of N elements. The task is to find the length of the longest subarray such that sum of the subarray is even.
Examples:

Input : N = 6, arr[] = {1, 2, 3, 2, 1, 4}
Output : 5
Explanation: In the example the subarray
in range [2, 6] has sum 12 which is even,
so the length is 5.

Input : N = 4, arr[] = {1, 2, 3, 2}
Output : 4

Approach: First check if the total sum of the array is even. If the total sum of the array is even then the answer will be N.
If the total sum of the array is not even, means it is ODD. So, the idea is to find an odd element from the array such that excluding that element and comparing the length of both parts of the array we can obtain the max length of the subarray with even sum.
It is obvious that the subarray with even sum will exist in range [1, x) or (x, N],
where 1 <= x <= N, and arr[x] is ODD.
Below is the implementation of above approach:

## C++

 // C++ implementation of the above approach   #include using namespace std;   // Function to find length of the longest // subarray such that sum of the // subarray is even int maxLength(int a[], int n) {     int sum = 0, len = 0;       // Check if sum of complete array is even     for (int i = 0; i < n; i++)         sum += a[i];       if (sum % 2 == 0) // total sum is already even         return n;       // Find an index i such the a[i] is odd     // and compare length of both halfs excluding     // a[i] to find max length subarray     for (int i = 0; i < n; i++) {         if (a[i] % 2 == 1)             len = max(len, max(n - i - 1, i));     }       return len; }   // Driver Code int main() {     int a[] = { 1, 2, 3, 2 };     int n = sizeof(a) / sizeof(a[0]);       cout << maxLength(a, n) << "\n";       return 0; }

## Java

 // Java implementation of the approach   class GFG {       // Function to find length of the longest     // subarray such that sum of the     // subarray is even     static int maxLength(int a[], int n)     {         int sum = 0, len = 0;           // Check if sum of complete array is even         for (int i = 0; i < n; i++)         {             sum += a[i];         }           if (sum % 2 == 0) // total sum is already even         {             return n;         }           // Find an index i such the a[i] is odd         // and compare length of both halfs excluding         // a[i] to find max length subarray         for (int i = 0; i < n; i++)         {             if (a[i] % 2 == 1)             {                 len = Math.max(len, Math.max(n - i - 1, i));             }         }           return len;     }       // Driver Code     public static void main(String[] args)     {         int a[] = {1, 2, 3, 2};         int n = a.length;         System.out.println(maxLength(a, n));       } }   // This code has been contributed by 29AjayKumar

## Python

 # Python3 implementation of the above approach   # Function to find Length of the longest # subarray such that Sum of the # subarray is even def maxLength(a, n):       Sum = 0     Len = 0       # Check if Sum of complete array is even     for i in range(n):         Sum += a[i]       if (Sum % 2 == 0): # total Sum is already even         return n       # Find an index i such the a[i] is odd     # and compare Length of both halfs excluding     # a[i] to find max Length subarray     for i in range(n):         if (a[i] % 2 == 1):             Len = max(Len, max(n - i - 1, i))       return Len   # Driver Code   a= [1, 2, 3, 2] n = len(a)   print(maxLength(a, n))   # This code is contributed by mohit kumar

## C#

 // C# implementation of the approach using System;   class GFG {           // Function to find length of the longest     // subarray such that sum of the     // subarray is even     static int maxLength(int []a, int n)     {         int sum = 0, len = 0;           // Check if sum of complete array is even         for (int i = 0; i < n; i++)         {             sum += a[i];         }           if (sum % 2 == 0) // total sum is already even         {             return n;         }           // Find an index i such the a[i] is odd         // and compare length of both halfs excluding         // a[i] to find max length subarray         for (int i = 0; i < n; i++)         {             if (a[i] % 2 == 1)             {                 len = Math.Max(len, Math.Max(n - i - 1, i));             }         }           return len;     }       // Driver Code     static public void Main ()     {         int []a = {1, 2, 3, 2};         int n = a.Length;         Console.WriteLine(maxLength(a, n));       } }   // This code has been contributed by ajit.



## Javascript



Output:

4

Time Complexity: O(N)

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