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Maximum length of subarray such that sum of the subarray is even
  • Difficulty Level : Medium
  • Last Updated : 04 Mar, 2021
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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 <bits/stdc++.h>
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.

PHP




<?php
//PHP implementation of the above approach
 
// Function to find length of the longest
// subarray such that sum of the
// subarray is even
function maxLength($a, $n)
{
    $sum = 0;
    $len = 0;
 
    // Check if sum of complete array is even
    for ($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 ($i = 0; $i < $n; $i++)
    {
        if ($a[$i] % 2 == 1)
            $len = max($len, $max($n - $i - 1, $i));
    }
 
    return $len;
}
 
// Driver Code
$a = array (1, 2, 3, 2 );
$n = count($a);
 
echo maxLength($a, $n) , "\n";
 
 
// This code is contributed by akt_mit.
?>

Javascript




<script>
 
// Javascript implementation of the above approach
 
// Function to find length of the longest
// subarray such that sum of the
// subarray is even
function maxLength(a, n)
{
    let sum = 0, len = 0;
 
    // Check if sum of complete array is even
    for (let 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 (let i = 0; i < n; i++) {
        if (a[i] % 2 == 1)
            len = Math.max(len, Math.max(n - i - 1, i));
    }
 
    return len;
}
 
// Driver Code
    let a = [ 1, 2, 3, 2 ];
    let n = a.length;
 
    document.write(maxLength(a, n) + "<br>");
 
     
// This code is contributed by Mayank Tyagi
 
</script>
Output: 
4

 

Time Complexity: O(N)
 

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