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

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

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

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. 
?> 

Output:

Time Complexity: O(N)

My Personal Notes

Suggest a Topic

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

C++

Java

Python

C#

PHP

Recommended Posts: