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

Last Updated : 19 Apr, 2024

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

Naive Approach

The idea is to find all subarrays and then find those subarrays whose sum of elements are even. After that choose the longest length of those subarrays.

Steps to implement-

Declare a variable ans with value 0 to store the final answer
Run two loops to find all subarrays
Find the sum of all elements of the subarray
When the sum of all elements of the subarray is even
- Then update ans as the maximum of ans and the length of that subarray

Code-

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 arr[], int N) {    //To store answer    int ans=0;        //Find all subarray    for(int i=0;i<N;i++){        //To store length of subarray        int length=0;        for(int j=i;j<N;j++){            //Increment the length            length++;                        //Boolean variable to tell whether sum             //of all elements are even or not            bool val=false;                        //To store sum of all elements of subarray            int sum=0;                        for(int k=i;k<=j;k++){                sum+=arr[k];            }                        //When sum of all elements are even            if(sum%2==0){                ans=max(ans,length);            }        }    }    return ans; }  // Driver Code int main() {     int arr[] = { 1, 2, 3, 2 };     int N = sizeof(arr) / sizeof(arr[0]);      cout << maxLength(arr, N) << "\n";      return 0; }

Java

// Java implementation of the above approach import java.util.*;  public class Main {      // Function to find length of the longest     // subarray such that sum of the     // subarray is even     static int maxLength(int[] arr, int N) {         // To store answer         int ans = 0;          // Find all subarray         for (int i = 0; i < N; i++) {             // To store length of subarray             int length = 0;             for (int j = i; j < N; j++) {                 // Increment the length                 length++;                  // Boolean variable to tell whether sum                 // of all elements are even or not                 boolean val = false;                  // To store sum of all elements of subarray                 int sum = 0;                  for (int k = i; k <= j; k++) {                     sum += arr[k];                 }                  // When sum of all elements is even                 if (sum % 2 == 0) {                     ans = Math.max(ans, length);                 }             }         }         return ans;     }      // Driver Code     public static void main(String[] args) {         int[] arr = {1, 2, 3, 2};         int N = arr.length;          System.out.println(maxLength(arr, N));     } }  // This code is contributed by Pushpesh Raj

Python3

# Python3 implementation of the above approach  # Function to find length of the longest subarray such  # that sum of the subarray is even def maxLength(arr):     # To store the answer     ans = 0      # Find all subarrays     for i in range(len(arr)):         # To store the length of subarray         length = 0         for j in range(i, len(arr)):             # Increment the length             length += 1              # Boolean variable to tell whether the sum              # of all elements is even or not             val = False              # To store the sum of all elements of subarray             subarray_sum = 0              for k in range(i, j + 1):                 subarray_sum += arr[k]              # When the sum of all elements is even             if subarray_sum % 2 == 0:                 ans = max(ans, length)      return ans  # Driver Code arr = [1, 2, 3, 2] print(maxLength(arr))

using System;  public class GFG {     // Function to find length of the longest     // subarray such that sum of the     // subarray is even     public static int MaxLength(int[] arr, int N)     {         // To store the answer         int ans = 0;          // Find all subarrays         for (int i = 0; i < N; i++)         {             // To store the length of subarray             int length = 0;             for (int j = i; j < N; j++)             {                 // Increment the length                 length++;                  // To store the sum of all elements of subarray                 int sum = 0;                  for (int k = i; k <= j; k++)                 {                     sum += arr[k];                 }                  // When the sum of all elements is even                 if (sum % 2 == 0)                 {                     ans = Math.Max(ans, length);                 }             }         }         return ans;     }      // Driver code     public static void Main(string[] args)     {         int[] arr = { 1, 2, 3, 2 };         int N = arr.Length;          Console.WriteLine(MaxLength(arr, N));     } }

JavaScript

function MaxLength(arr, N) {     // To store the answer     let ans = 0;      // Find all subarrays     for (let i = 0; i < N; i++) {         // To store the length of subarray         let length = 0;         for (let j = i; j < N; j++) {             // Increment the length             length++;              // To store the sum of all elements of subarray             let sum = 0;              for (let k = i; k <= j; k++) {                 sum += arr[k];             }              // When the sum of all elements is even             if (sum % 2 === 0) {                 ans = Math.max(ans, length);             }         }     }     return ans; }  // Driver code const arr = [1, 2, 3, 2]; const N = arr.length;  console.log(MaxLength(arr, N));

Output

Time Complexity: O(N³), because of two nested loops to find all subarray and third loop is to find the sum of all elements of subarray
Auxiliary Space: O(1), because no extra space has been used

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 halves 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# 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

<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>

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)
Auxiliary Space: O(1)

Sum of all odd length subarrays

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

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