Length of longest Subarray with equal number of odd and even elements

Given an integer array arr[], the task is to find the length of the longest subarray with an equal number of odd and even elements.

Examples:

Input: arr[] = {1, 2, 1, 2}
Output: 4 
Explanation:
Subarrays in the given array are – 
{{1}, {1, 2}, {1, 2, 1}, {1, 2, 1, 2}, {2}, {2, 1}, {2, 1, 2}, {1}, {1, 2}, {2}} 
where the length of the longest subarray with an equal number of even and odd elements is 4 – {1, 2, 1, 2}

Input: arr[] = {12, 4, 7, 8, 9, 2, 11, 0, 2, 13} 
Output: 8  

Naive Approach: Simple solution is to consider all subarrays one by one and check the count of even and odd elements in the subarray. Then find the maximum length of those subarray that contain an equal number of even elements and odd elements. 

Steps to implement-

• Initialize a variable ans to store the final answer
• Run two loops to find all subarrays and simultaneously find their length
• After that count number of odd elements and the number of even elements in that
• If that subarray contains an equal number of odd and even elements then 
  • Update ans with maximum of ans and the length of that subarray

Code-

C++

// C++ program to find the length
// of the longest sub-array with an
// equal number of odd and even elements
#include <bits/stdc++.h>
using namespace std;

// Function that returns the length of
// the longest sub-array with an equal
// number of odd and even elements
int maxSubarrayLength(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++;

            //To store count of even elements
            int even=0;

            //To store count of odd elements
            int odd=0;

            //Find the number of even elements and odd elements
            for(int k=i;k<=j;k++){
                if(arr[k]%2==0){even++;}
                else if(arr[k]%2==1){odd++;}
            }

            //When subarray contains equal number of odd
            //and even elements
            if(odd==even){
                ans=max(ans,length);
            }

        }
    }
    return ans;
}

// Driver Code
int main()
{
    int arr[] = { 12, 4, 7, 8, 9, 2,
                        11, 0, 2, 13 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << maxSubarrayLength(arr, n);

    return 0;
}

Java

import java.util.*;

public class Main {

    // Function that returns the length of
    // the longest sub-array with an equal
    // number of odd and even elements
    static int maxSubarrayLength(int[] arr, int N)
    {
        // To store answer
        int ans = 0;

        // Find all subarrays
        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++;

                // To store count of even elements
                int even = 0;

                // To store count of odd elements
                int odd = 0;

                // Find the number of even elements and odd
                // elements
                for (int k = i; k <= j; k++) {
                    if (arr[k] % 2 == 0) {
                        even++;
                    }
                    else if (arr[k] % 2 == 1) {
                        odd++;
                    }
                }

                // When subarray contains equal number of
                // odd and even elements
                if (odd == even) {
                    ans = Math.max(ans, length);
                }
            }
        }
        return ans;
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 };
        int n = arr.length;

        System.out.println(maxSubarrayLength(arr, n));
    }
}

Python

# Function that returns the length of
# the longest sub-array with an equal
# number of odd and even elements


def maxSubarrayLength(arr, N):
    # To store the answer
    ans = 0

    # Find all subarrays
    for i in range(N):
        # To store the length of subarray
        length = 0
        for j in range(i, N):
            # Increment the length
            length += 1

            # To store the count of even elements
            even = 0

            # To store the count of odd elements
            odd = 0

            # Find the number of even elements and odd elements
            for k in range(i, j + 1):
                if arr[k] % 2 == 0:
                    even += 1
                elif arr[k] % 2 == 1:
                    odd += 1

            # When the subarray contains an equal number of odd
            # and even elements
            if odd == even:
                ans = max(ans, length)

    return ans


# Driver Code
if __name__ == "__main__":
    arr = [12, 4, 7, 8, 9, 2, 11, 0, 2, 13]
    n = len(arr)

    print(maxSubarrayLength(arr, n))

# This code is contributed by Taranpreet Singh.

C#

using System;

class GFG
{
    // Function that returns the length of
    // the longest sub-array with an equal
    // number of odd and even elements
    static int MaxSubarrayLength(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 the subarray
            int length = 0;
            for (int j = i; j < N; j++)
            {
                // Increment the length
                length++;
                // To store count of even elements
                int even = 0;
                // To store count of odd elements
                int odd = 0;
                // Find the number of even elements and odd elements
                for (int k = i; k <= j; k++)
                {
                    if (arr[k] % 2 == 0)
                    {
                        even++;
                    }
                    else if (arr[k] % 2 == 1)
                    {
                        odd++;
                    }
                }
                // When subarray contains an equal number of odd
                // and even elements
                if (odd == even)
                {
                    ans = Math.Max(ans, length);
                }
            }
        }
        return ans;
    }
    // Driver Code
    static void Main()
    {
        int[] arr = { 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 };
        int n = arr.Length;
        Console.WriteLine(MaxSubarrayLength(arr, n));
    }
}

Javascript

// JavaScript program to find the length
// of the longest sub-array with an
// equal number of odd and even elements

// Function that returns the length of
// the longest sub-array with an equal
// number of odd and even elements
 function maxSubarrayLength(arr, N)
 {
    //To store answer
    let ans=0;

    //Find all subarray
    for(let i=0;i<N;i++){
        //To store length of subarray
        let length=0;
        for(let j=i;j<N;j++){
            //Increment the length
            length++;

            //To store count of even elements
            let even=0;

            //To store count of odd elements
            let odd=0;

            //Find the number of even elements and odd elements
            for(let k=i;k<=j;k++){
                if(arr[k]%2==0){even++;}
                else if(arr[k]%2==1){odd++;}
            }

            //When subarray contains equal number of odd
            //and even elements
            if(odd==even){
                ans=Math.max(ans,length);
            }

        }
    }
    return ans;
 }

// Driver Code
var arr = [12, 4, 7, 8, 9, 2,
                    11, 0, 2, 13 ];
var n = arr.length;
console.log(maxSubarrayLength(arr, n));

Output

8

Time Complexity: O(N³), because of two nested loops to find all subarrays and a third loop to find the number of odd and even elements in a subarray
Auxiliary Space: O(1), because no extra space has been used

Efficient Approach: The idea is to consider the odd elements as 1 and even elements as -1 and return the length of the longest sub-array with the sum equal to 0. The subarray with a given sum can be found using this method. 
Time Complexity: O(N) 
Below is the implementation of the above approach: 

C++

// C++ program to find the length
// of the longest sub-array with an
// equal number of odd and even elements
#include <bits/stdc++.h>
using namespace std;

// Function that returns the length of
// the longest sub-array with an equal
// number of odd and even elements
int maxSubarrayLength(int* A, int N)
{
    // Initialize variable to store result
    int maxLen = 0;

    // Initialize variable to store sum
    int curr_sum = 0;

    // Create an empty map to store
    // index of the sum
    unordered_map<int, int> hash;

    // Loop through the array
    for (int i = 0; i < N; i++) {
        if (A[i] % 2 == 0)
            curr_sum -= 1;
        else
            curr_sum += 1;

        // Check if number of even and
        // odd elements are equal
        if (curr_sum == 0)
            maxLen = max(maxLen, i + 1);

        // If curr_sum already exists in map
        // we have a subarray with 0 sum, i.e,
        // equal number of odd and even number
        if (hash.find(curr_sum) != hash.end())
            maxLen = max(maxLen,
                        i - hash[curr_sum]);

        // Store the index of the sum
        else
            hash[curr_sum] = i;
    }
    return maxLen;
}

// Driver Code
int main()
{
    int arr[] = { 12, 4, 7, 8, 9, 2,
                        11, 0, 2, 13 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << maxSubarrayLength(arr, n);

    return 0;
}

Java

// Java program to find the length
// of the longest sub-array with an
// equal number of odd and even elements
import java.util.*;

class GFG
{

// Function that returns the length of
// the longest sub-array with an equal
// number of odd and even elements
static int maxSubarrayLength(int []A, int N)
{
    // Initialize variable to store result
    int maxLen = 0;

    // Initialize variable to store sum
    int curr_sum = 0;

    // Create an empty map to store
    // index of the sum
    HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>();

    // Loop through the array
    for (int i = 0; i < N; i++)
    {
        if (A[i] % 2 == 0)
            curr_sum -= 1;
        else
            curr_sum += 1;

        // Check if number of even and
        // odd elements are equal
        if (curr_sum == 0)
            maxLen = Math.max(maxLen, i + 1);

        // If curr_sum already exists in map
        // we have a subarray with 0 sum, i.e,
        // equal number of odd and even number
        if (hash.containsKey(curr_sum))
            maxLen = Math.max(maxLen,
                        i - hash.get(curr_sum));

        // Store the index of the sum
        else {
            hash.put(curr_sum, i);
        }
    }
    return maxLen;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 12, 4, 7, 8, 9, 2,
                        11, 0, 2, 13 };
    int n = arr.length;

    System.out.print(maxSubarrayLength(arr, n));
}
}

// This code is contributed by 29AjayKumar

Python3

# Python3 program to find the length
# of the longest sub-array with an
# equal number of odd and even elements

# Function that returns the length of
# the longest sub-array with an equal
# number of odd and even elements
def maxSubarrayLength(A, N) :

    # Initialize variable to store result
    maxLen = 0;

    # Initialize variable to store sum
    curr_sum = 0;

    # Create an empty map to store
    # index of the sum
    hash = {};

    # Loop through the array
    for i in range(N) :
        if (A[i] % 2 == 0) :
            curr_sum -= 1;
        else :
            curr_sum += 1;

        # Check if number of even and
        # odd elements are equal
        if (curr_sum == 0) :
            maxLen = max(maxLen, i + 1);

        # If curr_sum already exists in map
        # we have a subarray with 0 sum, i.e,
        # equal number of odd and even number
        if curr_sum in hash :
            maxLen = max(maxLen, i - hash[curr_sum]);

        # Store the index of the sum
        else :
            hash[curr_sum] = i;

    return maxLen;

# Driver Code
if __name__ == "__main__" :

    arr = [ 12, 4, 7, 8, 9, 2, 11, 0, 2, 13 ];
    n = len(arr);

    print(maxSubarrayLength(arr, n));

# This code is contributed by AnkitRai01

C#

// C# program to find the length
// of the longest sub-array with an
// equal number of odd and even elements
using System;
using System.Collections.Generic;

class GFG
{

// Function that returns the length of
// the longest sub-array with an equal
// number of odd and even elements
static int maxSubarrayLength(int []A, int N)
{
    // Initialize variable to store result
    int maxLen = 0;

    // Initialize variable to store sum
    int curr_sum = 0;

    // Create an empty map to store
    // index of the sum
    Dictionary<int, int> hash = new Dictionary<int, int>();

    // Loop through the array
    for (int i = 0; i < N; i++)
    {
        if (A[i] % 2 == 0)
            curr_sum -= 1;
        else
            curr_sum += 1;

        // Check if number of even and
        // odd elements are equal
        if (curr_sum == 0)
            maxLen = Math.Max(maxLen, i + 1);

        // If curr_sum already exists in map
        // we have a subarray with 0 sum, i.e,
        // equal number of odd and even number
        if (hash.ContainsKey(curr_sum))
            maxLen = Math.Max(maxLen,
                        i - hash[curr_sum]);

        // Store the index of the sum
        else {
            hash.Add(curr_sum, i);
        }
    }
    return maxLen;
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 12, 4, 7, 8, 9, 2,
                        11, 0, 2, 13 };
    int n = arr.Length;
    Console.Write(maxSubarrayLength(arr, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript program to find the length
// of the longest sub-array with an
// equal number of odd and even elements

// Function that returns the length of
// the longest sub-array with an equal
// number of odd and even elements
function maxSubarrayLength(A, N)
{
    // Initialize variable to store result
    var maxLen = 0;

    // Initialize variable to store sum
    var curr_sum = 0;

    // Create an empty map to store
    // index of the sum
    var hash = new Map();

    // Loop through the array
    for (var i = 0; i < N; i++) {
        if (A[i] % 2 == 0)
            curr_sum -= 1;
        else
            curr_sum += 1;

        // Check if number of even and
        // odd elements are equal
        if (curr_sum == 0)
            maxLen = Math.max(maxLen, i + 1);

        // If curr_sum already exists in map
        // we have a subarray with 0 sum, i.e,
        // equal number of odd and even number
        if (hash.has(curr_sum))
            maxLen = Math.max(maxLen,
                        i - hash.get(curr_sum));

        // Store the index of the sum
        else
            hash.set(curr_sum, i);
    }
    return maxLen;
}

// Driver Code
var arr = [12, 4, 7, 8, 9, 2,
                    11, 0, 2, 13 ];
var n = arr.length;
document.write( maxSubarrayLength(arr, n));

</script>

Output

8

Time Complexity: O(n)
Auxiliary Space: O(n), where n is the size of the given array.