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, length of longest subarray with 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 and find the maximum out of those subarrays.
Time Complexity: O(N2)

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

8



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Improved By : AnkitRai01, 29AjayKumar