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

• Last Updated : 27 May, 2021

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:
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 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++

 `// C++ program to find the length``// of the longest sub-array with an``// equal number of odd and even elements``#include ``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);` `    ``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 hash = ``new` `HashMap();` `    ``// 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

 ``
Output:
`8` My Personal Notes arrow_drop_up