# Length of the longest alternating even odd subarray

• Difficulty Level : Easy
• Last Updated : 11 Jul, 2022

Given an array a[] of N integers, the task is to find the length of the longest Alternating Even Odd subarray present in the array.
Examples:

Input: a[] = {1, 2, 3, 4, 5, 7, 9}
Output:
Explanation:
The subarray {1, 2, 3, 4, 5} has alternating even and odd elements.
Input: a[] = {1, 3, 5}
Output:
Explanation:
There is no such alternating sequence possible.

Approach 1: Naive approach

Explanation : Consider every subarray and find the length of even and odd subarray

## C++

 `#include ``using` `namespace` `std;`` ` `// Function to find the longest subarray``int` `longestEvenOddSubarray(``int` `a[], ``int` `n)``{``    ``// Length of longest``    ``// alternating subarray``    ``int` `ans = 1;`` ` `    ``// Iterate in the array``    ``for` `(``int` `i = 0; i < n ; i++) {``        ``int` `cnt = 1;``        ``// Iterate for every  subarray``        ``for` `(``int` `j = i + 1; j < n; j++) {``            ``if` `((a[j - 1] % 2 == 0 && a[j] % 2 != 0)``                ``|| (a[j - 1] % 2 != 0 && a[j] % 2 == 0))``                ``cnt++;``            ``else``                ``break``;``        ``}``        ``// store max count``        ``ans = max(ans, cnt);``    ``}``    ``// Length of 'ans' can never be 1``    ``// since even odd has to occur in pair or more``    ``// so return 0 if ans = 1``    ``if` `(ans == 1)``        ``return` `0;``    ``return` `ans;``}`` ` `/* Driver code*/``int` `main()``{``    ``int` `a[] = { 1, 2, 3, 4, 5, 7, 8};`` ` `    ``int` `n = ``sizeof``(a) / ``sizeof``(a);`` ` `    ``cout << longestEvenOddSubarray(a, n);``    ``return` `0;``}`

## Java

 `// Java program for above approach``import` `java.io.*;``import` `java.util.ArrayList;``import` `java.util.List;`` ` `// Function to check if it is possible``// to make the array elements consecutive``public` `class` `GfG{`` ` `// Function to find the longest subarray``static` `int` `longestEvenOddSubarray(ArrayLista, ``int` `n){``     ` `    ``// Length of longest``    ``// alternating subarray``    ``int` `ans = ``1``;``     ` `    ``// Iterate in the array``    ``for` `(``int` `i = ``0``; i < n ; i++) {``        ``int` `cnt = ``1``;``         ` `        ``// Iterate for every subarray``        ``for` `(``int` `j = i + ``1``; j < n; j++) {``                ``if` `((a.get(j - ``1``)% ``2` `== ``0` `&& a.get(j) % ``2` `!= ``0``)``                    ``|| (a.get(j - ``1``) % ``2` `!= ``0` `&& a.get(j) % ``2` `== ``0``))``                    ``cnt++;``                ``else``                    ``break``;``        ``}``             ` `        ``// store max count``        ``ans = Math.max(ans, cnt);``    ``}``     ` `    ``//Length of 'ans' can never be 1``    ``// since even odd has to occur in pair or more``    ``// so return 0 if ans = 1``    ``if` `(ans == ``1``)``        ``return` `0``;``    ``return` `ans;``}``     ` `// Drivers code``public` `static` `void` `main(String args[])``{``    ``ArrayLista = ``new` `ArrayList(List.of(``1``, ``2``, ``3``, ``4``, ``5``, ``7``, ``8` `));``    ``int` `n = a.size();``    ``System.out.println(longestEvenOddSubarray(a, n));``}``}`` ` `// This code is contributed by shinjanpatra`

## Python3

 `import` `math`` ` `# Function to find the longest subarray``def` `longestEvenOddSubarray(a, n):``   ` `    ``# Length of longest``    ``# alternating subarray``    ``ans ``=` `1`` ` `    ``# Iterate in the array``    ``for` `i ``in` `range``(n):``        ``cnt ``=` `1``         ` `        ``# Iterate for every  subarray``        ``for` `j ``in` `range``(i ``+` `1``,n):``            ``if` `((a[j ``-` `1``] ``%` `2` `=``=` `0` `and` `a[j] ``%` `2` `!``=` `0``)``                ``or` `(a[j ``-` `1``] ``%` `2` `!``=` `0` `and` `a[j] ``%` `2` `=``=` `0``)):``                ``cnt ``=` `cnt``+``1``            ``else``:``                ``break``        ``# store max count``        ``ans ``=` `max``(ans, cnt)``              ` `    ``# Length of 'longest' can never be 1 since``    ``# even odd has to occur in pair or more ``    ``# so return 0 if longest = 1``    ``if``(ans ``=``=` `1``):``       ``return` `0``    ``return` `ans`` ` `# Driver code``a ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``7``, ``8` `]`` ` `n ``=` `len``(a)`` ` `print``(longestEvenOddSubarray(a, n))`` ` `# This code is contributed by shinjanpatra.`

## Javascript

 ``

Output

`5`

Approach 2: To solve the problem mentioned above we have to observe that the Sum of two even numbers is even, Sum of two odd numbers is even but sum of one even and one odd number is odd.

• Initially initialize cnt a counter to store the length as 1.
• Iterate among the array elements, check if consecutive elements has an odd sum.
• Increase the cnt by 1 if it has a odd sum.
• If it does not has a odd sum, then re-initialize cnt by 1.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the Length of the``// longest alternating even odd subarray`` ` `#include ``using` `namespace` `std;`` ` `// Function to find the longest subarray``int` `longestEvenOddSubarray(``int` `arr[], ``int` `n)``{``    ``// Length of longest``    ``// alternating subarray``     ` `   ``int` `count = 1;``    ``int` `maxcount = 1;``    ``for` `(``int` `i = 0; i < n - 1; i++)``    ``{``        ``if` `(arr[i] % 2 == 0 && arr[i + 1] % 2 != 0)``        ``{``            ``count++;``        ``}``        ``if` `(arr[i] % 2 != 0 && arr[i + 1] % 2 == 0)``        ``{``            ``count++;``        ``}``        ``if` `(arr[i] % 2 == 0 && arr[i + 1] % 2 == 0)``        ``{``            ``count = 1;``        ``}``        ``if` `(arr[i] % 2 != 0 && arr[i + 1] % 2 != 0)``        ``{``            ``count = 1;``        ``}``        ``maxcount = max(maxcount, count);``    ``}``     ``// Length of 'maxcount' can never be 1``    ``// since even odd has to occur in pair or more``    ``// so return 0 if maxcount = 1``    ``if` `(maxcount == 1)``        ``return` `0;``    ``return` `maxcount;``}`` ` `/* Driver code*/``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4, 5, 7, 8 };`` ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);`` ` `    ``cout << longestEvenOddSubarray(arr, n);``    ``return` `0;``}`

## Java

 `// Java program to find the Length of the``// longest alternating even odd subarray``import` `java.util.*;``class` `GFG{``     ` `// Function to find the longest subarray``static` `int` `longestEvenOddSubarray(``int` `a[], ``int` `n)``{``    ``// Length of longest``    ``// alternating subarray``    ``int` `longest = ``1``;``    ``int` `cnt = ``1``;`` ` `    ``// Iterate in the array``    ``for` `(``int` `i = ``0``; i < n - ``1``; i++) ``    ``{`` ` `        ``// increment count if consecutive``        ``// elements has an odd sum``        ``if` `((a[i] + a[i + ``1``]) % ``2` `== ``1``) ``        ``{``            ``cnt++;``        ``}``        ``else` `        ``{``            ``// Store maximum count in longest``            ``longest = Math.max(longest, cnt);`` ` `            ``// Reinitialize cnt as 1 consecutive``            ``// elements does not have an odd sum``            ``cnt = ``1``;``        ``}``    ``}`` ` `    ``// Length of 'longest' can never be 1``    ``// since even odd has to occur in pair or more``    ``// so return 0 if longest = 1``    ``if` `(longest == ``1``)``        ``return` `0``;`` ` `    ``return` `Math.max(cnt, longest);``}`` ` `// Driver code``public` `static` `void` `main(String[] args)``{``    ``int` `a[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``7``, ``8` `};`` ` `    ``int` `n = a.length;`` ` `    ``System.out.println(longestEvenOddSubarray(a, n));``}``}`` ` `// This code is contributed by offbeat`

## Python3

 `# Python3 program to find the length of the ``# longest alternating even odd subarray`` ` `# Function to find the longest subarray ``def` `longestEvenOddSubarray(arr, n):``     ` `    ``# Length of longest ``    ``# alternating subarray``    ``longest ``=` `1``    ``cnt ``=` `1``     ` `    ``# Iterate in the array``    ``for` `i ``in` `range``(n ``-` `1``):``         ` `        ``# Increment count if consecutive ``        ``# elements has an odd sum``        ``if``((arr[i] ``+` `arr[i ``+` `1``]) ``%` `2` `=``=` `1``):``            ``cnt ``=` `cnt ``+` `1``        ``else``:``             ` `            ``# Store maximum count in longest``            ``longest ``=` `max``(longest, cnt)``             ` `            ``# Reinitialize cnt as 1 consecutive ``            ``# elements does not have an odd sum``            ``cnt ``=` `1``             ` `    ``# Length of 'longest' can never be 1 since``    ``# even odd has to occur in pair or more ``    ``# so return 0 if longest = 1``    ``if``(longest ``=``=` `1``):``       ``return` `0``         ` `    ``return` `max``(cnt, longest)`` ` `# Driver Code ``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``7``, ``8` `]``n ``=` `len``(arr)`` ` `print``(longestEvenOddSubarray(arr, n))`` ` `# This code is contributed by skylags`

## C#

 `// C# program to find the Length of the``// longest alternating even odd subarray``using` `System;`` ` `class` `GFG{``     ` `// Function to find the longest subarray``static` `int` `longestEvenOddSubarray(``int``[] a, ``int` `n)``{``     ` `    ``// Length of longest``    ``// alternating subarray``    ``int` `longest = 1;``    ``int` `cnt = 1;``     ` `    ``// Iterate in the array``    ``for``(``int` `i = 0; i < n - 1; i++) ``    ``{``         ` `       ``// Increment count if consecutive``       ``// elements has an odd sum``       ``if` `((a[i] + a[i + 1]) % 2 == 1) ``       ``{``           ``cnt++;``       ``}``       ``else``       ``{``            ` `           ``// Store maximum count in longest``           ``longest = Math.Max(longest, cnt);``            ` `           ``// Reinitialize cnt as 1 consecutive``           ``// elements does not have an odd sum``           ``cnt = 1;``       ``}``    ``}``     ` `    ``// Length of 'longest' can never be 1``    ``// since even odd has to occur in pair ``    ``// or more so return 0 if longest = 1``    ``if` `(longest == 1)``        ``return` `0;``    ``return` `Math.Max(cnt, longest);``}`` ` `// Driver code``static` `void` `Main()``{``         ` `    ``int``[] a = { 1, 2, 3, 4, 5, 7, 8 };``    ``int` `n = a.Length;``         ` `    ``Console.WriteLine(longestEvenOddSubarray(a, n));``}``}`` ` `// This code is contributed by divyeshrabadiya07`

## Javascript

 ``

Approach 3 – By simply storing the nature of previous element we encounter( odd or even) and comparing it with the next element.

Steps –

1. Store nature of first element in a variable, i.e. if it is even or odd. (this variable will store nature of previous element)
2. Now iterate over the array from index 1 to n – 1 and keep checking if the nature(odd/even) of current index is not same to as that of previous number.
3. Keep storing the max_length of subarray whenever the length of subarray exceeds value to previously found max subarray.
4. Return the found max length of subarray.

## C++

 `// C++ code to find longest subarray of alternating even and odds `` ` `#include ``using` `namespace` `std;`` ` `int` `maxEvenOdd(``int` `arr[], ``int` `n)``{``    ``if` `(n == 0)``        ``return` `0;`` ` `    ``int` `maxLength = 0;``    ``bool` `prevOdd = arr % 2; ``// stroring the nature of first element, if remainder = 1, it is odd``    ``int` `curLength = 1;`` ` `    ``for` `(``int` `i = 1; i < n; i++)``    ``{``        ``if` `(arr[i] % 2 != prevOdd) ``// everytime we check if previous element has opposite even/odd nature or not``            ``curLength++;``        ``else``            ``curLength = 1; ``// reset value when pattern is broken`` ` `        ``if` `(curLength > maxLength) ``// changing value  when new maximum subaaray is found``            ``maxLength = curLength;`` ` `        ``prevOdd = arr[i] % 2; ``// updating even/odd nature of prev number encountered everytime``    ``}`` ` `    ``return` `maxLength;``}`` ` `int` `main()``{``    ``int` `arr[] = {1,2,3,4,5,3,7,2,9,4}; ``//longest subarray should be 1 2 3 4 5 , therefore length = 5``    ``int` `n = ``sizeof``(arr)/``sizeof``(``int``);``      ``cout << ``"Length of longest subarray of even and odds is : "` `<< maxEvenOdd(arr, n);``   ` `    ``return` `0;``}`` ` `//this code is contributed by Anshit Bansal`

Output

`Length of longest subarray of even and odds is : 5`

Time Complexity – Since we need to iterate over whole array once-> O(n)
Auxiliary Space – No extra space was used -> O(1)

Related Topic: Subarrays, Subsequences, and Subsets in Array

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