# Length of the longest alternating even odd subarray

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.

## Naive approach:

The idea is to consider every subarray and find the length of even and odd subarrays.

Follow the steps below to solve the problem:

• Iterate for every subarray from i = 0
• Make a nested loop, iterate from j = i + 1
• Now, check if a[j – 1] is even and a[j] is odd or a[j – 1] is odd and a[j] is even then increment count
• Maintain an answer variable which calculates max count so far

Below is the implementation of the above approach:

## 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[0]);` `    ``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(ArrayList 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.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[])``    ``{``        ``ArrayList a = ``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.`

## C#

 `// C# program for above approach``using` `System;``using` `System.Collections.Generic;` `// 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(List<``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 = 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)``  ``{``    ``List<``int``> a = ``new` `List<``int``>{ 1, 2, 3, 4, 5, 7, 8 };``    ``int` `n = a.Count;``    ``Console.WriteLine(longestEvenOddSubarray(a, n));``  ``}``}` `// This code is contributed by phasing17`

## Javascript

 ``

Output
`5`

Time Complexity: O(N2), Iterating over every subarray therefore N2 are possible
Auxiliary Space: O(1)

## Length of the longest alternating even odd subarray by Checking Parity of Sum:

Observe that the Sum of two even numbers is even, the Sum of two odd numbers is even but the sum of one even and one odd number is odd.

Follow the steps below to solve the problem:

• Initially initialize cnt a counter to store the length as 1.
• Iterate among the array elements, and check if consecutive elements have an odd sum.
• Increase the cnt by 1 if it has an odd sum.
• If it does not has an odd sum, then re-initialize cnt by 1.
• The function should return at least value 1 if there are elements in the array, because there can always be a subarray with length 1 which has either odd or even element.

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 be 1 as well because we want length of longest subarray ``    ``// which has alternate even-odd or vice-versa elements. It is not mentioned that ``    ``// they have to occur in pair.``    ``return` `maxcount;``}` `/* Driver code*/``int` `main()``{``    ``int` `arr[] = { 1, 2, 3, 4, 5, 7, 8 };` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``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``        ``longest = Math.max(longest, cnt);``        ``if` `(longest == ``1``)``            ``return` `0``;` `        ``return` `longest;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `a[] = { ``1``, ``2``, ``3``, ``4``, ``5``, ``6``, ``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

 ``

Output
`5`

Time Complexity: O(N), Traversing over the array one time.
Auxiliary Space: O(1)

## Length of the longest alternating even odd subarray by Storing the previous element

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

Follow the steps below to solve the problem:

• Initialize a variable maxLength to 0, to keep the track of maximum length of the alternating subarray obtained.
• Initialize a variable currLen to 1 considering first element as the part of alternating subarray.
• Starting with element at index 1, compare every element with it’s previous. If there nature are different, increment the currLen variable.
• Otherwise, reset the currLen to 1 again so that, this current element is considered in new alternating subarray.
• Keep storing the max length of subarray in maxLength before resetting the currLen.
• Return the found max length of subarray.

Below is the implementation of above approach:

## 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;` `    ``int` `currLen = 1;` `    ``for` `(``int` `i = 1; i < n; i++) {``        ``// everytime we check if previous``        ``// element has opposite even/odd``        ``// nature or not``        ``if` `(arr[i] % 2 != arr[i-1] % 2)``            ``currLen++;``        ``else``        ``{``            ``// store max in maxLength``            ``maxLength = max(maxLength, currLen);``            ``// reset value when pattern is broken``            ``currLen = 1;``        ``}``    ``}``      ``// since, even-odd should occur in pair``    ``if``(maxLength == 1)``        ``return` `0;``      ``// if the pair is in last``      ``if``(currLen > 1){``        ``maxLength = max(maxLength , currLen);``    ``}``    ``return` `maxLength;``}` `// Driver Code``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 and improved by Aniket Raj`

## Java

 `// Java code to find longest subarray ``// of alternating even and odds``import` `java.util.*;` `class` `GFG {` `  ``public` `static` `int` `maxEvenOdd(``int``[] arr, ``int` `n)``  ``{``    ``if` `(n == ``0``)``      ``return` `0``;` `    ``int` `maxLength = ``0``;` `    ``// storing the nature of first element, if``    ``// remainder = 1, it is odd``    ``int` `prevOdd = arr[``0``] % ``2``;``    ``int` `curLength = ``1``;``    ``for` `(``int` `i = ``1``; i < n; i++) ``    ``{` `      ``// everytime we check if previous``      ``// element has opposite even/odd``      ``// nature or not``      ``if` `(arr[i] % ``2` `!= prevOdd)``        ``curLength++;``      ``else` `        ``// reset value when pattern is broken``        ``curLength = ``1``;` `      ``// changing value  when new maximum``      ``// subarray is found``      ``if` `(curLength > maxLength)``        ``maxLength = curLength;` `      ``// updating even/odd nature of prev``      ``// number encountered everytime``      ``prevOdd = arr[i] % ``2``;``    ``}` `    ``return` `maxLength;``  ``}` `  ``static` `public` `void` `main(String[] args)``  ``{` `    ``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 = arr.length;``    ``System.out.print(``      ``"Length of longest subarray of even and odds is : "``);``    ``System.out.print(maxEvenOdd(arr, n));``  ``}``}` `// This code is contributed by phasing17`

## Python3

 `# Python3 code to find longest subarray of alternating even and``# odds``def` `maxEvenOdd(arr, n):` `    ``if` `(n ``=``=` `0``):``        ``return` `0``;` `    ``maxLength ``=` `0``;``    ` `    ``# storing the nature of first element, if``    ``# remainder = 1, it is odd``    ``prevOdd ``=` `arr[``0``] ``%` `2``;``    ``curLength ``=` `1``;` `    ``for` `i ``in` `range``(``1``, n):``        ``# everytime we check if previous``        ``# element has opposite even/odd``        ``# nature or not``        ``if` `(arr[i] ``%` `2` `!``=` `prevOdd):``            ``curLength``+``=``1``;``        ``else``:``            ``# reset value when pattern is broken``            ``curLength ``=` `1``;``        ``# changing value  when new maximum``        ``# subarray is found``        ``if` `(curLength > maxLength):``            ``maxLength ``=` `curLength;` `        ``# updating even/odd nature of prev``        ``# number encountered everytime``        ``prevOdd ``=` `arr[i] ``%` `2``;``    ``return` `maxLength;` `# Driver Code``arr ``=` `[ ``1``, ``2``, ``3``, ``4``, ``5``, ``3``, ``7``, ``2``, ``9``, ``4` `];` `# longest subarray should be 1 2 3 4 5 , therefore``# length = 5``n ``=` `len``(arr);``print``(``"Length of longest subarray of even and odds is :"``, maxEvenOdd(arr, n));` `# This code is contributed by phasing17`

## C#

 `// C# code to find longest subarray ``// of alternating even and odds``using` `System;` `public` `class` `GFG {` `  ``public` `static` `int` `maxEvenOdd(``int``[] arr, ``int` `n)``  ``{``    ``if` `(n == 0)``      ``return` `0;` `    ``int` `maxLength = 0;` `    ``// storing the nature of first element, if``    ``// remainder = 1, it is odd``    ``int` `prevOdd = arr[0] % 2;``    ``int` `curLength = 1;``    ``for` `(``int` `i = 1; i < n; i++) ``    ``{` `      ``// everytime we check if previous``      ``// element has opposite even/odd``      ``// nature or not``      ``if` `(arr[i] % 2 != prevOdd)``        ``curLength++;``      ``else` `        ``// reset value when pattern is broken``        ``curLength = 1;` `      ``// changing value  when new maximum``      ``// subarray is found``      ``if` `(curLength > maxLength)``        ``maxLength = curLength;` `      ``// updating even/odd nature of prev``      ``// number encountered everytime``      ``prevOdd = arr[i] % 2;``    ``}` `    ``return` `maxLength;``  ``}` `  ``static` `public` `void` `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 = arr.Length;``    ``Console.Write(``      ``"Length of longest subarray of even and odds is : "``);``    ``Console.Write(maxEvenOdd(arr, n));``  ``}``}` `// This code is contributed by akashish__`

## Javascript

 `// JS code to find longest subarray of alternating even and``// odds` `function` `maxEvenOdd(arr, n)``{``    ``if` `(n == 0)``        ``return` `0;` `    ``let maxLength = 0;``    ``// storing the nature of first element, if``    ``// remainder = 1, it is odd``    ``let prevOdd = arr[0] % 2;``    ``let curLength = 1;` `    ``for` `(``var` `i = 1; i < n; i++) {``        ``// everytime we check if previous``        ``// element has opposite even/odd``        ``// nature or not``        ``if` `(arr[i] % 2 != prevOdd)``            ``curLength++;``        ``else``            ``// reset value when pattern is broken``            ``curLength = 1;``        ``// changing value  when new maximum``        ``// subarray is found``        ``if` `(curLength > maxLength)``            ``maxLength = curLength;` `        ``// updating even/odd nature of prev``        ``// number encountered everytime``        ``prevOdd = arr[i] % 2;``    ``}` `    ``return` `maxLength;``}` `// Driver Code``let arr = [ 1, 2, 3, 4, 5, 3, 7, 2, 9, 4 ];` `// longest subarray should be 1 2 3 4 5 , therefore``// length = 5``let n = arr.length;``console.log(``"Length of longest subarray of even and odds is : "` `+ ``         ``maxEvenOdd(arr, n));`  `// this code is contributed by phasing17`

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

Time Complexity: O(N), Since we need to iterate over the whole array once
Auxiliary Space: O(1)

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