Length of the longest alternating even odd subarray

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: 5 Explanation: The subarray {1, 2, 3, 4, 5} has alternating even and odd elements.Input: a[] = {1, 3, 5} Output: 0 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++
Java
Python3
Javascript
#include <iostream>using namespace std;  // Function to find the longest subarrayint 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 program for above approachimport java.io.*;import java.util.ArrayList;import java.util.List;  // Function to check if it is possible// to make the array elements consecutivepublic class GfG{  // Function to find the longest subarraystatic int longestEvenOddSubarray(ArrayList<Integer>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 codepublic static void main(String args[]){    ArrayList<Integer>a = new ArrayList<Integer>(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
import math  # Function to find the longest subarraydef 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 codea = [ 1, 2, 3, 4, 5, 7, 8 ]  n = len(a)  print(longestEvenOddSubarray(a, n))  # This code is contributed by shinjanpatra.
<script>  // Function to find the longest subarrayfunction longestEvenOddSubarray(a, n){      // Length of longest    // alternating subarray    let ans = 1;      // Iterate in the array    for (let i = 0; i < n ; i++) {        let cnt = 1;                  // Iterate for every  subarray        for (let 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;}  /* Driver code*/  let a = [ 1, 2, 3, 4, 5, 7, 8 ];  let n = a.length;  document.write(longestEvenOddSubarray(a, n),"</br>");  // This code is contributed by shinjanpatra.</script>
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++
Java
Python3
C#
Javascript
// C++ program to find the Length of the// longest alternating even odd subarray  #include <bits/stdc++.h>using namespace std;  // Function to find the longest subarrayint 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[0]);      cout << longestEvenOddSubarray(arr, n);    return 0;}
// Java program to find the Length of the// longest alternating even odd subarrayimport java.util.*;class GFG{      // Function to find the longest subarraystatic 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 codepublic 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 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# program to find the Length of the// longest alternating even odd subarrayusing System;  class GFG{      // Function to find the longest subarraystatic 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 codestatic 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
<script>// JavaScript program to find the Length of the // longest alternating even odd subarray   // Function to find the longest subarray function longestEvenOddSubarray(a, n) {       // Length of longest     // alternating subarray     let longest = 1;     let cnt = 1;       // Iterate in the array     for (let 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*/    let a = [ 1, 2, 3, 4, 5, 7, 8 ];     let n = a.length;     document.write(longestEvenOddSubarray(a, n));   // This code is contributed by Surbhi Tyagi.</script>
Approach 3 – By simply storing the nature of previous element we encounter( odd or even) and comparing it with the next element.
Steps – 
Store nature of first element in a variable, i.e. if it is even or odd. (this variable will store nature of previous element)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.Keep storing the max_length of subarray whenever the length of subarray exceeds value to previously found max subarray.Return the found max length of subarray.
Store nature of first element in a variable, i.e. if it is even or odd. (this variable will store nature of previous element)
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.
Keep storing the max_length of subarray whenever the length of subarray exceeds value to previously found max subarray.
Return the found max length of subarray.
C++
// C++ code to find longest subarray of alternating even and odds   #include <iostream>using namespace std;  int maxEvenOdd(int arr[], int n){    if (n == 0)        return 0;      int maxLength = 0;    bool prevOdd = arr[0] % 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
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
skylags
offbeat
divyeshrabadiya07
surbhityagi15
sooda367
anushikasethh
ad10
shinjanpatra
anshitbansal
gourangamaity15
subarray
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