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Length of the longest alternating even odd subarray

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  • 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 <iostream>
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<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 code
public 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

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




<script>
  
// Function to find the longest subarray
function 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>

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 <bits/stdc++.h>
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[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
    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




<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 – 

  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 <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

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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