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Length of the longest alternating even odd subarray

  • Difficulty Level : Easy
  • Last Updated : 08 Jun, 2021

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: 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 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 = 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*/
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 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>
Output: 
5

 

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