Length of the longest ZigZag subarray of the given array

Given an array containing n numbers. The problem is to find the length of the longest zigZag subarray such that every element in the subarray should be in form a < b > c < d > e < f.. Time Complexity should be O(n).

Examples:

Input : arr[] = {1, 5, 0, 4, 3}
Output : 5
The subarray is {1, 5, 0, 4, 3}

Input : arr[] = {12, 13, 1, 5, 4, 7, 8, 10, 10, 11}
Output : 4
The subarray is {12, 13, 1, 5, 4, 7}

Input : arr[] = {1, 2, 3, 4, 5}
Output : 2
The subarray is {1, 2} or {2, 3} or {4, 5}

Approach:

Initially initialize cnt as 1.
Iterate among the array elements, check if elements are in form a < b > c < d > e < f.. .
If true Increase the cnt by 1.
If it is not in form a < b > c < d > e < f.. , then re-initialize cnt by 1.

// C++ implementation to find the length of 
// the longest zigzag subarray
#include <bits/stdc++.h>

using namespace std;

// Function to find the length of the longest zigzag contiguous subarray
int lenOfLongZigZagArr(int a[], int n) 
{ 
    // 'maxLen' to store the length of the longest zigzag subarray
    // 'len' to store the current length of the zigzag subarray
    int maxLen = 1, len = 1; 
    
    // Traverse the array from the first element
    for (int i = 0; i < n - 1; i++) 
    { 
        if (i % 2 == 0 && (a[i] < a[i + 1]))  
            len++;
        else if (i % 2 == 1 && (a[i] > a[i + 1]))
            len++;
        else
        { 
            // Check if 'maxLen' is less than the length of the current zigzag subarray
            // If true, update 'maxLen'
            if (maxLen < len)   
                maxLen = len; 
                
            // Reset 'len' to 1 as a new zigzag subarray starts
            len = 1;     
        }    
    } 
    
    // Compare the length of the last zigzag subarray with 'maxLen'
    if (maxLen < len) 
        maxLen = len; 
    
    // Return the maximum length
    return maxLen; 
} 

// Driver program to test the above function
int main() 
{ 
    int arr[] = {1, 5, 0, 4, 3}; 
    int n = sizeof(arr) / sizeof(arr[0]); 
    cout << "Length = " << lenOfLongZigZagArr(arr, n); 
    return 0;    
}

Output: