Length of the longest ZigZag subarray of the given array

Given an array arr[] containing n numbers, the task 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

Examples:

Input: arr[] = {12, 13, 1, 5, 4, 7, 8, 10, 10, 11} 
Output: 6 
Explanation:
The subarray is {12, 13, 1, 5, 4, 7} whose length is 6 and is in zigzag fashion.

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

Approach: To solve the problem mentioned above following steps are followed: 

• Initially initialize cnt, a counter 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.

Below is the implementation of the above approach:  

C++

// C++ implementation to find
// the length of longest zigzag
// subarray of the given array

#include <bits/stdc++.h>
using namespace std;

// Function to find the length of
// longest zigZag contiguous subarray
int lenOfLongZigZagArr(int a[], int n)
{

    // 'max' to store the length
    // of longest zigZag subarray
    int max = 1,

        // 'len' to store the lengths
        // of longest zigZag subarray
        // at different instants of time
        len = 1;

    // Traverse the array from the beginning
    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 'max' length
            // is less than the length
            // of the current zigzag subarray.
            // If true, then update 'max'
            if (max < len)
                max = len;

            // Reset 'len' to 1
            // as from this element,
            // again the length of the
            // new zigzag subarray
            // is being calculated
            len = 1;
        }
    }

    // comparing the length of the last
    // zigzag subarray with 'max'
    if (max < len)
        max = len;

    // Return required maximum length
    return max;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);

    cout << lenOfLongZigZagArr(arr, n);

    return 0;
}

Java

// Java implementation to find
// the length of longest zigzag
// subarray of the given array
import java.io.*;
import java.util.*;
class GFG {

// Function to find the length of
// longest zigZag contiguous subarray
static int lenOfLongZigZagArr(int a[], int n)
{
    // 'max' to store the length
    // of longest zigZag subarray
    int max = 1,

    // 'len' to store the lengths
    // of longest zigZag subarray
    // at different instants of time
    len = 1;

    // Traverse the array from the beginning
    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 'max' length
            // is less than the length
            // of the current zigzag subarray.
            // If true, then update 'max'
            if (max < len)
                max = len;

            // Reset 'len' to 1
            // as from this element,
            // again the length of the
            // new zigzag subarray
            // is being calculated
            len = 1;
        }
    }

    // comparing the length of the last
    // zigzag subarray with 'max'
    if (max < len)
        max = len;

    // Return required maximum length
    return max;
}

// Driver Code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;

    System.out.println(lenOfLongZigZagArr(arr, n));
}
}

// This code is contributed by coder001

Python3

# Python3 implementation to find the
# length of longest zigzag subarray
# of the given array

# Function to find the length of
# longest zigZag contiguous subarray
def lenOfLongZigZagArr(a, n):

    # '_max' to store the length
    # of longest zigZag subarray
    _max = 1

    # '_len' to store the lengths
    # of longest zigZag subarray
    # at different instants of time
    _len = 1

    # Traverse the array from the beginning
    for i in range(n - 1):

        if i % 2 == 0 and a[i] < a[i + 1]:
            _len += 1

        elif i % 2 == 1 and a[i] > a[i + 1]:
            _len += 1

        else:

            # Check if '_max' length is less than
            # the length of the current zigzag
            # subarray. If true, then update '_max'
            if _max < _len:
                _max = _len

            # Reset '_len' to 1 as from this element,
            # again the length of the new zigzag
            # subarray is being calculated
            _len = 1

    # Comparing the length of the last
    # zigzag subarray with '_max'
    if _max < _len:
        _max = _len

    # Return required maximum length
    return _max

# Driver code
arr = [ 1, 2, 3, 4, 5 ]
n = len(arr)

print(lenOfLongZigZagArr(arr, n))

# This code is contributed by divyamohan123

C#

// C# implementation to find
// the length of longest zigzag
// subarray of the given array
using System;

class GFG{

// Function to find the length of
// longest zigZag contiguous subarray
static int lenOflongZigZagArr(int []a, int n)
{

    // 'max' to store the length
    // of longest zigZag subarray
    int max = 1,

    // 'len' to store the lengths
    // of longest zigZag subarray
    // at different instants of time
    len = 1;

    // Traverse the array from the beginning
    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 'max' length
           // is less than the length
           // of the current zigzag subarray.
           // If true, then update 'max'
           if (max < len)
               max = len;

           // Reset 'len' to 1
           // as from this element,
           // again the length of the
           // new zigzag subarray
           // is being calculated
           len = 1;
       }
    }

    // Comparing the length of the last
    // zigzag subarray with 'max'
    if (max < len)
        max = len;

    // Return required maximum length
    return max;
}

// Driver Code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;

    Console.WriteLine(lenOflongZigZagArr(arr, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// Javascript implementation to find
// the length of longest zigzag
// subarray of the given array

// Function to find the length of
// longest zigZag contiguous subarray
function lenOfLongZigZagArr( a, n)
{

    // 'max' to store the length
    // of longest zigZag subarray
    var max = 1,

    // 'len' to store the lengths
    // of longest zigZag subarray
    // at different instants of time
    len = 1;

    // Traverse the array from the beginning
    for (var 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 'max' length
            // is less than the length
            // of the current zigzag subarray.
            // If true, then update 'max'
            if (max < len)
                max = len;

            // Reset 'len' to 1
            // as from this element,
            // again the length of the
            // new zigzag subarray
            // is being calculated
            len = 1;
        }
    }

    // comparing the length of the last
    // zigzag subarray with 'max'
    if (max < len)
        max = len;

    // Return required maximum length
    return max;
}

// Driver code
var arr = [ 1, 2, 3, 4, 5 ];
var n = arr.length;
document.write( lenOfLongZigZagArr(arr, n));

</script>

Output:

2

Complexity Analysis :

Time Complexity – O(n), where n is the length of the array.

Auxiliary Space – O(1), no extra space required.