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++ 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 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 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# 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 |
<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> |
2
Complexity Analysis :
Time Complexity – O(n), where n is the length of the array.
Auxiliary Space – O(1), no extra space required.