Given an array containing n numbers. The problem is to find the length of the longest contiguous subarray in a circular manner such that every element in the subarray is strictly greater than its previous element in the same subarray. Time Complexity should be O(n).
Examples:
Input : arr[] = {2, 3, 4, 5, 1} Output : 5 {2, 3, 4, 5, 1} is the subarray if we circularly start from the last element and then take the first four elements. This will give us an increasing subarray {1, 2, 3, 4, 5} in a circular manner. Input : arr[] = {2, 3, 8, 4, 6, 7, 10, 12, 9, 1} Output : 5
Method 1 (Using extra space): Make a temp[] array of size 2*n. Copy the elements of arr[] in temp[] two times. Now, find length of Longest increasing subarray in temp[].
Method 2 (Without using extra space):
Following are the steps:
- If n == 1, return 1.
- Find length of longest increasing subarray starting with first element of arr[]. Let its length be startLen.
- Starting from the next element where the first increasing subarray ends, find the length of the longest increasing subarray. Let it be max.
- Consider the length of the increasing subarray that ends with the last element of arr[]. Let it be endLen.
- If arr[n-1] < arr[0], then endLen = endLen + startLen.
- Finally, return maximum of (max, endLen, startLen).
Implementation:
C++
// C++ implementation to find length of longest // increasing circular subarray #include <bits/stdc++.h> using namespace std;
// function to find length of longest // increasing circular subarray int longlenCircularSubarr( int arr[], int n)
{ // if there is only one element
if (n == 1)
return 1;
// 'startLen' stores the length of the longest
// increasing subarray which starts from
// first element
int startLen = 1, i;
int len = 1, max = 0;
// finding the length of the longest
// increasing subarray starting from
// first element
for (i = 1; i < n; i++) {
if (arr[i - 1] < arr[i])
startLen++;
else
break ;
}
if (max < startLen)
max = startLen;
// traverse the array index (i+1)
for ( int j = i + 1; j < n; j++) {
// if current element if greater than previous
// element, then this element helps in building
// up the previous increasing subarray encountered
// so far
if (arr[j - 1] < arr[j])
len++;
else {
// check if 'max' length is less than the length
// of the current increasing 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 increasing
// subarray is being calculated
len = 1;
}
}
// if true, then add length of the increasing
// subarray ending at last element with the
// length of the increasing subarray starting
// from first element - This is done for
// circular rotation
if (arr[n - 1] < arr[0])
len += startLen;
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if (max < len)
max = len;
return max;
} // Driver program to test above int main()
{ int arr[] = { 2, 3, 4, 5, 1 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << "Length = "
<< longlenCircularSubarr(arr, n);
return 0;
} |
Java
// Java implementation to find length // of longest increasing circular subarray class Circular
{ // function to find length of longest
// increasing circular subarray
public static int longlenCircularSubarr( int arr[],
int n)
{
// if there is only one element
if (n == 1 )
return 1 ;
// 'startLen' stores the length of the
// longest increasing subarray which
// starts from first element
int startLen = 1 , i;
int len = 1 , max = 0 ;
// finding the length of the longest
// increasing subarray starting from
// first element
for (i = 1 ; i < n; i++) {
if (arr[i - 1 ] < arr[i])
startLen++;
else
break ;
}
if (max < startLen)
max = startLen;
// traverse the array index (i+1)
for ( int j = i + 1 ; j < n; j++) {
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered so far
if (arr[j - 1 ] < arr[j])
len++;
else {
// check if 'max' length is less than
// the length of the current increasing
// 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 increasing
// subarray is being calculated
len = 1 ;
}
}
// if true, then add length of the increasing
// subarray ending at last element with the
// length of the increasing subarray starting
// from first element - This is done for
// circular rotation
if (arr[n - 1 ] < arr[ 0 ])
len += startLen;
// check if 'max' length is less than the
// length of the current increasing subarray.
// If true, then update 'max'
if (max < len)
max = len;
return max;
}
// driver code
public static void main(String[] args)
{
int arr[] = { 2 , 3 , 4 , 5 , 1 };
int n = 5 ;
System.out.print( "Length = " +
longlenCircularSubarr(arr, n));
}
} // This code is contributed by rishabh_jain |
Python3
# Python3 implementation to find length # of longest increasing circular subarray # function to find length of longest # increasing circular subarray def longlenCircularSubarr (arr, n):
# if there is only one element
if n = = 1 :
return 1
# 'startLen' stores the length of the
# longest increasing subarray which
# starts from first element
startLen = 1
len = 1
max = 0
# finding the length of the longest
# increasing subarray starting from
# first element
for i in range ( 1 , n):
if arr[i - 1 ] < arr[i]:
startLen + = 1
else :
break
if max < startLen:
max = startLen
# traverse the array index (i+1)
for j in range (i + 1 , n):
# if current element if greater than
# previous element, then this element
# helps in building up the previous
# increasing subarray encountered
# so far
if arr[j - 1 ] < arr[j]:
len + = 1
else :
# check if 'max' length is less
# than the length of the current
# increasing 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 increasing subarray is
# being calculated
len = 1
# if true, then add length of the increasing
# subarray ending at last element with the
# length of the increasing subarray starting
# from first element - This is done for
# circular rotation
if arr[n - 1 ] < arr[ 0 ]:
len + = startLen
# check if 'max' length is less than the
# length of the current increasing subarray.
# If true, then update 'max'
if max < len :
max = len
return max
# Driver code to test above arr = [ 2 , 3 , 4 , 5 , 1 ]
n = len (arr)
print ( "Length = " ,longlenCircularSubarr(arr, n))
# This code is contributed by "Sharad_Bhardwaj". |
C#
// C# implementation to find length // of longest increasing circular subarray using System;
public class GFG {
// function to find length of longest
// increasing circular subarray
static int longlenCircularSubarr( int [] arr, int n)
{
// if there is only one element
if (n == 1)
return 1;
// 'startLen' stores the length of the longest
// increasing subarray which starts from
// first element
int startLen = 1, i;
int len = 1, max = 0;
// finding the length of the longest
// increasing subarray starting from
// first element
for (i = 1; i < n; i++) {
if (arr[i - 1] < arr[i])
startLen++;
else
break ;
}
if (max < startLen)
max = startLen;
// traverse the array index (i+1)
for ( int j = i + 1; j < n; j++) {
// if current element if greater than previous
// element, then this element helps in building
// up the previous increasing subarray encountered
// so far
if (arr[j - 1] < arr[j])
len++;
else {
// check if 'max' length is less than the length
// of the current increasing 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 increasing
// subarray is being calculated
len = 1;
}
}
// if true, then add length of the increasing
// subarray ending at last element with the
// length of the increasing subarray starting
// from first element - This is done for
// circular rotation
if (arr[n - 1] < arr[0])
len += startLen;
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if (max < len)
max = len;
return max;
}
// Driver program to test above
static public void Main()
{
int [] arr = { 2, 3, 4, 5, 1 };
int n = arr.Length;
Console.WriteLine( "Length = " +
longlenCircularSubarr(arr, n));
// Code
}
} // This code is contributed by vt_m. |
PHP
<?php // PHP implementation to find length of longest // increasing circular subarray // function to find length of longest // increasing circular subarray function longlenCircularSubarr(& $arr , $n )
{ // if there is only one element
if ( $n == 1)
return 1;
// 'startLen' stores the length of the longest
// increasing subarray which starts from
// first element
$startLen = 1;
$len = 1;
$max = 0;
// finding the length of the longest
// increasing subarray starting from
// first element
for ( $i = 1; $i < $n ; $i ++)
{
if ( $arr [ $i - 1] < $arr [ $i ])
$startLen ++;
else
break ;
}
if ( $max < $startLen )
$max = $startLen ;
// traverse the array index (i+1)
for ( $j = $i + 1; $j < $n ; $j ++)
{
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered
// so far
if ( $arr [ $j - 1] < $arr [ $j ])
$len ++;
else
{
// check if 'max' length is less than
// the length of the current increasing
// 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 increasing
// subarray is being calculated
$len = 1;
}
}
// if true, then add length of the increasing
// subarray ending at last element with the
// length of the increasing subarray starting
// from first element - This is done for
// circular rotation
if ( $arr [ $n - 1] < $arr [0])
$len += $startLen ;
// check if 'max' length is less than the length
// of the current increasing subarray. If true,
// then update 'max'
if ( $max < $len )
$max = $len ;
return $max ;
} // Driver Code $arr = array ( 2, 3, 4, 5, 1 );
$n = sizeof( $arr );
echo "Length = " . longlenCircularSubarr( $arr , $n );
// This code is contributed by ita_c ?> |
Javascript
<script> // Javascript implementation to find length // of longest increasing circular subarray // function to find length of longest
// increasing circular subarray
function longlenCircularSubarr(arr,n)
{
// if there is only one element
if (n == 1)
return 1;
// 'startLen' stores the length of the
// longest increasing subarray which
// starts from first element
let startLen = 1, i;
let len = 1, max = 0;
// finding the length of the longest
// increasing subarray starting from
// first element
for (i = 1; i < n; i++) {
if (arr[i - 1] < arr[i])
startLen++;
else
break ;
}
if (max < startLen)
max = startLen;
// traverse the array index (i+1)
for (let j = i + 1; j < n; j++) {
// if current element if greater than
// previous element, then this element
// helps in building up the previous
// increasing subarray encountered so far
if (arr[j - 1] < arr[j])
len++;
else {
// check if 'max' length is less than
// the length of the current increasing
// 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 increasing
// subarray is being calculated
len = 1;
}
}
// if true, then add length of the increasing
// subarray ending at last element with the
// length of the increasing subarray starting
// from first element - This is done for
// circular rotation
if (arr[n - 1] < arr[0])
len += startLen;
// check if 'max' length is less than the
// length of the current increasing subarray.
// If true, then update 'max'
if (max < len)
max = len;
return max;
}
// driver code
let arr=[2, 3, 4, 5, 1 ];
let n = 5;
document.write( "Length = " +
longlenCircularSubarr(arr, n));
// This code is contributed by avanitrachhadiya2155
</script> |
Output
Length = 5
Time Complexity: O(n).