Given an array arr[], the task is to remove at most one element and calculate the maximum length of strictly increasing subarray.
Examples:
Input: arr[] = {1, 2, 5, 3, 4}
Output: 4
After deleting 5, the resulting array will be {1, 2, 3, 4}
and the maximum length of its strictly increasing subarray is 4.Input: arr[] = {1, 2}
Output: 2
The complete array is already strictly increasing.
Approach:
- Create two arrays pre[] and pos[] of size N.
- Iterate over the input array arr[] from (0, N) to find out the contribution of the current element arr[i] in the array till now [0, i) and update the pre[] array if it contributes to the strictly increasing subarray.
- Iterate over the input array arr[] from [N – 2, 0] to find out the contribution of the current element arr[j] in the array till now (N, j) and update the pos[] array if arr[j] contributes in the longest increasing subarray.
- Calculate the maximum length of the strictly increasing subarray without removing any element.
- Iterate over the array pre[] and pos[] to find out the contribution of the current element by excluding that element.
- Maintain a variable ans to find the maximum found till now.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
// Function to return the maximum length of strictly // increasing subarray after removing atmost one element int maxIncSubarr( int a[], int n)
{ // Create two arrays pre and pos
int pre[n] = { 0 };
int pos[n] = { 0 };
pre[0] = 1;
pos[n - 1] = 1;
int l = 0;
// Find out the contribution of the current element in
// array[0, i] and update pre[i]
for ( int i = 1; i < n; i++) {
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current element in
// array[N - 1, i] and update pos[i]
l = 1;
for ( int i = n - 2; i >= 0; i--) {
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the strictly
// increasing subarray without removing any element
int ans = 0;
l = 1;
for ( int i = 1; i < n; i++) {
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = max(ans, l);
}
// Calculate the maximum length of the strictly
// increasing subarray after removing the current
// element
for ( int i = 1; i <= n - 2; i++)
if (a[i - 1] < a[i + 1])
ans = max(pre[i - 1] + pos[i + 1], ans);
return ans;
} // Driver code int main()
{ int arr[] = { 1, 2 };
int n = sizeof (arr) / sizeof ( int );
cout << maxIncSubarr(arr, n);
return 0;
} |
C
// C implementation of the approach #include <stdio.h> // Find maximum between two numbers. int max( int num1, int num2)
{ return (num1 > num2) ? num1 : num2;
} // Function to return the maximum length of strictly // increasing subarray after removing atmost one element int maxIncSubarr( int a[], int n)
{ // Create two arrays pre and pos
int pre[n];
int pos[n];
for ( int i = 0; i < n; i++)
pre[i] = 0;
for ( int i = 0; i < n; i++)
pos[i] = 0;
pre[0] = 1;
pos[n - 1] = 1;
int l = 0;
// Find out the contribution of the current element in
// array[0, i] and update pre[i]
for ( int i = 1; i < n; i++) {
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current element in
// array[N - 1, i] and update pos[i]
l = 1;
for ( int i = n - 2; i >= 0; i--) {
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the strictly
// increasing subarray without removing any element
int ans = 0;
l = 1;
for ( int i = 1; i < n; i++) {
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = max(ans, l);
}
// Calculate the maximum length of the strictly
// increasing subarray after removing the current
// element
for ( int i = 1; i <= n - 2; i++)
if (a[i - 1] < a[i + 1])
ans = max(pre[i - 1] + pos[i + 1], ans);
return ans;
} // Driver code int main()
{ int arr[] = { 1, 2 };
int n = sizeof (arr) / sizeof ( int );
printf ( "%d" , maxIncSubarr(arr, n));
return 0;
} // This code is contributed by Sania Kumari Gupta |
Java
// Java implementation of the approach import java.io.*;
class GFG
{ // Function to return the maximum length of
// strictly increasing subarray after
// removing atmost one element
static int maxIncSubarr( int a[], int n)
{
// Create two arrays pre and pos
int pre[] = new int [n] ;
int pos[] = new int [n] ;
pre[ 0 ] = 1 ;
pos[n - 1 ] = 1 ;
int l = 0 ;
// Find out the contribution of the current
// element in array[0, i] and update pre[i]
for ( int i = 1 ; i < n; i++)
{
if (a[i] > a[i - 1 ])
pre[i] = pre[i - 1 ] + 1 ;
else
pre[i] = 1 ;
}
// Find out the contribution of the current
// element in array[N - 1, i] and update pos[i]
l = 1 ;
for ( int i = n - 2 ; i >= 0 ; i--)
{
if (a[i] < a[i + 1 ])
pos[i] = pos[i + 1 ] + 1 ;
else
pos[i] = 1 ;
}
// Calculate the maximum length of the
// strictly increasing subarray without
// removing any element
int ans = 0 ;
l = 1 ;
for ( int i = 1 ; i < n; i++)
{
if (a[i] > a[i - 1 ])
l++;
else
l = 1 ;
ans = Math.max(ans, l);
}
// Calculate the maximum length of the
// strictly increasing subarray after
// removing the current element
for ( int i = 1 ; i <= n - 2 ; i++)
{
if (a[i - 1 ] < a[i + 1 ])
ans = Math.max(pre[i - 1 ] +
pos[i + 1 ], ans);
}
return ans;
}
// Driver code
public static void main (String[] args)
{
int arr[] = { 1 , 2 };
int n = arr.length;
System.out.println(maxIncSubarr(arr, n));
}
} // This code is contributed by AnkitRai01 |
Python3
# Python implementation of the approach # Function to return the maximum length of # strictly increasing subarray after # removing atmost one element def maxIncSubarr(a, n):
# Create two arrays pre and pos
pre = [ 0 ] * n;
pos = [ 0 ] * n;
pre[ 0 ] = 1 ;
pos[n - 1 ] = 1 ;
l = 0 ;
# Find out the contribution of the current
# element in array[0, i] and update pre[i]
for i in range ( 1 , n):
if (a[i] > a[i - 1 ]):
pre[i] = pre[i - 1 ] + 1 ;
else :
pre[i] = 1 ;
# Find out the contribution of the current
# element in array[N - 1, i] and update pos[i]
l = 1 ;
for i in range (n - 2 , - 1 , - 1 ):
if (a[i] < a[i + 1 ]):
pos[i] = pos[i + 1 ] + 1 ;
else :
pos[i] = 1 ;
# Calculate the maximum length of the
# strictly increasing subarray without
# removing any element
ans = 0 ;
l = 1 ;
for i in range ( 1 , n):
if (a[i] > a[i - 1 ]):
l + = 1 ;
else :
l = 1 ;
ans = max (ans, l);
# Calculate the maximum length of the
# strictly increasing subarray after
# removing the current element
for i in range ( 1 , n - 1 ):
if (a[i - 1 ] < a[i + 1 ]):
ans = max (pre[i - 1 ] + pos[i + 1 ], ans);
return ans;
# Driver code if __name__ = = '__main__' :
arr = [ 1 , 2 ];
n = len (arr);
print (maxIncSubarr(arr, n));
# This code is contributed by PrinciRaj1992 |
C#
// C# implementation of the approach using System;
class GFG
{ // Function to return the maximum length of
// strictly increasing subarray after
// removing atmost one element
static int maxIncSubarr( int []a, int n)
{
// Create two arrays pre and pos
int []pre = new int [n] ;
int []pos = new int [n] ;
pre[0] = 1;
pos[n - 1] = 1;
int l = 0;
// Find out the contribution of the current
// element in array[0, i] and update pre[i]
for ( int i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current
// element in array[N - 1, i] and update pos[i]
l = 1;
for ( int i = n - 2; i >= 0; i--)
{
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the
// strictly increasing subarray without
// removing any element
int ans = 0;
l = 1;
for ( int i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = Math.Max(ans, l);
}
// Calculate the maximum length of the
// strictly increasing subarray after
// removing the current element
for ( int i = 1; i <= n - 2; i++)
{
if (a[i - 1] < a[i + 1])
ans = Math.Max(pre[i - 1] +
pos[i + 1], ans);
}
return ans;
}
// Driver code
public static void Main()
{
int []arr = {1, 2};
int n = arr.Length;
Console.WriteLine(maxIncSubarr(arr, n));
}
} // This code is contributed by AnkitRai01 |
Javascript
<script> // Javascript implementation of the approach // Function to return the maximum length of // strictly increasing subarray after // removing atmost one element function maxIncSubarr(a, n)
{ // Create two arrays pre and pos
let pre = new Array(n);
let pos = new Array(n);
pre.fill(0);
pos.fill(0);
pre[0] = 1;
pos[n - 1] = 1;
let l = 0;
// Find out the contribution of the current
// element in array[0, i] and update pre[i]
for (let i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
pre[i] = pre[i - 1] + 1;
else
pre[i] = 1;
}
// Find out the contribution of the current
// element in array[N - 1, i] and update pos[i]
l = 1;
for (let i = n - 2; i >= 0; i--)
{
if (a[i] < a[i + 1])
pos[i] = pos[i + 1] + 1;
else
pos[i] = 1;
}
// Calculate the maximum length of the
// strictly increasing subarray without
// removing any element
let ans = 0;
l = 1;
for (let i = 1; i < n; i++)
{
if (a[i] > a[i - 1])
l++;
else
l = 1;
ans = Math.max(ans, l);
}
// Calculate the maximum length of the
// strictly increasing subarray after
// removing the current element
for (let i = 1; i <= n - 2; i++)
{
if (a[i - 1] < a[i + 1])
ans = Math.max(pre[i - 1] +
pos[i + 1], ans);
}
return ans;
} // Driver code let arr = [ 1, 2 ]; let n = arr.length; document.write(maxIncSubarr(arr, n)); // This code is contributed by rameshtravel07 </script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(N)