# Maximum length of Strictly Increasing Sub-array after removing at most one element

• Difficulty Level : Medium
• Last Updated : 09 Apr, 2021

Given an array arr[], the task is to remove at most one element and calculate the maximum length of strictly increasing subarray.

Examples:

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Input: arr[] = {1, 2, 5, 3, 4}
Output:
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:
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 in 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 stricly 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 ``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``    ``// stricly 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;``}`

## Java

 `// Java implementation of the approach``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``        ``// stricly 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``    ``# stricly 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``        ``// stricly 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

 ``
Output:
`2`

Time Complexity: O(N)

Space Complexity: O(N)

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