# Check whether an array can be made strictly increasing by modifying atmost one element

Given an array arr[] of positive integers, the task is to find whether it is possible to make this array strictly increasing by modifying atmost one element.

Examples:

Input: arr[] = {2, 4, 8, 6, 9, 12}
Output: Yes
By modifying 8 to 5, array will become strictly increasing.
i.e. {2, 4, 5, 6, 9, 12}

Input: arr[] = {10, 5, 2}
Output: No

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: For every element arr[i], if it is greater than both arr[i – 1] and arr[i + 1] or it is smaller than both arr[i – 1] and arr[i + 1] then arr[i] needs to be modified.
i.e. arr[i] = (arr[i – 1] + arr[i + 1]) / 2. If after modification, arr[i] = arr[i – 1] or arr[i] = arr[i + 1] then the array cannot be made strictly increasing without affecting more than a single element. Else count all such modifications, if the count of modifications in the end is less than or equal to 1 then print “Yes” else print “No”.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach ` `#include ` `using` `namespace` `std; ` ` `  `// Function that returns true if arr[] ` `// can be made strictly increasing after ` `// modifying at most one element ` `bool` `check(``int` `arr[], ``int` `n) ` `{ ` `    ``// To store the number of modifications ` `    ``// required to make the array ` `    ``// strictly increasing ` `    ``int` `modify = 0; ` ` `  `    ``// Check whether the first element needs ` `    ``// to be modify or not ` `    ``if` `(arr > arr) { ` `        ``arr = arr / 2; ` `        ``modify++; ` `    ``} ` ` `  `    ``// Loop from 2nd element to the 2nd last element ` `    ``for` `(``int` `i = 1; i < n - 1; i++) { ` ` `  `        ``// Check whether arr[i] needs to be modified ` `        ``if` `((arr[i - 1] < arr[i] && arr[i + 1] < arr[i]) ` `            ``|| (arr[i - 1] > arr[i] && arr[i + 1] > arr[i])) { ` ` `  `            ``// Modifying arr[i] ` `            ``arr[i] = (arr[i - 1] + arr[i + 1]) / 2; ` ` `  `            ``// Check if arr[i] is equal to any of ` `            ``// arr[i-1] or arr[i+1] ` `            ``if` `(arr[i] == arr[i - 1] || arr[i] == arr[i + 1]) ` `                ``return` `false``; ` ` `  `            ``modify++; ` `        ``} ` `    ``} ` ` `  `    ``// Check whether the last element needs ` `    ``// to be modify or not ` `    ``if` `(arr[n - 1] < arr[n - 2]) ` `        ``modify++; ` ` `  `    ``// If more than 1 modification is required ` `    ``if` `(modify > 1) ` `        ``return` `false``; ` ` `  `    ``return` `true``; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 2, 4, 8, 6, 9, 12 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``if` `(check(arr, n)) ` `        ``cout << ``"Yes"` `<< endl; ` `    ``else` `        ``cout << ``"No"` `<< endl; ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the approach ` `class` `GFG { ` ` `  `    ``// Function that returns true if arr[] ` `    ``// can be made strictly increasing after ` `    ``// modifying at most one element ` `    ``static` `boolean` `check(``int` `arr[], ``int` `n) ` `    ``{ ` `        ``// To store the number of modifications ` `        ``// required to make the array ` `        ``// strictly increasing ` `        ``int` `modify = ``0``; ` ` `  `        ``// Check whether the first element needs ` `        ``// to be modify or not ` `        ``if` `(arr[``0``] > arr[``1``]) { ` `            ``arr[``0``] = arr[``1``] / ``2``; ` `            ``modify++; ` `        ``} ` ` `  `        ``// Loop from 2nd element to the 2nd last element ` `        ``for` `(``int` `i = ``1``; i < n - ``1``; i++) { ` ` `  `            ``// Check whether arr[i] needs to be modified ` `            ``if` `((arr[i - ``1``] < arr[i] && arr[i + ``1``] < arr[i]) ` `                ``|| (arr[i - ``1``] > arr[i] && arr[i + ``1``] > arr[i])) { ` ` `  `                ``// Modifying arr[i] ` `                ``arr[i] = (arr[i - ``1``] + arr[i + ``1``]) / ``2``; ` ` `  `                ``// Check if arr[i] is equal to any of ` `                ``// arr[i-1] or arr[i+1] ` `                ``if` `(arr[i] == arr[i - ``1``] || arr[i] == arr[i + ``1``]) ` `                    ``return` `false``; ` ` `  `                ``modify++; ` `            ``} ` `        ``} ` ` `  `        ``// Check whether the last element needs ` `        ``// to be modify or not ` `        ``if` `(arr[n - ``1``] < arr[n - ``2``]) ` `            ``modify++; ` ` `  `        ``// If more than 1 modification is required ` `        ``if` `(modify > ``1``) ` `            ``return` `false``; ` ` `  `        ``return` `true``; ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String[] args) ` `    ``{ ` ` `  `        ``int``[] arr = { ``2``, ``4``, ``8``, ``6``, ``9``, ``12` `}; ` `        ``int` `n = arr.length; ` ` `  `        ``if` `(check(arr, n)) ` `            ``System.out.print(``"Yes"``); ` `        ``else` `            ``System.out.print(``"No"``); ` `    ``} ` `} `

## C#

 `// C# implementation of the approach  ` `using` `System; ` ` `  `class` `GFG ` `{  ` ` `  `    ``// Function that returns true if arr[]  ` `    ``// can be made strictly increasing after  ` `    ``// modifying at most one element  ` `    ``static` `bool` `check(``int` `[]arr, ``int` `n)  ` `    ``{  ` `        ``// To store the number of modifications  ` `        ``// required to make the array  ` `        ``// strictly increasing  ` `        ``int` `modify = 0;  ` ` `  `        ``// Check whether the first element needs  ` `        ``// to be modify or not  ` `        ``if` `(arr > arr)  ` `        ``{  ` `            ``arr = arr / 2;  ` `            ``modify++;  ` `        ``}  ` ` `  `        ``// Loop from 2nd element to the 2nd last element  ` `        ``for` `(``int` `i = 1; i < n - 1; i++)  ` `        ``{  ` ` `  `            ``// Check whether arr[i] needs to be modified  ` `            ``if` `((arr[i - 1] < arr[i] && arr[i + 1] < arr[i])  ` `                ``|| (arr[i - 1] > arr[i] && arr[i + 1] > arr[i]))  ` `            ``{  ` ` `  `                ``// Modifying arr[i]  ` `                ``arr[i] = (arr[i - 1] + arr[i + 1]) / 2;  ` ` `  `                ``// Check if arr[i] is equal to any of  ` `                ``// arr[i-1] or arr[i+1]  ` `                ``if` `(arr[i] == arr[i - 1] || arr[i] == arr[i + 1])  ` `                    ``return` `false``;  ` ` `  `                ``modify++;  ` `            ``}  ` `        ``}  ` ` `  `        ``// Check whether the last element needs  ` `        ``// to be modify or not  ` `        ``if` `(arr[n - 1] < arr[n - 2])  ` `            ``modify++;  ` ` `  `        ``// If more than 1 modification is required  ` `        ``if` `(modify > 1)  ` `            ``return` `false``;  ` ` `  `        ``return` `true``;  ` `    ``}  ` ` `  `    ``// Driver code  ` `    ``public` `static` `void` `Main()  ` `    ``{  ` ` `  `        ``int``[] arr = { 2, 4, 8, 6, 9, 12 };  ` `        ``int` `n = arr.Length;  ` ` `  `        ``if` `(check(arr, n))  ` `            ``Console.WriteLine(``"Yes"``);  ` `        ``else` `            ``Console.WriteLine(``"No"``);  ` `    ``}  ` `}  ` ` `  `// This code is contributed by AnkitRai01 `

## Python 3

 `# Python 3 implementation of above approach ` ` `  `# Function that returns true if arr[] ` `# can be made strictly increasing after ` `# modifying at most one element ` `def` `check( arr, n): ` ` `  `    ``# To store the number of modifications ` `    ``# required to make the array ` `    ``# strictly increasing ` `    ``modify ``=` `0` ` `  `    ``# Check whether the first element needs ` `    ``# to be modify or not ` `    ``if` `(arr[``0``] > arr[``1``]) : ` `        ``arr[``0``] ``=` `arr[``1``] ``/``/` `2` `        ``modify``+``=``1` `     `  ` `  `    ``# Loop from 2nd element to the 2nd last element ` `    ``for` `i ``in` `range` `( ``1``, n ``-` `1``): ` ` `  `        ``# Check whether arr[i] needs to be modified ` `        ``if` `((arr[i ``-` `1``] < arr[i] ``and` `arr[i ``+` `1``] < arr[i]) ` `            ``or` `(arr[i ``-` `1``] > arr[i] ``and` `arr[i ``+` `1``] > arr[i])): ` ` `  `            ``# Modifying arr[i] ` `            ``arr[i] ``=` `(arr[i ``-` `1``] ``+` `arr[i ``+` `1``]) ``/``/` `2` ` `  `            ``# Check if arr[i] is equal to any of ` `            ``# arr[i-1] or arr[i+1] ` `            ``if` `(arr[i] ``=``=` `arr[i ``-` `1``] ``or` `arr[i] ``=``=` `arr[i ``+` `1``]): ` `                ``return` `False` ` `  `            ``modify``+``=``1` `         `  ` `  `    ``# Check whether the last element needs ` `    ``# to be modify or not ` `    ``if` `(arr[n ``-` `1``] < arr[n ``-` `2``]): ` `        ``modify``+``=``1` ` `  `    ``# If more than 1 modification is required ` `    ``if` `(modify > ``1``): ` `        ``return` `False` ` `  `    ``return` `True` ` `  `# Driver code ` `if` `__name__ ``=``=` `"__main__"``: ` `    ``arr ``=` `[ ``2``, ``4``, ``8``, ``6``, ``9``, ``12` `] ` `    ``n ``=` `len``(arr) ` ` `  `    ``if` `(check(arr, n)): ` `        ``print` `( ``"Yes"``) ` `    ``else``: ` `        ``print` `(``"No"``) ` ` `  `# This code is contributed by ChitraNayal     `

Output:

```Yes
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01, chitranayal