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# Check whether an array can be made strictly decreasing by modifying at most one element

Given an array arr[] of positive integers, the task is to find whether it is possible to make this array strictly decreasing by modifying at most one element.
Examples:

Input: arr[] = {12, 9, 10, 5, 2}
Output: Yes
{12, 11, 10, 5, 2} is one of the valid solutions.
Input: arr[] = {1, 2, 3, 4}
Output: No

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 + 1] then the array cannot be made strictly decreasing without affecting at most one 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 the array``// can be made strictly decreasing``// with at most one change``bool` `check(vector<``int``> arr, ``int` `n)``{` `    ``// To store the number of modifications``    ``// required to make the array``    ``// strictly decreasing``    ``int` `modify = 0;` `    ``// Check whether the last element needs``    ``// to be modify or not``    ``if` `(arr[n - 1] >= arr[n - 2]) {``        ``arr[n - 1] = arr[n - 2] - 1;``        ``modify++;``    ``}` `    ``// Check whether the first element needs``    ``// to be modify or not``    ``if` `(arr[0] <= arr[1]) {``        ``arr[0] = arr[1] + 1;``        ``modify++;``    ``}` `    ``// Loop from 2nd element to the 2nd last element``    ``for` `(``int` `i = n - 2; i > 0; 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;``            ``modify++;` `            ``// 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``;``        ``}``    ``}` `    ``// If more than 1 modification is required``    ``if` `(modify > 1)``        ``return` `false``;` `    ``return` `true``;``}` `// Driver code``int` `main()``{` `    ``vector<``int``> arr = { 10, 5, 11, 2 };``    ``int` `n = arr.size();` `    ``if` `(check(arr, n))``        ``cout << ``"Yes"``;``    ``else``        ``cout << ``"No"``;` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``class` `GFG {` `    ``// Function that returns true if the array``    ``// can be made strictly decreasing``    ``// with at most one change``    ``public` `static` `boolean` `check(``int``[] arr, ``int` `n)``    ``{` `        ``// To store the number of modifications``        ``// required to make the array``        ``// strictly decreasing``        ``int` `modify = ``0``;` `        ``// Check whether the last element needs``        ``// to be modify or not``        ``if` `(arr[n - ``1``] >= arr[n - ``2``]) {``            ``arr[n - ``1``] = arr[n - ``2``] - ``1``;``            ``modify++;``        ``}` `        ``// Check whether the first element needs``        ``// to be modify or not``        ``if` `(arr[``0``] <= arr[``1``]) {``            ``arr[``0``] = arr[``1``] + ``1``;``            ``modify++;``        ``}` `        ``// Loop from 2nd element to the 2nd last element``        ``for` `(``int` `i = n - ``2``; i > ``0``; 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``;``                ``modify++;` `                ``// 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``;``            ``}``        ``}` `        ``// If more than 1 modification is required``        ``if` `(modify > ``1``)``            ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int``[] arr = { ``10``, ``5``, ``11``, ``3` `};``        ``int` `n = arr.length;` `        ``if` `(check(arr, n))``            ``System.out.print(``"Yes"``);``        ``else``            ``System.out.print(``"No"``);``    ``}``}`

## Python3

 `# Python3 implementation of the approach` `# Function that returns true if the array``# can be made strictly decreasing``# with at most one change``def` `check(arr, n):` `    ``modify ``=` `0``    ` `    ``# Check whether the last element needs``    ``# to be modify or not``    ``if` `(arr[n ``-` `1``] >``=` `arr[n ``-` `2``]):``        ``arr[n``-``1``] ``=` `arr[n``-``2``] ``-` `1``        ``modify ``+``=` `1``    ` `    ``# Check whether the first element needs``    ``# to be modify or not``    ``if` `(arr[``0``] <``=` `arr[``1``]):``        ``arr[``0``] ``=` `arr[``1``] ``+` `1``        ``modify ``+``=` `1` `    ``# Loop from 2nd element to the 2nd last element``    ``for` `i ``in` `range``(n``-``2``, ``0``, ``-``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``            ``modify ``+``=` `1``            ` `            ``# 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`  `    ``# If more than 1 modification is required``    ``if` `(modify > ``1``):``        ``return` `False` `    ``return` `True` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``10``, ``5``, ``11``, ``3``]``    ``n ``=` `len``(arr)` `    ``if` `(check(arr, n)):``        ``print``(``"Yes"``)``    ``else``:``        ``print``(``"No"``)`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `    ``// Function that returns true if the array``    ``// can be made strictly decreasing``    ``// with at most one change``    ``public` `static` `bool` `check(``int``[]arr, ``int` `n)``    ``{` `        ``// To store the number of modifications``        ``// required to make the array``        ``// strictly decreasing``        ``int` `modify = 0;` `        ``// Check whether the last element needs``        ``// to be modify or not``        ``if` `(arr[n - 1] >= arr[n - 2])``        ``{``            ``arr[n - 1] = arr[n - 2] - 1;``            ``modify++;``        ``}` `        ``// Check whether the first element needs``        ``// to be modify or not``        ``if` `(arr[0] <= arr[1])``        ``{``            ``arr[0] = arr[1] + 1;``            ``modify++;``        ``}` `        ``// Loop from 2nd element to the 2nd last element``        ``for` `(``int` `i = n - 2; i > 0; 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;``                ``modify++;` `                ``// 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``;``            ``}``        ``}` `        ``// If more than 1 modification is required``        ``if` `(modify > 1)``            ``return` `false``;` `        ``return` `true``;``    ``}` `    ``// Driver code``    ``static` `public` `void` `Main ()``    ``{``        ``int``[]arr = { 10, 5, 11, 3 };``        ``int` `n = arr.Length;` `        ``if` `(check(arr, n))``            ``Console.Write(``"Yes"``);``        ``else``            ``Console.Write(``"No"``);``    ``}``}` `// This code is contributed by ajit.`

## Javascript

 ``

Output:

`Yes`

Time Complexity: O(N)
Auxiliary Space: O(1)