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 <bits/stdc++.h>
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

<script>
    // Javascript implementation of the approach

    // Function that returns true if the array
    // can be made strictly decreasing
    // with at most one change
    function check(arr, n)
    {

        // To store the number of modifications
        // required to make the array
        // strictly decreasing
        let 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 (let 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] = parseInt((arr[i - 1] + arr[i + 1]) / 2, 10);
                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;
    }

    let arr = [ 10, 5, 11, 3 ];
    let n = arr.length;

    if (check(arr, n))
      document.write("Yes");
    else
      document.write("No");

</script>

Output:

Yes

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