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 

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 <iostream>
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[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
int main()
{
    int arr[] = { 2, 4, 8, 6, 9, 12 };
    int n = sizeof(arr) / sizeof(arr[0]);

    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[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()
    {

        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

Javascript

<script>


// Javascript implementation of the approach

// Function that returns true if arr[]
// can be made strictly increasing after
// modifying at most one element
function check(arr, n)
{
    // To store the number of modifications
    // required to make the array
    // strictly increasing
    var 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 (var 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
var arr = [ 2, 4, 8, 6, 9, 12 ];
var n = arr.length;
if (check(arr, n))
    document.write( "Yes" );
else
    document.write( "No" );

// This code is contributed by noob2000.
</script>

Output:

Yes

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