Reduce N to 0 or less by given X and Y operations

Given three integers N, X, and Y, the task is to check if it is possible to reduce N to 0 or less by the following operations: 

• Update N to ?N/2? + 10, at most X times
• Update N to N - 10, at most Y times.

Example:

Input: N = 100, X = 3, Y = 4
Output: Yes
Explanation:
Update N = 100 to ?100/2? + 10 = 60.
Update N = 60 to 60 ? 10 = 50.
Update N = 50 to ?50/2? + 10 = 35.
Update N = 35 to ?35/2? + 10 = 27.
Update N = 27 to 27 ? 10 = 17. 
Update N = 17 to 17 ? 10 = 7.
Update N = 7 to 7 ? 10 = ?3.

Input: N = 50, X = 1, Y = 2
Output: No

Approach: This problem can be solved based on the following observations:

• Case 1: If ?N/2? + 10 >= N, then perform only second operation as the first operation will increase the value N.
• Case 2: If first operation is performed followed by the second operation then the result is:

Operation 1: N = ?N/2? + 10 
Operation 2: (?N/2? + 10) - 10 = ?N/2?

• 
  The value of N is reduced to (?N/2?).
• Case 3: If second operation is performed followed by the first operation then the result is:

Operation 2: N = N - 10 
Operation 1: ?(N - 10)/2? + 10 = (?N/2? - 5 + 10) = (?N/2? + 5)

• 
  The value of N is reduced to (?N/2? + 5).

From the above two observations, if N > N/2 +10, then it can be concluded that, to reduce the value of N to 0 or less, the first operation must be performed before the second operation to decrease the value of N.
Follow the steps below to solve the problem:

1. Update the value of N to ?N/2? + 10 if N > N/2 + 10 and X > 0.
2. Update the value of N to N – 10 at most Y times.
3. Check if the updated value of N is less than equal to 0 or not.
4. If N ? 0, then print "Yes". Otherwise, print "No".

Below is the implementation of the above approach:

C++

// C++ Program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to check if N can be
// reduced to <= 0 or not
bool NegEqu(int N, int X, int Y)
{

    // Update N to N/2 + 10 at most X times
    while (X-- and N > N/2 + 10) {
        N = N / 2 + 10;
    }

    // Update N to N - 10 Y times
    while (Y--) {
        N = N - 10;
    }

    if (N <= 0)
        return true;

    return false;
}

// Driver Code
int main()
{
    int N = 100;
    int X = 3;
    int Y = 4;

    if (NegEqu(N, X, Y)) {
        cout << "Yes";
    }
    else {
        cout << "No";
    }
}

Java

// Java program to implement
// the above approach
class GFG{

// Function to check if N can be
// reduced to <= 0 or not
static boolean NegEqu(int N, int X, int Y)
{
    
    // Update N to N/2 + 10 at most X times
    while (X > 0 && (N > N / 2 + 10))
    {
        N = N / 2 + 10;
        X -= 1;
    }

    // Update N to N - 10 Y times
    while (Y > 0)
    {
        N = N - 10;
        Y -= 1;
    }

    if (N <= 0)
        return true;

    return false;
}

// Driver Code
public static void main(String[] args)
{
    int N = 100;
    int X = 3;
    int Y = 4;

    if (NegEqu(N, X, Y)) 
    {
        System.out.println("Yes");
    }
    else
    {
        System.out.println("No");
    }
}
}

// This code is contributed by jana_sayantan

Python3

# Python3 program to implement
# the above approach

# Function to check if N can be
# reduced to <= 0 or not
def NegEqu(N, X, Y):

    # Update N to N/2 + 10 at most X times
    while (X and N > N // 2 + 10):
        N = N // 2 + 10
        X -= 1

    # Update N to N - 10 Y times
    while (Y):
        N = N - 10
        Y -= 1

    if (N <= 0):
        return True

    return False

# Driver Code
if __name__ == '__main__':
    
    N = 100
    X = 3
    Y = 4

    if (NegEqu(N, X, Y)):
        print("Yes")
    else:
        print("No")

# This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to check if N can be
// reduced to <= 0 or not
public static bool NegEqu(int N, int X, 
                          int Y)
{
    
    // Update N to N/2 + 10 at most X times
    while (X > 0 && (N > N / 2 + 10))
    {
        N = N / 2 + 10;
        X -= 1;
    }

    // Update N to N - 10 Y times
    while (Y > 0)
    {
        N = N - 10;
        Y -= 1;
    }

    if (N <= 0)
        return true;

    return false;
}
    
// Driver Code 
public static void Main(String[] args) 
{ 
    int N = 100;
    int X = 3;
    int Y = 4;

    if (NegEqu(N, X, Y)) 
    { 
        Console.WriteLine("Yes"); 
    } 
    else
    { 
        Console.WriteLine("No"); 
    } 
} 
} 

// This code is contributed by jana_sayantan

JavaScript

<script>

// JavaScript implementation of the above approach

// Function to check if N can be
// reduced to <= 0 or not
function NegEqu(N, X, Y)
{
      
    // Update N to N/2 + 10 at most X times
    while (X > 0 && (N > N / 2 + 10))
    {
        N = N / 2 + 10;
        X -= 1;
    }
  
    // Update N to N - 10 Y times
    while (Y > 0)
    {
        N = N - 10;
        Y -= 1;
    }
  
    if (N <= 0)
        return true;
  
    return false;
}

// Driver code
        
    let N = 100;
    let X = 3;
    let Y = 4;
  
    if (NegEqu(N, X, Y)) 
    {
        document.write("Yes");
    }
    else
    {
        document.write("No");
    } 
  
  // This code is contributed by code_hunt.
</script>

Output:

Yes

 Time Complexity: O(X + Y)
Auxiliary Space:O(1)

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