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

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  • Last Updated : 20 Apr, 2021

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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