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:
- Update the value of N to ?N/2? + 10 if N > N/2 + 10 and X > 0.
- Update the value of N to N – 10 at most Y times.
- Check if the updated value of N is less than equal to 0 or not.
- If N ? 0, then print “Yes”. Otherwise, print “No”.
Below is the implementation of the above approach:
// 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 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 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# 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 |
<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> |
Yes
Time Complexity: O(X + Y)
Auxiliary Space:O(1)