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++
// 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 notbool 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 Codeint 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 approachclass GFG{// Function to check if N can be// reduced to <= 0 or notstatic 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 Codepublic 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 notdef 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 Codeif __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 approachusing System;class GFG{// Function to check if N can be// reduced to <= 0 or notpublic 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 notfunction 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)
