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Check if N can be obtained from 0 using at most X increment and at most Y doubling operations

  • Last Updated : 21 Feb, 2022

Given three integers N, X and Y, the task is to check if N can be obtained from 0 using the following operations: 

  • The current integer can be incremented by one (i.e., x = x + 1) and this can be done at most X times.
  • Double the current integer (i.e., x = 2 * x) and this can be done at most Y times.

Return false if the number cannot be reached.

Examples:

Input: N = 24, X = 6 ,Y = 2
Output: true
Explanation: Initially, a = 0
Increment once so a = 1
Increment once so a = 2
Increment once so a = 3
Increment once so a = 4
Increment once so a = 5
Increment once so a = 6
Double once so a = 12
Double again so a = 24

Input: N = 4, X = 2 ,Y = 0
Output: false

 

Approach: The question can also be considered in the exact opposite scenario, where N is given and we need to reduce it to 0 by using two operations :

  • Dividing target by 2 as many as maxDoubles times
  • Decrementing target by 1 as many as maxadd times.

Use recursive function in this way to check if 0 can be obtained or not. In each recursive scenario decrement 1 and call the function again. And if the number is even divide it by 2 and call the recursive function. If 0 can be obtained for within the given limit of moves return true. Otherwise, return false.

Below is the implementation of the above approach: 

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if N is reached
bool checktogetTarget(int N, int X, int Y)
{
    // If N is already zero return true
    if (N == 0)
        return true;
     
    // If can't double and increment,
    // just return false
    if (Y == 0 && X == 0)
        return false;
    int temp = N;
    while (1) {
         
        // If N is not divisible by 2,
        // then just decrement it by 1
        if (temp % 2 == 0 && Y != 0) {
            temp = temp / 2;
            Y--;
        }
        else if (X != 0) {
            temp--;
            X--;
        }
        if (temp == 0)
            break;
        if (Y == 0 && X == 0)
            break;
    }
   
    // if temp becomes 0 after
    // performing operation
    if (temp == 0) {
        return true;
    }
    else {
        return false;
    }
}
 
// Driver Code
int main()
{
    int N = 24;
    int X = 6;
    int Y = 2;
    bool ans = checktogetTarget(N, X, Y);
    if (ans)
        cout << "true";
    else
        cout << "false";
    return 0;
}

Java




// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Function to check if N is reached
  static Boolean checktogetTarget(int N, int X, int Y)
  {
    // If N is already zero return true
    if (N == 0)
      return true;
 
    // If can't double and increment,
    // just return false
    if (Y == 0 && X == 0)
      return false;
    int temp = N;
    while (true) {
 
      // If N is not divisible by 2,
      // then just decrement it by 1
      if (temp % 2 == 0 && Y != 0) {
        temp = temp / 2;
        Y--;
      }
      else if (X != 0) {
        temp--;
        X--;
      }
      if (temp == 0)
        break;
      if (Y == 0 && X == 0)
        break;
    }
 
    // if temp becomes 0 after
    // performing operation
    if (temp == 0) {
      return true;
    }
    else {
      return false;
    }
  }
 
  // Driver Code
  public static void main (String[] args) {
    int N = 24;
    int X = 6;
    int Y = 2;
    Boolean ans = checktogetTarget(N, X, Y);
    if (ans)
      System.out.print("true");
    else
      System.out.print("false");
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3




# Python code for the above approach
 
# Function to check if N is reached
def checktogetTarget(N, X, Y):
   
    # If N is already zero return true
    if (N == 0):
        return True;
 
    # If can't double and increment,
    # just return false
    if (Y == 0 and X == 0):
        return False;
    temp = N;
    while (1):
 
        # If N is not divisible by 2,
        # then just decrement it by 1
        if (temp % 2 == 0 and Y != 0):
            temp = temp / 2;
            Y -= 1
        elif (X != 0):
            temp -= 1
            X -= 1
        if (temp == 0):
            break;
        if (Y == 0 and X == 0):
            break;
 
    # if temp becomes 0 after
    # performing operation
    if (temp == 0):
        return True
    else:
        return False
 
# Driver Code
 
N = 24;
X = 6;
Y = 2;
ans = checktogetTarget(N, X, Y);
if (ans):
    print("true");
else:
    print("false");
 
# This code is contributed by Saurabh Jaiswal

C#




// C# program for above approach
using System;
class GFG
{
 
  // Function to check if N is reached
  static bool checktogetTarget(int N, int X, int Y)
  {
 
    // If N is already zero return true
    if (N == 0)
      return true;
 
    // If can't double and increment,
    // just return false
    if (Y == 0 && X == 0)
      return false;
    int temp = N;
    while (true) {
 
      // If N is not divisible by 2,
      // then just decrement it by 1
      if (temp % 2 == 0 && Y != 0) {
        temp = temp / 2;
        Y--;
      }
      else if (X != 0) {
        temp--;
        X--;
      }
      if (temp == 0)
        break;
      if (Y == 0 && X == 0)
        break;
    }
 
    // if temp becomes 0 after
    // performing operation
    if (temp == 0) {
      return true;
    }
    else {
      return false;
    }
  }
 
  // Driver Code
  public static void Main()
  {
    int N = 24;
    int X = 6;
    int Y = 2;
    bool ans = checktogetTarget(N, X, Y);
    if (ans)
      Console.Write("true");
    else
      Console.Write("false");
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript




<script>
       // JavaScript code for the above approach
 
       // Function to check if N is reached
       function checktogetTarget(N, X, Y) {
           // If N is already zero return true
           if (N == 0)
               return true;
 
           // If can't double and increment,
           // just return false
           if (Y == 0 && X == 0)
               return false;
           let temp = N;
           while (1) {
 
               // If N is not divisible by 2,
               // then just decrement it by 1
               if (temp % 2 == 0 && Y != 0) {
                   temp = temp / 2;
                   Y--;
               }
               else if (X != 0) {
                   temp--;
                   X--;
               }
               if (temp == 0)
                   break;
               if (Y == 0 && X == 0)
                   break;
           }
 
           // if temp becomes 0 after
           // performing operation
           if (temp == 0) {
               return true;
           }
           else {
               return false;
           }
       }
 
       // Driver Code
 
       let N = 24;
       let X = 6;
       let Y = 2;
       let ans = checktogetTarget(N, X, Y);
       if (ans)
           document.write("true");
       else
           document.write("false");
 
        // This code is contributed by Potta Lokesh
   </script>

 
 

Output
true

 

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

 


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