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First number to leave an odd remainder after repetitive division by 2

  • Last Updated : 14 Sep, 2021

Given two integers A and B, the task is to print the integer among the two, which will be converted to an odd integer by a smaller number of divisions by 2. If both the numbers get converted to an odd integer after the same number of operations, print -1. 

Examples:  

Input: A = 10 and B = 8 
Output: 10 
Explanation: 
Step 1: A/2 = 5, B/2 = 4 
Hence, A is first number to be converted to an odd integer.

Input: A = 20 and B = 12 
Output: -1 
Explanation: 
Step 1: A/2 = 10, B/2 = 6 
Step 2: A/2 = 5, B/2 = 3 
Hence, A and B are converted to an odd integer at the same time. 
 

Naive Approach: 
The simplest approach to solve the problem is as follows: 



  • Check if any of the two given numbers are even or odd. If one of them is odd, print that number.
  • If both are odd, print -1.
  • Otherwise, keep dividing both of them by 2 at each step and check if any of them is converted to an odd integer or not. If one of them is converted, print the initial value of that number. If both are converted at the same time print -1.

Below is the implementation of the above approach: 

C++




// C++ program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the first number
// to to be converted to an odd value
int odd_first(int a, int b)
{
     
    // Initial values
    int true_a = a;
    int true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
         
// Driver code
int main()
{
    int a = 10;
    int b = 8;
 
    cout << odd_first(a, b);
}
 
// This code is contributed by code_hunt
Java



// Java program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
import java.util.*;
import java.lang.Math;
import java.io.*;
 
class GFG{
     
// Function to return the first number
// to to be converted to an odd value
static int odd_first(int a, int b)
{
     
    // Initial values
    int true_a = a;
    int true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
     
// Driver code
public static void main(String[] args)
{
    int a = 10;
    int b = 8;
 
    System.out.print(odd_first(a, b));
}
}
 
// This code is contributed by code_hunt
Python3



# Python3 program to find the first
# number to be converted to an odd
# integer by repetitive division by 2
 
# Function to return the first number
# to to be converted to an odd value
def odd_first(a, b):
 
  # Initial values
  true_a = a
  true_b = b
 
  # Perform repetitive divisions by 2
  while(a % 2 != 1 and b % 2 != 1):
    a = a//2
    b = b//2
 
  # If both become odd at same step
  if a % 2 == 1 and b % 2 == 1:
    return -1
 
  # If a is first to become odd
  elif a % 2 == 1:
    return true_a
 
  # If b is first to become odd
  else:
    return true_b
 
# Driver Code
a, b = 10, 8
print(odd_first(a, b))
C#



// C# program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
using System;
 
class GFG{
     
// Function to return the first number
// to to be converted to an odd value
static int odd_first(int a, int b)
{
     
    // Initial values
    int true_a = a;
    int true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
     
// Driver code
public static void Main()
{
    int a = 10;
    int b = 8;
 
    Console.Write(odd_first(a, b));
}
}
 
// This code is contributed by sanjoy_62
Javascript



<script>
 
// Javascript program to find the first
// number to be converted to an odd
// integer by repetitive division by 2
 
// Function to return the first number
// to to be converted to an odd value
function odd_first(a, b)
{
     
    // Initial values
    var true_a = a;
    var true_b = b;
 
    // Perform repetitive divisions by 2
    while(a % 2 != 1 && b % 2 != 1)
    {
        a = a / 2;
        b = b / 2;
    }
 
    // If both become odd at same step
    if (a % 2 == 1 && b % 2 == 1)
        return -1;
 
    // If a is first to become odd
    else if (a % 2 == 1)
        return true_a;
 
    // If b is first to become odd
    else
        return true_b;
}
 
// Driver code
var a = 10, b = 8;
 
document.write(odd_first(a, b));
 
// This code is contributed by Ankita saini
    
</script>
Output: 
10
 
Time complexity: O(log(min(a, b))) 
Auxiliary Space: O(1)
Efficient Approach: 
Follow the steps below to optimize the above approach:  
  • Dividing a number by 2 is equivalent to perform the right shift on that number.
  • Hence, the number with the Least Significant Set Bit among the two will be the first to be converted to an odd integer.
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 return the position
// least significant set bit
int getFirstSetBitPos(int n)
{
    return log2(n & -n) + 1;
}
 
// Function return the first number
// to be converted to an odd integer
int oddFirst(int a, int b)
{
 
    // Stores the positions of the
    // first set bit
    int steps_a = getFirstSetBitPos(a);
    int steps_b = getFirstSetBitPos(b);
 
    // If both are same
    if (steps_a == steps_b) {
        return -1;
    }
 
    // If A has the least significant
    // set bit
    if (steps_a > steps_b) {
        return b;
    }
 
    // Otherwise
    if (steps_a < steps_b) {
        return a;
    }
}
 
// Driver code
int main()
{
    int a = 10;
    int b = 8;
 
    cout << oddFirst(a, b);
} 
Java



// Java program implementation
// of the approach
import java.util.*;
import java.lang.Math;
import java.io.*;
 
class GFG{
     
// Function to return the position
// least significant set bit
static int getFirstSetBitPos(int n)
{
    return (int)(Math.log(n & -n) /
                 Math.log(2));
}
 
// Function return the first number
// to be converted to an odd integer
static int oddFirst(int a, int b)
{
     
    // Stores the positions of the
    // first set bit
    int steps_a = getFirstSetBitPos(a);
    int steps_b = getFirstSetBitPos(b);
 
    // If both are same
    if (steps_a == steps_b)
    {
        return -1;
    }
 
    // If A has the least significant
    // set bit
    else if (steps_a > steps_b)
    {
        return b;
    }
 
    // Otherwise
    else
    {
        return a;
    }
}
     
// Driver code
public static void main(String[] args)
{
    int a = 10;
    int b = 8;
 
    System.out.print(oddFirst(a, b));
}
}
 
// This code is contributed by code_hunt
Python3



# Python3 program to implement the
# above approach
from math import log
 
# Function to return the position
# least significant set bit
def getFirstSetBitPos(n):
    return log(n & -n, 2) + 1
 
# Function return the first number
# to be converted to an odd integer
def oddFirst(a, b):
 
    # Stores the positions of the
    # first set bit
    steps_a = getFirstSetBitPos(a)
    steps_b = getFirstSetBitPos(b)
 
    # If both are same
    if (steps_a == steps_b):
        return -1
 
    # If A has the least significant
    # set bit
    if (steps_a > steps_b):
        return b
 
    # Otherwise
    if (steps_a < steps_b):
        return a
 
# Driver code
if __name__ == '__main__':
     
    a = 10
    b = 8
     
    print(oddFirst(a, b))
 
# This code is contributed by mohit kumar 29
C#



// C# program implementation
// of the approach
using System;
 
class GFG{
     
// Function to return the position
// least significant set bit
static int getFirstSetBitPos(int n)
{
    return (int)(Math.Log(n & -n) /
                 Math.Log(2));
}
 
// Function return the first number
// to be converted to an odd integer
static int oddFirst(int a, int b)
{
     
    // Stores the positions of the
    // first set bit
    int steps_a = getFirstSetBitPos(a);
    int steps_b = getFirstSetBitPos(b);
 
    // If both are same
    if (steps_a == steps_b)
    {
        return -1;
    }
 
    // If A has the least significant
    // set bit
    else if (steps_a > steps_b)
    {
        return b;
    }
 
    // Otherwise
    else
    {
        return a;
    }
}
     
// Driver code
public static void Main()
{
    int a = 10;
    int b = 8;
 
    Console.Write(oddFirst(a, b));
}
}
 
// This code is contributed by sanjoy_62
Javascript



<script>
 
// Javascript program implementation
// of the approach
 
// Function to return the position
// least significant set bit
function getFirstSetBitPos(n)
{
    return (Math.log(n & -n) /
                 Math.log(2));
}
  
// Function return the first number
// to be converted to an odd integer
function oddFirst(a, b)
{
      
    // Stores the positions of the
    // first set bit
    let steps_a = getFirstSetBitPos(a);
    let steps_b = getFirstSetBitPos(b);
  
    // If both are same
    if (steps_a == steps_b)
    {
        return -1;
    }
  
    // If A has the least significant
    // set bit
    else if (steps_a > steps_b)
    {
        return b;
    }
  
    // Otherwise
    else
    {
        return a;
    }
}
   
 
// Driver Code
     
       let a = 10;
    let b = 8;
  
    document.write(oddFirst(a, b));
         
</script>
Output: 
10
 
Time complexity: O(1) 
Auxiliary Space: O(1)
 



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Article Contributed By :
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deepanshu_rustagi
@deepanshu_rustagi
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