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

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:

    Python3

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    # 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))

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    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 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++

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    // 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);
    }

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    Python3

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    # 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

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    Output:

    10
    

    Time complexity: O(1)
    Auxiliary Space: O(1)

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