# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

 `# 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)) `

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++

 `// C++ Program to implement the ` `// above approach ` `#include ` `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); ` `} `

## 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 `

Output:

```10
```

Time complexity: O(1)
Auxiliary Space: O(1) My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : mohit kumar 29