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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 ``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

 ``
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 ``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

 ``
Output:
`10`

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

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