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:
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, 8print(odd_first(a, b)) |
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 <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); } |
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 |
10
Time complexity: O(1)
Auxiliary Space: O(1)
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Improved By : mohit kumar 29
