26. Minimum Operations

The problem can be found at the following link: Question Link

My Approach

In this problem, we just need to count the '0' and '1' bits in the given number 'n'. We add one for each '1' bit and perform multiple operations for all the bits.

To find the minimum operations for a given number n, I perform the following steps:

1. Initialize a variable out to -1.

2. While n is not zero:

   • If n is odd (i.e., n % 2 is 1), increment out.

   • Divide n by 2.

   • Always increment out.

3. Return the value of out.

This approach efficiently counts the minimum operations required to reach 1 from n by either subtracting 1 or dividing by 2.

Time and Auxiliary Space Complexity

• Time Complexity: O(log n) - The while loop iterates until n becomes 0, which typically takes log base 2 of n steps.

• Auxiliary Space Complexity: O(1) - The algorithm uses a constant amount of extra space.

Code (C++)

class Solution {
public:
    int minOperation(int n) {
        int out = -1;
        while (n) {
            if (n % 2)
                ++out;
            n /= 2;
            ++out;
        }
        return out;
    }
};

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