LeetCode 1611: Minimum One Bit Operations to Make Integers Zero

Explanation:

To solve this problem, we can use a backtracking approach. We start with the given integer n and try to reach 0 by applying the allowed operations. The key insight here is that the most significant bit (MSB) of n determines the number of steps required to reach 0. We can recursively apply the operations to zero out the MSB of n.

1. If the MSB is 0, we can simply drop that bit and continue with the rest of the number.
2. If the MSB is 1, we need to flip it to 0. To do this, we recursively zero out the rest of the bits below MSB.

By considering the position of the MSB, we can determine the minimum number of operations needed to reach 0.

Time Complexity: The time complexity of this approach is O(log n) where n is the given integer.

Space Complexity: The space complexity is O(log n) due to the recursive calls.

:

class Solution {
    public int minimumOneBitOperations(int n) {
        return helper(n, true);
    }
    
    private int helper(int n, boolean isMSB) {
        if (n == 0) return 0;
        
        if (isMSB) {
            int msb = 1;
            while ((msb << 1) <= n) {
                msb <<= 1;
            }
            return helper(n ^ msb, false) + 1 + helper(msb >> 1, true);
        } else {
            return helper(n, true);
        }
    }
}