Problem

Given an integer n, you must transform it into 0 using the following operations any number of times:

  • Change the rightmost (0th) bit in the binary representation of n.
  • Change the ith bit in the binary representation of n if the (i-1)th bit is set to 1 and the (i-2)th through 0th bits are set to 0.

Return the minimum number of operations to transform n into 0.

Examples

Example 1

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Input: n = 3
Output: 2
Explanation: The binary representation of 3 is "11".
"11" -> "01" with the 2nd operation since the 0th bit is 1.
"01" -> "00" with the 1st operation.

Example 2

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Input: n = 6
Output: 4
Explanation: The binary representation of 6 is "110".
"110" -> "010" with the 2nd operation since the 1st bit is 1 and 0th through 0th bits are 0.
"010" -> "011" with the 1st operation.
"011" -> "001" with the 2nd operation since the 0th bit is 1.
"001" -> "000" with the 1st operation.

Constraints

  • 0 <= n <= 10^9

Solution

Method 1 – Recursive Bit Manipulation (Gray Code)

Intuition

The problem is related to the minimum number of operations to convert n to 0 using Gray code properties. The operation count for each bit is recursive: flip the highest set bit, then recursively solve for the rest.

Approach

  1. If n == 0, return 0.
  2. Find the highest set bit position k in n.
  3. The answer is (1 « (k+1)) - 1 - minimumOneBitOperations(n ^ (1 « k)).
  4. Recursively solve for n ^ (1 « k).

Code

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class Solution {
public:
    int minimumOneBitOperations(int n) {
        if (n == 0) return 0;
        int k = 31 - __builtin_clz(n);
        return (1 << (k+1)) - 1 - minimumOneBitOperations(n ^ (1 << k));
    }
};
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func minimumOneBitOperations(n int) int {
    if n == 0 {
        return 0
    }
    k := 0
    for i := 31; i >= 0; i-- {
        if n&(1<<i) > 0 {
            k = i
            break
        }
    }
    return (1<<(k+1)) - 1 - minimumOneBitOperations(n^(1<<k))
}
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class Solution {
    public int minimumOneBitOperations(int n) {
        if (n == 0) return 0;
        int k = 31 - Integer.numberOfLeadingZeros(n);
        return (1 << (k+1)) - 1 - minimumOneBitOperations(n ^ (1 << k));
    }
}
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class Solution {
    fun minimumOneBitOperations(n: Int): Int {
        if (n == 0) return 0
        val k = 31 - Integer.numberOfLeadingZeros(n)
        return (1 shl (k+1)) - 1 - minimumOneBitOperations(n xor (1 shl k))
    }
}
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class Solution:
    def minimumOneBitOperations(self, n: int) -> int:
        if n == 0:
            return 0
        k = n.bit_length() - 1
        return (1 << (k+1)) - 1 - self.minimumOneBitOperations(n ^ (1 << k))
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impl Solution {
    pub fn minimum_one_bit_operations(n: i32) -> i32 {
        if n == 0 {
            return 0;
        }
        let k = 31 - n.leading_zeros();
        (1 << (k+1)) - 1 - Solution::minimum_one_bit_operations(n ^ (1 << k))
    }
}
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class Solution {
    minimumOneBitOperations(n: number): number {
        if (n === 0) return 0;
        let k = 31;
        while (k >= 0 && ((n >> k) & 1) === 0) k--;
        return (1 << (k+1)) - 1 - this.minimumOneBitOperations(n ^ (1 << k));
    }
}

Complexity

  • ⏰ Time complexity: O(log n) — Each recursion reduces the highest bit.
  • 🧺 Space complexity: O(log n) — For recursion stack.