RUST GYM Leetcode 1611 Minimum One Bit Operations to Make Integers Zero Rust Solution

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1611. Minimum One Bit Operations to Make Integers Zero

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

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

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

Example 1:

Input: n = 0
Output: 0

Example 2:

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 3:

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.

Example 4:

Input: n = 9
Output: 14

Example 5:

Input: n = 333
Output: 393

Constraints:

0 <= n <= 10⁹

Rust Solution

struct Solution;

impl Solution {
    fn minimum_one_bit_operations(n: i32) -> i32 {
        if n <= 1 {
            n
        } else {
            let mut bit = 0;
            while 1 << bit <= n {
                bit += 1;
            }
            let m = n ^ (1 << (bit - 1));
            let k = (1 << bit) - 1;
            k - Solution::minimum_one_bit_operations(m)
        }
    }
}

#[test]
fn test() {
    let n = 0;
    let res = 0;
    assert_eq!(Solution::minimum_one_bit_operations(n), res);
    let n = 3;
    let res = 2;
    assert_eq!(Solution::minimum_one_bit_operations(n), res);
    let n = 6;
    let res = 4;
    assert_eq!(Solution::minimum_one_bit_operations(n), res);
}

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