On a broken calculator that has a number showing on its display, we can perform two operations:
Initially, the calculator is displaying the number X.
Return the minimum number of operations needed to display the number Y.
Example 1:
Input: X = 2, Y = 3 Output: 2 Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.
Example 2:
Input: X = 5, Y = 8 Output: 2 Explanation: Use decrement and then double {5 -> 4 -> 8}.
Example 3:
Input: X = 3, Y = 10 Output: 3 Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.
Example 4:
Input: X = 1024, Y = 1 Output: 1023 Explanation: Use decrement operations 1023 times.
Note:
1 <= X <= 10^91 <= Y <= 10^9struct Solution;
impl Solution {
fn broken_calc(x: i32, mut y: i32) -> i32 {
let mut res = 0;
while y > x {
if y % 2 == 0 {
y /= 2;
} else {
y += 1;
}
res += 1;
}
res + x - y
}
}
#[test]
fn test() {
let x = 2;
let y = 3;
let res = 2;
assert_eq!(Solution::broken_calc(x, y), res);
let x = 5;
let y = 8;
let res = 2;
assert_eq!(Solution::broken_calc(x, y), res);
let x = 3;
let y = 10;
let res = 3;
assert_eq!(Solution::broken_calc(x, y), res);
let x = 1024;
let y = 1;
let res = 1023;
assert_eq!(Solution::broken_calc(x, y), res);
}