leetcode 991 - Broken Calculator

https://leetcode.com/problems/broken-calculator/

On a broken calculator that has a number showing on its display, we can perform two operations:

Double: Multiply the number on the display by 2, or;
Decrement: Subtract 1 from the number on the display.
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 <= 10^9
1 <= Y <= 10^9

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class Solution {
public:
int brokenCalc(int X, int Y) {
int ops = 0;

while (X < Y) {
ops++;

if (Y % 2 == 0)
Y = Y / 2;
else
Y++;
}

return ops + (X - Y);
}
};