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
ไปฃ็
Approach #1 Work Backwards
Time O(logY) && Space O(1)
classSolution {publicintbrokenCalc(int X,int Y) {int ans =0;while (Y > X) { ans++;if (Y %2==1) { Y++; } else { Y /=2; } }return ans + X - Y; }}