Broken Calculator

LeetCode Q 991 - 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

Solution

Intuition
Instead of multiplying by 2 or subtracting 1 from X, we could divide by 2 (when Y is even) or add 1 to Y.
The motivation for this is that it turns out we always greedily divide by 2:

• If say Y is even, then if we perform 2 additions and one division, we could instead perform one division and one addition for less operations [(Y+2) / 2 vs Y/2 + 1].
• If say Y is odd, then if we perform 3 additions and one division, we could instead perform 1 addition, 1 division, and 1 addition for less operations [(Y+3) / 2 vs (Y+1) / 2 + 1].

Algorithm
While Y is larger than X, add 1 if it is odd, else divide by 2. After, we need to do X - Y additions to reach X.

Code:

public int brokenCalc(int X, int Y) {

	int res = 0;

	while (Y > X) {
		Y = Y % 2 == 0 ? Y / 2 : Y + 1;
		res++;
	}

	return res + X - Y;
}