Formatted question description: https://leetcode.ca/all/991.html

991. Broken Calculator

Level

Medium

Description

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. 1 <= X <= 10^9
  2. 1 <= Y <= 10^9

Solution

It is difficult to figure out how X becomes Y directly, but the problem can be solved in the opposite direction, which is try to make Y become X after some operations. To make Y become X, two operations are allowed. The first operation is half, which is divide Y by 2, which is only allowed when Y is even, and the second operation is increment, which is add 1 to Y, which is allowed for all values of Y.

When X > Y or Y is odd, obviously only increment can be used on Y. When X < Y and Y is even, both half and increment can be used on Y. Obviously using half costs fewer operations than using increment. Even if for the case where X * 2 > Y it is better to use half first, since using half first and increments next costs 1 + X - Y / 2 operations, and using increments and half first (first increase Y to X * 2 then divide Y by 2) costs X * 2 - Y + 1, and obviously 1 + X - Y / 2 < X * 2 - Y + 1.

class Solution {
    public int brokenCalc(int X, int Y) {
        int count = 0;
        while (X < Y) {
            if (Y % 2 == 0)
                Y /= 2;
            else
                Y++;
            count++;
        }
        return count + X - Y;
    }
}

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