991. Broken Calculator

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.

Constraints

1. 1 <= X <= 10^9

2. 1 <= Y <= 10^9

Approach

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].

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: X = 2, Y = 3

Output: 2

Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.

Example 2

Example 3

Example 4

Solutions

Solution 1

/**
 * Time complexity : O(logY)
 * Space complexity : O(1)
 */

class Solution {
    public int brokenCalc(int X, int Y) {
        int result = 0;
        while(Y > X) {
            result++;
            if(Y%2 == 1) {
                Y++;
            } else {
                Y /= 2;
            }
        }
        return result + X - Y;
    }
}

Follow up