100 Days of Code #Day3

As I am running a bit late today, I planned to solve the today’s leet code challenge itself.

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}.

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

At the base level this problem looks complex but we can understand it by breaking down the cases

1. X can perform only two operations multiply by 2 or decrement by 1. So if Y is less than X only Decrement operation can be performed
2. Another edge case is that if x and y are equal we should return 0.
3. And if X<Y continue dividing the Y by 2 and increment the number of steps
4. But we might reach a situation where X*2 > Y since we are consciously decreasing Y by division.
5. Here we have two options, One is to multiply x and 2 and find the difference of Y and X
6. Or Decrement X until we reach the value which multiplied by 2 gives Y
7. Return the minimal of 2 .
8. Return the steps

Live Execution:

#100DaysofCode #Day3 #Challenge