|
| 1 | +#!/usr/bin/python3 |
| 2 | +""" |
| 3 | +On a broken calculator that has a number showing on its display, we can perform |
| 4 | +two operations: |
| 5 | +
|
| 6 | +Double: Multiply the number on the display by 2, or; |
| 7 | +Decrement: Subtract 1 from the number on the display. |
| 8 | +Initially, the calculator is displaying the number X. |
| 9 | +
|
| 10 | +Return the minimum number of operations needed to display the number Y. |
| 11 | +
|
| 12 | +
|
| 13 | +
|
| 14 | +Example 1: |
| 15 | +
|
| 16 | +Input: X = 2, Y = 3 |
| 17 | +Output: 2 |
| 18 | +Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}. |
| 19 | +Example 2: |
| 20 | +
|
| 21 | +Input: X = 5, Y = 8 |
| 22 | +Output: 2 |
| 23 | +Explanation: Use decrement and then double {5 -> 4 -> 8}. |
| 24 | +Example 3: |
| 25 | +
|
| 26 | +Input: X = 3, Y = 10 |
| 27 | +Output: 3 |
| 28 | +Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}. |
| 29 | +Example 4: |
| 30 | +
|
| 31 | +Input: X = 1024, Y = 1 |
| 32 | +Output: 1023 |
| 33 | +Explanation: Use decrement operations 1023 times. |
| 34 | +
|
| 35 | +
|
| 36 | +Note: |
| 37 | +
|
| 38 | +1 <= X <= 10^9 |
| 39 | +1 <= Y <= 10^9 |
| 40 | +""" |
| 41 | + |
| 42 | + |
| 43 | +classSolution: |
| 44 | +defbrokenCalc(self,X:int,Y:int)->int: |
| 45 | +""" |
| 46 | + greedy + work backward |
| 47 | +
|
| 48 | + If Y is odd, we can do only Y = Y + 1 |
| 49 | + If Y is even, if we plus 1 to Y, then Y is odd, we need to plus another 1. |
| 50 | + And because (Y + 1 + 1) / 2 = (Y / 2) + 1, 3 operations are more than 2. |
| 51 | + We always choose Y / 2 if Y is even. |
| 52 | + """ |
| 53 | +t=0 |
| 54 | +whileY>X: |
| 55 | +ifY%2==0: |
| 56 | +Y//=2 |
| 57 | +else: |
| 58 | +Y+=1 |
| 59 | +t+=1 |
| 60 | + |
| 61 | +returnt+X-Y |
| 62 | + |
| 63 | +defbrokenCalc_TLE(self,X:int,Y:int)->int: |
| 64 | +""" |
| 65 | + BFS |
| 66 | + """ |
| 67 | +q= [X] |
| 68 | +t=0 |
| 69 | +has_larger=False |
| 70 | +whileq: |
| 71 | +cur_q= [] |
| 72 | +foreinq: |
| 73 | +ife==Y: |
| 74 | +returnt |
| 75 | + |
| 76 | +cur=e*2 |
| 77 | +ifcur>=1: |
| 78 | +ifcur>Yandnothas_larger: |
| 79 | +has_larger=True |
| 80 | +cur_q.append(cur) |
| 81 | +elifcur<=Y: |
| 82 | +cur_q.append(cur) |
| 83 | + |
| 84 | +cur=e-1 |
| 85 | +ifcur>=1: |
| 86 | +cur_q.append(cur) |
| 87 | +q=cur_q |
| 88 | +t+=1 |
| 89 | + |
| 90 | +raise |
| 91 | + |
| 92 | + |
| 93 | +if__name__=="__main__": |
| 94 | +assertSolution().brokenCalc(2,3)==2 |