Monday, August 8, 2016

Google – Get Some Beer


Google – Get Some Beer
有三台接啤酒的机器,分别是small, medium, large。三种size的机器按一次分别能distribute一定范围的啤酒,比如
small的能接[100, 150]ml
medium的能接[200, 300]ml
large的能接[300, 350]ml
问题是如果一个顾客拿一个范围在[minVolume, maxVolume]的杯子去接啤酒,意思就是必须接minVolume的啤酒,但不能超过maxVolume。
[Mistake]
这题其实和dp没有什么关系。本来看着挺像coin change的问题,想着用三维dp来解决,lookup[i][x][y]就表示用前i个machine是否能接到范围在[x, y]的啤酒。
但根据上面的例子,当杯子范围在[300, 400]的时候是无法用small和medium来接的,但dp的结果依然会是true。因为如果按一次medium,剩余量就是
300 – 200 = 100,
400 – 300 = 100,
而[100, 100]的范围按一次small是有可能接到的,所以会返回true。
[Solution]
其实这道题有点类似于Subset的问题,可以这么考虑,
先想清楚哪些范围是一定能接的,那么列出所有能接的范围,如果杯子不在这些范围里面,那就是接不了。
那么问题来了,哪些一定能接?
无非就是:
1 small, 2 small, … n small
1 small + 1 medium, 1 small + 2 medium … 1 small + n medium
2 small + 1 medium, 2 small + 2 medium … 2 small + n medium

n small + 1 medium, n small + 2 medium … n small + n medium


上面就是一个subset问题,用backtracking做就好。

class BeerMachine {
  int min;
  int max;
  BeerMachine(int min, int max) {
    this.min = min;
    this.max = max;
  }
}
class Solution {
  public boolean canGetBeer(BeerMachine[] machines, int minVolume, int maxVolume) {
    if (maxVolume < machines[0].min) {
      return false;
    }
    return backtracking(machines, 0, minVolume, maxVolume, 0, 0);
  }
  private boolean backtracking(BeerMachine[] machines, int pos, int minVolume, int maxVolume, int tmpMin, int tmpMax) {
    if (tmpMin == minVolume && maxVolume == tmpMax) {//??
      return true;
    }
    if (tmpMin > minVolume) {
      return false;
    }
    for (int i = pos; i < machines.length; i++) {
      tmpMin += machines[i].min;
      tmpMax += machines[i].max;
      if (backtracking(machines, i, minVolume, maxVolume, tmpMin, tmpMax)) {
        return true;
      }
      tmpMin -= machines[i].min;
      tmpMax -= machines[i].max;
      i++;
    }
    return false;
  }
}
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