Google Interview Question for Software Engineer / Developers


Country: United States
Interview Type: Written Test




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0
of 0 vote

This is one of the latest Google Foobar questions. It's unlikely to come up in an interview setting. The answer involves a fair amount of graph theory and a fast implementation.

- talumbau January 03, 2017 | Flag Reply
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Here is a brute-ish solution. Traverse the graph, and maintain a "bunny state", consisting of the bunnies rescued, and the time remaining. At each node, store the state in a hash table. Or, if the same state was previously encountered with an equal or better time remaining, terminate that branch of the search.

With a maximum of 5 bunnies, I don't know why "a fair amount of graph theory and a fast implementation" is necessary.

const int cMaxBunnies = 5;

union bunnyState
{
   struct  
   {
      int8_t bRescued[cMaxBunnies];
      int8_t location;
   };
   int64_t hashVal;
};

typedef std::vector<int> rowType;

void Rescue_Recurse(const std::vector<rowType>& mtx, rowType& result, const bunnyState& state, std::unordered_map<int64_t, int>& bestStates, 
int loc, int tm)
{
   bunnyState newState = state;
   newState.location = loc;
   if (loc == mtx.size() - 1)
   {
      int rescued = std::count(state.bRescued, state.bRescued + cMaxBunnies, 1);
      if ((tm >= 0) && (rescued > (int)result.size()))
      {
         result.clear();
         for (int i = 0; i < cMaxBunnies; i++)
            if (state.bRescued[i])
               result.push_back(i);
      }
   }
   else if (loc > 0)
      newState.bRescued[loc-1] = 1;  
   auto curVal = bestStates.find(newState.hashVal);
   if ((curVal == bestStates.end()) || curVal->second < tm)
   {
      bestStates[newState.hashVal] = tm;
      for (int i = 0; i < (int)mtx[loc].size(); i++)
         Rescue_Recurse(mtx, result, newState, bestStates, i, tm - mtx[loc][i]);
   }
}

rowType Do_Rescue(const std::vector<rowType>& mtx, int tm)
{
   std::vector<int> result;
   std::unordered_map<int64_t, int> bestStates;
   Rescue_Recurse(mtx, result, {0}, bestStates, 0, tm);
   printf("{");
   for (int i: result)
      printf("%d,", i);
   printf("}\n");
   return result;
}

int main()
{
   Do_Rescue({ {0, 1, 1, 1, 1}, {1, 0, 1, 1, 1}, {1, 1, 0, 1, 1}, {1, 1, 1, 0, 1}, {1, 1, 1, 1, 0} }, 3); //output: {0,1,}
   Do_Rescue({ {0, 2, 2, 2, -1}, {9, 0, 2, 2, -1}, {9, 3, 0, 2, -1}, {9, 3, 2, 0, -1}, {9, 3, 2, 2, 0} }, 1); //output: {1,2}
}

- tjcbs2 January 03, 2017 | Flag Reply
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of 0 vote

I got this question in the foobar challenge, I solved it using Edmonds' blossom-contraction algorithm for maximum carnality matching in general graphs , it was not easy ...

- geek January 14, 2017 | Flag Reply
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0
of 0 vote

hey can you elaborate it more like how the blossom-contraction algorithm can be used for this problem.

- someone February 21, 2017 | Flag Reply
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0
of 0 vote

What is the importance of time_limit in this question

- kob August 22, 2017 | Flag Reply


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