## Google Interview Question for Software Engineers

• 3

Country: United States

Comment hidden because of low score. Click to expand.
1
of 1 vote

Simply, since each query has a count, we build array of ranges and each range represent a query, then we use rand()%countSum , where countSum is the sum of all counts for all queries, after that we do binary search to see this fits in which range and we return the query that represent this range !

here is the code, O(N)

``````class Range : public pair<int,int>{
public:
int queryIndex;

Range(int f, int s, int qindex=0) : pair<int, int>(f, s), queryIndex(qindex){}

};

int binarySearch(vector< Range >& rangesList,int value, int L, int R){

if (L <= R)
{

int m = (L + R) / 2;

if (rangesList[m].first <= value && value <= rangesList[m].second)
return m;
else
if (value > rangesList[m].second){
return binarySearch(rangesList, value, m + 1, R);
}
else{
return binarySearch(rangesList, value, L, m - 1);
}
}

return -1;
}

string getRandomQuery(const vector< pair<string, int> >& queries){

vector< Range > rangesList;

int rangeCount = 0;

int countSum = 0;

for (int i = 0; i < queries.size(); i++){
Range range(rangeCount, rangeCount + queries[i].second - 1, i);
rangesList.push_back(range);

countSum += queries[i].second;

rangeCount += queries[i].second;
}

int randomValue = rand() % countSum;

int index = binarySearch(rangesList, randomValue, 0, rangesList.size() - 1);

return queries[index].first;
}``````

Comment hidden because of low score. Click to expand.
0

Overall, it's good. But you don't have to keep ranges - it's enough to have just a regular int array Bounds, so that Bounds[0] = Counts[0] and Bounds[i + 1] = Bounds[i] + Counts[i + 1]. Index in that array found with BinSearch is an index of the result query.

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````input -
query - count
q1 	- 	3
q2 	-	5
q3	-	2
sum = 10; ( sum of all counts)

for(int i=0;i<k;i++) {
int r = rand()%sum;
if( !( r/count1) )
select = q1; count1--;
else if ( !(r/count1+count2) )
select= q2; count2--;
else
select= q3; count3--;
sum--;
print select;
}``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

If I understand the answer properly, we will be getting a frequency of each query and we will use that as weighting for which query to return. In Python, I have made the assumption that the data is formatted as a list of tuples where the tuple contains the query and the count. E.g. [('dog', 5), ('cat', 7), ('cow', 2)]

``````from random import randint

def rand_query(queries):
data = []
x = 0
for query in queries:
data.append((query, x, x+query[1]))
x += (query[1] + 1)

max = sum(x[1] for x in queries)
rand = randint(0, max)

for entry in data:
if rand >= entry[1] and rand <= entry[2]:
return entry[0][0]

if '__main__' == __name__:
print(rand_query([('dog', 5), ('cat', 7), ('cow', 2)]))``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

``````function QueryGenerator(queries) {

//calculate total of all scores
var sum = queries
.map(function(x){return x[1];})
.reduce(function(a,x) { return a+x;});

var score = {};
var scoreSum = 0;

return function getQuery() {

while (true) {

if (scoreSum === sum) {
//invalidate score
score = {};
scoreSum = 0;
}

var rand = Math.round(Math.random() * (queries.length - 1));

var query = queries[rand][0];
var freq = queries[rand][1];

if (score[query] === undefined) {
score[query] = 1;

scoreSum++;
return query;
}
else if (score[query] < freq) {
score[query]++;

scoreSum++;
return query;
}
}
}
}``````

Comment hidden because of low score. Click to expand.
0

Here's test

``````var queries = [
['One', 1],
['Two', 2],
['Three', 3]
];

var generator = QueryGenerator(queries);
//verify
var stats = {};
for(var i =0; i < 1000; i++) {
var q = generator();
if(stats[q] === undefined) stats[q] = 0;
stats[q]++;
}
console.log(stats);``````

Comment hidden because of low score. Click to expand.
0
of 0 vote

similar to shuffling a deck of cards. However, in this case you have counts associated with each 'card/query'.

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