Uber Interview Question for Software Developers


Country: India
Interview Type: Written Test




Comment hidden because of low score. Click to expand.
2
of 2 vote

// ZoomBA 
/* 
A replacement of '*' -> '(' implies +1 
A replacement of '*' -> ')' implies -1 
At any point, if the count is < 0, drop the path.
Finally, if the count is == 0, accept the path.
As we have three options, the choices can be made in ternary,
Thus, given k choice locations, there are 3 ** k options
Hence, we can reduce the recursion to iteration.  
Using a mixture of imp/dec because of readability  
*/

def count_options( s ){
  num_stars = sum( s.value ) -> { $.item == _'*' ? 1 : 0  }
  options = set() ; len = #|s|
  map = { _'0' : '',  _'1' : '(' , _'2' : ')' , _'(' : 1, _')' : -1 }
  for ( select_pattern : [0: 3 ** num_stars ] ){
    star_count = -1 
    ter_pattern = str( select_pattern , 3 ) 
    ter_pattern = '0' ** ( #|ter_pattern | - num_stars ) + ter_pattern 
    transformed = fold( s.value ,'' ) -> { 
      $.prev + ( $.item == _'*' ? map[ter_pattern[ star_count += 1 ]] :  $.item )
    }
    continue( transformed @ options ) // avoid unnecessary stuff
    res = sum ( transformed.value ) -> { 
      break( $.prev < 0 ){ -len } ; map[$.item] 
    }
    if ( res == 0 ){ options += transformed }
  }
  println( str( options,'\n' ) )
  size( options ) // return size of options 
}
println( count_options( '(*(*)*)' ) )

- NoOne October 29, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

The given sample cases has issues. The possible combinations are wrong.
for first test case:
((()))
(()())
(())
(())()
()(())
()()()
For second test case:
(((())))
((()()))
((()))
((()))()
(()())
(())
(())()
()((()))
()(())
()(())()
()()
()()()
()()()()

- akshaydhule9 December 06, 2016 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Following is basic recursive working code :

#include <iostream>
#include <set>
#include <stack>

using namespace std;

bool balancedPara(string s)
{
	stack<char> para;
	if(s.size()==0)
		return true;
	para.push(s[0]);
	int i=1;
	while(i<s.size())
	{
		if(s[i] == '(')
			para.push(s[i]);
		else if(s[i] == ')')
		{
			 if(!para.empty())
				 para.pop();
			 else
				 return false;
		}
		i++;
	}

	if(para.empty())
		return true;
	else
		return false;
}

void buildstring(string s, string input ,int pos, int count,set<string> &combs)
{
	if(pos<0)
	{
		if(balancedPara(s) == true)
		{
			combs.insert(s);
		}
		return;
	}
	else
	{

		int end = pos;
		while(input[pos] != '*' && pos>=0)
			pos--;
		int len = end - pos;
		if(count == 0)
		{
			string newS = input.substr(pos+1, len) + s;
			buildstring(newS, input,pos-1, count, combs);
		}
		else
		{
			string newS = input.substr(pos+1, len)+ '(' + s;
			buildstring(newS, input,pos-1, count-1, combs);
			string newS_ = input.substr(pos+1, len)+ ')' + s;
			buildstring(newS_, input,pos-1, count-1, combs);
			string newS__ = input.substr(pos+1, len) + s;
			buildstring(newS__, input,pos-1, count, combs);
		}
	}

}

void stringcombs(string input, set<string> &combs)
{
	int count =0;
	for(int i=0;i<input.size();i++)
	{
		if(input[i] == '*')
			count++;
	}
	string s = "";
	buildstring(s,input,input.size()-1, count,combs);
}

void combinations(string input)
{
	set<string> comb;
	stringcombs(input, comb);
	set<string> ::iterator it;
	for(it = comb.begin();it!= comb.end();it++)
	{
		cout<<*it<<endl;
	}
}

int main() {
	string input = "*(*(**)*";
	combinations(input);
	return 0;
}

- akshaydhule9 December 06, 2016 | Flag Reply


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