Amazon Interview Question for SDE-2s


Country: India




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

just do the symbolic multiplication by following the recursion

rec(res, i) = [
		rec(res + {a}, i-1) for all a in arr2d[i] 	if i >= 0
		res 								if i < 0
	    ]
result = {rec({}, len(arr2d))}

the code

#include <vector>
#include <iostream>

using namespace std;

void symbolicMulRec(const vector<vector<char>>& input, vector<char>& current, size_t& resCtr)
{
	if (current.size() == input.size()) { // base case
		if (current.size() == 0) return; // empty reesult, input.size() == 0
		cout << (resCtr++ == 0 ? "{" : ",{") << current[0];
		for (size_t i = 1; i < current.size(); ++i) {
			cout << "," << current[i];
		}
		cout << "}";
	} else {
		current.push_back(' ');
		for (char c : input[current.size() - 1]) {
			current.back() = c;
			symbolicMulRec(input, current, resCtr);
		}
		current.pop_back();
	}
}

void symbolicMul(const vector<vector<char>>& input) 
{
	cout << "{";
	vector<char> current;
	size_t resCtr = 0;
	symbolicMulRec(input, current, resCtr);
	cout << "}" << endl;
}

int main()
{
	symbolicMul({});
	symbolicMul({ {}, {} });
	symbolicMul({ { 'a', 'b'}, {'c', 'd'} });
	symbolicMul({ { 'a', 'b' },{ 'c', 'd' }, {'e', 'f'} });
	/* output
	{}
	{}
	{{a,c},{a,d},{b,c},{b,d}}
	{{a,c,e},{a,c,f},{a,d,e},{a,d,f},{b,c,e},{b,c,f},{b,d,e},{b,d,f}}
	*/
}

- ChrisK September 13, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

int[][] multiplied;
	int increment = 0;
public int[][] MultiplyRecursion(int[][] toMultiply) {
		multiplied = new int[(((toMultiply.length * toMultiply.length) - toMultiply.length) / 2)*4][2];
		
		for (int i = 0; i < toMultiply.length; i++) {
				RecuAdd(i,i+1, 0,toMultiply);
		}
		return multiplied;
	}
	
	public void RecuAdd(int i, int m, int n, int[][] toMultiply) {
		
		if (m == toMultiply.length) {
			m = i+1;
			n=n+1;
		}
		
		if (n==2 || i == toMultiply.length-1) 
			return;
		
		multiplied[increment] = new int[]{toMultiply[i][n],toMultiply[m][0]};
		increment+=1;
		multiplied[increment] = new int[]{toMultiply[i][n],toMultiply[m][1]};
		increment+=1;
		
		RecuAdd(i,m+1,n,toMultiply);
	}

- loki.cse September 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Similar idea:

function mult(left, rest, index) {
  if (rest.length === index) {
    console.log(left);
    return;
  }
  
  var current = rest[index];
  for(var i = 0; i < current.length; ++i) {
    left.push(current[i]);
    mult(left, rest, index + 1);
    left.pop();
  }
}


var tests = [
  [],
  
  [
    ['a', 'b'],
    ['c', 'd']
  ],
  
  [
    ['a', 'b'],
    ['c', 'd'],
    ['e', 'f']
  ],
  
];

tests.forEach(function(test) {
  mult([], test, 0);
  //0: []
  //1: ["a", "c"] ["a", "d"] ["b", "c"] ["b", "d"]
  //2: ["a", "c", "e"] ["a", "c", "f"]["a", "d", "e"]["a", "d", "f"]["b", "c", "e"]["b", "c", "f"]["b", "d", "e"]["b", "d", "f"]
})

- tnutty2k8 September 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Here is a non recursive solution:

function mult(input) {
  var left = input.length ? input.shift() : [];
  
  while(input.length) {
    var temp = [];    
    for(var i = 0; i < left.length; ++i) {
      for(var j = 0; j < input[0].length; ++j) {
        var item = (left[i] + input[0][j]);
        temp.push(item);
      }
    }
    
    input.shift();
    left = temp;
  }
  return left;
}

var tests = [
  [],
  
  [
    ['a', 'b'],
    ['c', 'd']
  ],
  
  [
    ['a', 'b'],
    ['c', 'd'],
    ['e', 'f']
  ],
  
];

tests.forEach(function(test) {
  console.log( mult(test) );
});

- tnutty2k8 September 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

/* The recursive formulation is too easy 
 So, let's improve on that.
 Observe the problem can be recursively formulated
 as cardinal product of two lists, left one, is a conjugate
 (items themselves may be list), while right ine is a primitive one.
 Thus, simple for loop suffices, which then yields the left list
 for the next iteration.
 */
def _cross_( conjugate, primitive ){
  result = list()
  for ( l : conjugate ){
    for ( r : primitive ){
        // concatenate these lists
        row = list(l)
        row += r 
        result.add(row)   
    }
  }
  result // return them 
}

def cross( arr_of_arr ){
  if ( size(arr_of_arr ) <= 0 ) return [] // fixed 
  cur_index = 0
  // generate left list, conjugate 
  left_list = list ( arr_of_arr[cur_index] ) as { list( $.o ) }
  cur_index += 1
  while ( cur_index < size(arr_of_arr) ){
    left_list = _cross_(left_list,arr_of_arr[cur_index] )
    cur_index += 1
  }
  left_list // return... and we are good 
}

// sample test case
a = [  [ 'a' , 'b' ] , [ 'c' , 'd' ] ]
r = cross( a )
println(r)

- NoOne September 16, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Multiplication using expansion, written in C++98.

#include <vector>
#include <iostream>

using namespace std;

void display(vector<vector<char> > result){
	cout << "{ ";
	for(int i = 0; i < result.size(); i++){
		cout << "{ ";
		for(int j = 0; j < result[i].size(); j++)
			(j != result[i].size() - 1) ? cout << result[i][j] << ", " : cout << result[i][j] << " }";
		(i != result.size() - 1) ? cout << ", " : cout << " ";
	}
	cout << "}\n";
}

void expand(vector<vector<char> >& original, int ext){
	if(ext == 0)
		return;
	int size = original.size();
	for(int j = 0; j < ext; j ++)
		for (int i = 0; i < size; i++)
			original.push_back(original[i]);
}

void multiply(vector<char> set, vector<vector<char> >& result){
	if(result.empty()){
		for(int i = 0; i < set.size(); i++){
			vector<char> re = vector<char>();
			re.push_back(set[i]);
			result.push_back(re);
		}
	}
	else{
		int size = result.size();
		expand(result, set.size() - 1);
		for(int i = 0; i < set.size(); i++)
			for(int j = 0; j < size; j++)
				result[i * size + j].push_back(set[i]);
	}
}

void multiply(vector<vector<char> > charset, vector<vector<char> >& result){
	if(charset.empty())
		return;
	for (int i = 0; i < charset.size(); i++)
		multiply(charset[i], result);
}

int main(){
	vector<vector<char> > charset = vector<vector<char> >();
	vector<char> set1 = vector<char>();
	set1.push_back('a');
	set1.push_back('b');
	vector<char> set2 = vector<char>();
	set2.push_back('c');
	set2.push_back('d');
	vector<char> set3 = vector<char>();
	set3.push_back('e');
	set3.push_back('f');
	charset.push_back(set1);
	charset.push_back(set2);
	charset.push_back(set3);
	vector<vector<char> > result = vector<vector<char> >();
	multiply(charset, result);
	display(result);
	return 0;
}

- Alireza October 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <vector>
#include <iostream>

using namespace std;

void display(vector<vector<char> > result){
	cout << "{ ";
	for(int i = 0; i < result.size(); i++){
		cout << "{ ";
		for(int j = 0; j < result[i].size(); j++)
			(j != result[i].size() - 1) ? cout << result[i][j] << ", " : cout << result[i][j] << " }";
		(i != result.size() - 1) ? cout << ", " : cout << " ";
	}
	cout << "}\n";
}

void expand(vector<vector<char> >& original, int ext){
	if(ext == 0)
		return;
	int size = original.size();
	for(int j = 0; j < ext; j ++)
		for (int i = 0; i < size; i++)
			original.push_back(original[i]);
}

void multiply(vector<char> set, vector<vector<char> >& result){
	if(result.empty()){
		for(int i = 0; i < set.size(); i++){
			vector<char> re = vector<char>();
			re.push_back(set[i]);
			result.push_back(re);
		}
	}
	else{
		int size = result.size();
		expand(result, set.size() - 1);
		for(int i = 0; i < set.size(); i++)
			for(int j = 0; j < size; j++)
				result[i * size + j].push_back(set[i]);
	}
}

void multiply(vector<vector<char> > charset, vector<vector<char> >& result){
	if(charset.empty())
		return;
	for (int i = 0; i < charset.size(); i++)
		multiply(charset[i], result);
}

int main(){
	vector<vector<char> > charset = vector<vector<char> >();
	vector<char> set1 = vector<char>();
	set1.push_back('a');
	set1.push_back('b');
	vector<char> set2 = vector<char>();
	set2.push_back('c');
	set2.push_back('d');
	vector<char> set3 = vector<char>();
	set3.push_back('e');
	set3.push_back('f');
	charset.push_back(set1);
	charset.push_back(set2);
	charset.push_back(set3);
	vector<vector<char> > result = vector<vector<char> >();
	multiply(charset, result);
	display(result);
	return 0;

}

- Alireza October 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <vector>
#include <iostream>

using namespace std;

void display(vector<vector<char> > result){
cout << "{ ";
for(int i = 0; i < result.size(); i++){
cout << "{ ";
for(int j = 0; j < result[i].size(); j++)
(j != result[i].size() - 1) ? cout << result[i][j] << ", " : cout << result[i][j] << " }";
(i != result.size() - 1) ? cout << ", " : cout << " ";
}
cout << "}\n";
}

void expand(vector<vector<char> >& original, int ext){
if(ext == 0)
return;
int size = original.size();
for(int j = 0; j < ext; j ++)
for (int i = 0; i < size; i++)
original.push_back(original[i]);
}

void multiply(vector<char> set, vector<vector<char> >& result){
if(result.empty()){
for(int i = 0; i < set.size(); i++){
vector<char> re = vector<char>();
re.push_back(set[i]);
result.push_back(re);
}
}
else{
int size = result.size();
expand(result, set.size() - 1);
for(int i = 0; i < set.size(); i++)
for(int j = 0; j < size; j++)
result[i * size + j].push_back(set[i]);
}
}

void multiply(vector<vector<char> > charset, vector<vector<char> >& result){
if(charset.empty())
return;
for (int i = 0; i < charset.size(); i++)
multiply(charset[i], result);
}

int main(){
vector<vector<char> > charset = vector<vector<char> >();
vector<char> set1 = vector<char>();
set1.push_back('a');
set1.push_back('b');
vector<char> set2 = vector<char>();
set2.push_back('c');
set2.push_back('d');
vector<char> set3 = vector<char>();
set3.push_back('e');
set3.push_back('f');
charset.push_back(set1);
charset.push_back(set2);
charset.push_back(set3);
vector<vector<char> > result = vector<vector<char> >();
multiply(charset, result);
display(result);
return 0;
}

- Anonymous October 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include <vector>
#include <iostream>

using namespace std;

void display(vector<vector<char> > result){
	cout << "{ ";
	for(int i = 0; i < result.size(); i++){
		cout << "{ ";
		for(int j = 0; j < result[i].size(); j++)
			(j != result[i].size() - 1) ? cout << result[i][j] << ", " : cout << result[i][j] << " }";
		(i != result.size() - 1) ? cout << ", " : cout << " ";
	}
	cout << "}\n";
}

void expand(vector<vector<char> >& original, int ext){
	if(ext == 0)
		return;
	int size = original.size();
	for(int j = 0; j < ext; j ++)
		for (int i = 0; i < size; i++)
			original.push_back(original[i]);
}

void multiply(vector<char> set, vector<vector<char> >& result){
	if(result.empty()){
		for(int i = 0; i < set.size(); i++){
			vector<char> re = vector<char>();
			re.push_back(set[i]);
			result.push_back(re);
		}
	}
	else{
		int size = result.size();
		expand(result, set.size() - 1);
		for(int i = 0; i < set.size(); i++)
			for(int j = 0; j < size; j++)
				result[i * size + j].push_back(set[i]);
	}
}

void multiply(vector<vector<char> > charset, vector<vector<char> >& result){
	if(charset.empty())
		return;
	for (int i = 0; i < charset.size(); i++)
		multiply(charset[i], result);
}

int main(){
	vector<vector<char> > charset = vector<vector<char> >();
	vector<char> set1 = vector<char>();
	set1.push_back('a');
	set1.push_back('b');
	vector<char> set2 = vector<char>();
	set2.push_back('c');
	set2.push_back('d');
	vector<char> set3 = vector<char>();
	set3.push_back('e');
	set3.push_back('f');
	charset.push_back(set1);
	charset.push_back(set2);
	charset.push_back(set3);
	vector<vector<char> > result = vector<vector<char> >();
	multiply(charset, result);
	display(result);
	return 0;

}

- Anonymous October 04, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

I assume that
if multiplying {{a,b}, {c,d}} = {{a,c}, {a,d}, {b,c}, {b,d}}
then, multiplying {{a,b}, {c,d}, {e,f}} = multiplying {{a,c}, {a,d}, {b,c}, {b,d}} by {e,f} = {a,e}, {a,f}, {c,e}, {c,f}, {a,e}, {a,f}, {d,e}, {d,f}, {b,e}, {b,f}, {c,e}, {c,f}, {b,e}, {b,f}, {d,e}, {d,f}

void Multiply(vector<vector<char>> const &a, vector<vector<char>> &out, char c1 = 0, int idx = 0)
{
	if (idx < 0 ||
		idx >= a.size() ||
		(c1 == 0 && idx != 0))
	{
		return;
	}
	for (char c2 : a[idx]) {
		if (c1 != 0 &&
			idx == a.size() - 1)
		{
			out.push_back({c1, c2});
		}
		Multiply(a, out, c1, idx + 1);
		Multiply(a, out, c2, idx + 1);
	}
}

- Alex November 03, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
void main()
{
int i,j,m,n,k,l;
printf("enter the no of subarray do you want\n\n");
scanf("%d",&m);
printf("enter the elements do you want to insert in every subset\n");
scanf("%d",&n);
char first[m][n];

for(i=0;i<m;i++)
{
printf("start entering elements in the sub array %d:",i+1);
for(j=0;j<n;j++)
{
scanf("%s",&first[i][j]);

}
}
printf("\n");
for(i=0;i<m;i++)
{
printf("\nentered elements in the sub array %d:",i+1);
printf("{");
for(j=0;j<n;j++)
{
printf("%c",first[i][j]);
if(j<n-1)
printf(",");
}
printf("}");
}
printf("\n\n");

for(i=0;i<1;i++)
{
printf("{");
for(j=0;j<n;j++)
{
for(k=i+1;k<m;k++)
{
for(l=0;l<n;l++)
{
printf("{%c,%c}",first[i][j],first[k][l]);
}
}
}
printf("}");
}
}

- prashant sharma November 03, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
void main()
{
 int i,j,m,n,k,l;
 printf("enter the no of subarray do you want\n\n");
 scanf("%d",&m);
 printf("enter the elements do you want to insert in every subset\n");
 scanf("%d",&n);
 char first[m][n];

 for(i=0;i<m;i++)
    {
        printf("start entering elements in the sub array %d:",i+1);
        for(j=0;j<n;j++)
        {
        scanf("%s",&first[i][j]);

        }
        }
        printf("\n");
for(i=0;i<m;i++)
    {
        printf("\nentered elements in the sub array %d:",i+1);
        printf("{");
        for(j=0;j<n;j++)
        {
        printf("%c",first[i][j]);
        if(j<n-1)
            printf(",");
        }
        printf("}");
        }
        printf("\n\n");

        for(i=0;i<1;i++)
        {
            printf("{");
            for(j=0;j<n;j++)
            {
              for(k=i+1;k<m;k++)
              {
                  for(l=0;l<n;l++)
                    {
                printf("{%c,%c}",first[i][j],first[k][l]);
              }
              }
            }
            printf("}");
        }
}

- prashant sharma November 03, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

#include<stdio.h>
void main()
{
int i,j,m,n,k,l;
printf("enter the no of subarray do you want\n\n");
scanf("%d",&m);
printf("enter the elements do you want to insert in every subset\n");
scanf("%d",&n);
char first[m][n];

for(i=0;i<m;i++)
{
printf("start entering elements in the sub array %d:",i+1);
for(j=0;j<n;j++)
{
scanf("%s",&first[i][j]);

}
}
printf("\n");
for(i=0;i<m;i++)
{
printf("\nentered elements in the sub array %d:",i+1);
printf("{");
for(j=0;j<n;j++)
{
printf("%c",first[i][j]);
if(j<n-1)
printf(",");
}
printf("}");
}
printf("\n\n");

for(i=0;i<1;i++)
{
printf("{");
for(j=0;j<n;j++)
{
for(k=i+1;k<m;k++)
{
for(l=0;l<n;l++)
{
printf("{%c,%c}",first[i][j],first[k][l]);
}
}
}
printf("}");
}
}

- prashant sharma November 03, 2017 | Flag Reply


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