## Adobe Interview Question for Software Engineers

• 0

Country: United States

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

The approach is simple:
match the head and for the tail do a swap for desired character to left.

I could not prove it would be lease

``````#include <iostream>
#include <map>
#include <cstring>
using namespace std;

void swap(char * convert , int j)
{
char temp = convert[j];
convert[j] = convert[j-1];
convert[j-1] = temp;
}

int transformCost(char * master, char * convert , int i)
{
int matchcounter = 0;
int cost = 0;
while(matchcounter < i)
{
if(master[matchcounter] == convert[matchcounter])
{
matchcounter++;
continue;
}
char m = master[matchcounter];
bool found = false;
int matchpos = -1;
for(int j = matchcounter ; j < i; j++)
{
if(convert[j] == m)
{
matchpos = j;
found = true;
break;
}
}
if(found)
{
swap(convert,matchpos);
cost++;
}
else
{
return -1;
}
}
cout<<"Master  : " << master<<endl;
cout<<"Convert : " << convert<<endl;
return cost;
}

int main()
{
char a[] = "101010101010";
char b[] = "111111000000";
cout<<transformCost(a,b, strlen(a))<<endl;
return 0;
}``````

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

Could you provide reasoning for it?

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

``````int SwapCount(int A[], int B[], int size)
{
int j = 0;
int minSwapCount = 0;
for(int i =0; i < size; i++)
{
if(A[i] == 1)
{

while(B[j] !=1)
{
j++;
}
if(i > j)
minSwapCount += i - j;
else
minSwapCount += j - i;
j++;
}
}

return minSwapCount;

}
int main()
{
int A[] = {1,0,0,1};
int B[] = {0,1,1,0};
printf("SwapCount = %d", SwapCount(9, 6));
return 0;
}``````

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

``````int minSwaps(int A[], int B[], int n)
{
assert(A != NULL && B != NULL && n > 0);

int ans = 0, rb = 1;

for (int i = 0; i < n; ++i)
{
if (A[i] != B[i])
{
if (rb <= i) rb = i + 1;

while (rb < n && A[rb] != B[i]) ++rb;

if (rb >= n) throw logic_error("there is no solution");

ans += rb - i; A[rb] = A[i]; ++rb;
}
}

return ans;
}``````

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

``````void findMInSwapBinary(int binArr1[n],int binArr2[n])
{
int minSwaps =0;
for(int i=0;i<n;i++)
{
if(binArr1[i] ^ binArr2[i]==1)
minSwaps++;
}
return;
}``````

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

Your logic does not work on this.
eg. arr1[]={1,1,1,0,0,0}
arr2[]={1,0,1,0,1,0}
You output would result 2

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

Only 2 swaps are required.
can u explain why it's 3?

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

sv, I think you are missing the part that says you can only swap adjacent elements.

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

O(1) solution .....

``````int numOfSwaps(int x, int y){
int tmp = x ^ y;
return NumberOfSetBits(tmp);
}

int NumberOfSetBits(int i)
{
i = i - ((i >> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}``````

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

O(1) solution ....

``````int countSwaps(int x, int y){
int tmp = x ^ y;
return numberOfSetBits(tmp);
}

int numberOfSetBits(int i)
{
i = i - ((i >> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}``````

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

Ooops submitted two times, pls ignore first one.

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

sv, The answer would have been 2 if we were not given the condition of adjacent swapping only. Since we can only swap two adjacent elements, minimum swaps will be 3.

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

yes, thats right..i missed that condition...

The solution I have given would work without swapping.
we can xor the numbers and find the bits which are set in the result.
We can just toggle those bits in the respective arrays for converting them.

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

``````void findMInSwapBinary(int binArr1[n],int binArr2[n])
{
int minSwaps =0;
for(int i=0;i<n;i++)
{
if(binArr1[i] ^ binArr2[i]==1)
minSwaps++;
}
return;``````

}

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

above function is wrong. For example in case of 1,0,0,1 & 0,1,1,0, this function will return 4 but the answer should be 2.

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

XOR the numbers, then see how many 1s there are in this value

Reasoning:
We know that a ^ a = 0 , so 1 ^ 1 = 0 and 0 ^ 0 = 0
But, 1 ^ 0 = 1 and 0 ^1 = 1, so we see that we can find differences by xoring

Some code:

``````int numSwaps(int a, int b) {

int c = a ^ b;
int output = 0;

while (c != 0) {

out += c & 1;
c =>> 1;
}

return out;
}``````

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

This one works, but given the answer size there might be an easier solution or this is not an interview question.

``````public static void main(String[] args) throws Exception {
Random rseed = new Random();
int seed = rseed.nextInt(100);
System.out.println("seed = " + seed);
Random r = new Random(seed);
int size = 10;
int[] orig;
int[] dest = new int[size];
for (int i = 0; i < dest.length; i++) {
dest[i] = r.nextInt(3)==1? 1 : 0;
}
orig = dest.clone();
for (int i = 0; i < dest.length-1; i++) {
int tmp = orig[i];
int destIndx = i + r.nextInt(size-i-1);
orig[i] = orig[destIndx];
orig[destIndx] = tmp;
}

//        orig = new int[]{1, 0, 0, 0, 1, 0, 0, 0, 0, 1};
//        dest = new int[]{0, 0, 0, 0, 0, 1, 0, 1, 0, 1};

System.out.println("Orig" + Arrays.toString(orig));
System.out.println("Dest"+Arrays.toString(dest));
System.out.println();
System.out.println("Best "+findMinSwaps(orig, dest));

}

private static int findMinSwaps(int[] orig, int[] dest) throws Exception {
int leftDest=0,
leftOrigin=0,
rightOrigin=orig.length-1,
rightDest=orig.length-1;

int swaps = 0;
while(!Arrays.equals(orig,dest)){
//scan left to right
while(leftDest < dest.length && (dest[leftDest]==0 || orig[leftDest]==1) )
leftDest++;
while(leftOrigin< orig.length && (orig[leftOrigin]==0 || dest[leftOrigin]==1))
leftOrigin++;
if(leftDest != dest.length-1 && leftOrigin != dest.length-1 )
swaps+= multipleSwaps(orig,leftOrigin,leftDest);

//scan right to left
while(rightDest>0 && (dest[rightDest]==0 || orig[rightDest]==1))
rightDest--;
while(rightOrigin>0 && (orig[rightOrigin]==0 || dest[rightOrigin]==1) )
rightOrigin--;
if(rightDest != 0 && rightOrigin != 0)
swaps+= multipleSwaps(orig,rightOrigin,rightDest);

leftDest++;
leftOrigin++;
rightDest--;
rightOrigin--;
}
return swaps;
}

private static int multipleSwaps(int[] orig, int from, int to) throws Exception {
if(from == to)
return 0;
int direCtion = (to-from)/Math.abs(to-from);
int swaps =0;
while(from != to){
if(orig[from+direCtion]==1){
swaps+=multipleSwaps(orig,from+direCtion,to);
swap(orig,from,from+direCtion);
System.out.println("Orig" +Arrays.toString(orig));
return swaps+1;
} else {
swap(orig,from,from+direCtion);
System.out.println("Orig" +Arrays.toString(orig));
from+=direCtion;
swaps++;
}
}
return swaps;
}

static void swap(int[] data, int indx1, int indx2) throws Exception {
if(Math.abs(indx1-indx2)!= 1)
throw new Exception("Wrong indexes " + indx1 + " " + indx2);
int tmp = data[indx1];
data[indx1] = data[indx2];
data[indx2] = tmp;
}``````

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