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Microsoft Interview Question for Software Engineer / Developers


Country: United States
Interview Type: In-Person



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

input array (a) and flag array are the members of class. Below is the code for counting both zero groups and one groups.

private void CountOneGroups()
    {
        flag = new boolean[a.length][a[0].length];
        for(int i = 0; i < a.length; i++)
            for(int j = 0; j < a[i].length; j++)
                flag[i][j] = false;
        int zeroCount, oneCount;
        zeroCount = 0;
        oneCount = 0;
        for(int i = 0; i < a.length; i++)
            for(int j = 0; j < a[i].length; j++)
                if(flag[i][j] != true)
                {
                    if(a[i][j] == 0)
                        zeroCount++;
                    else
                        oneCount++;
                    FourConnected(i, j, a[i][j]);
                }
        System.out.println("Zero Group Count : " + zeroCount);
        System.out.println("One Group Count : " + oneCount);
    }
    private void FourConnected(int i, int j, int value)
    {
        if(!flag[i][j])
        {
            flag[i][j] = true;
            if(i - 1 >= 0 && a[i - 1][j] == value)
                FourConnected(i - 1, j, value);
            if(j - 1 >= 0 && a[i][j - 1] == value)
                FourConnected(i, j - 1, value);
            if(i + 1 < a.length && a[i + 1][j] == value)
                FourConnected(i + 1, j, value);
            if(j + 1 < a[0].length && a[i][j + 1] == value)
                FourConnected(i, j + 1, value);
        }
    }

- Nani November 03, 2011 | Flag Reply
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0
of 0 votes

Thank you Nani. It is simple one. I think instread of creating a new boolean 2D Array, we might play around the existing array with different values like -1 for visited 0s and -2 for 1s and count the 1 group. What you think?.

- JobHunter November 05, 2011 | Flag
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0
of 0 votes

beautiful thinking.

- baghel November 07, 2011 | Flag
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0
of 0 votes

if input is
0 0 0 0
1 1 0 1
0 1 1 1

will the group of 1's is 1 or 3??

- Appu November 13, 2011 | Flag
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0
of 0 votes

The Algo provided by Nani will return 1 in your case, Appu

And I think this is also expected from the question's algo

- Skataria December 01, 2011 | Flag
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0
of 0 votes

cool soln!

- ashwin December 08, 2011 | Flag
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0
of 0 vote

<pre lang="" line="1" title="CodeMonkey35139" class="run-this">class Cluster {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int[][] matrix = { { 1, 0, 1 }, { 1, 0, 1 }, { 0, 1, 1 } };
int cluster = dfs(matrix);
System.out.print("Cluster:" + cluster);
}

private static int dfs(int[][] matrix) {
// TODO Auto-generated method stub
int cluster = 0;
printMatrix(matrix);
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == 1) {
dfsVisit(matrix, i, j);
cluster++;
}
}
}
printMatrix(matrix);
return cluster;
}

private static void dfsVisit(int[][] matrix, int i, int j) {
// TODO Auto-generated method stub
if (i < 0 || i >= matrix.length)
return;
if (j < 0 || j >= matrix[0].length)
return;
if (matrix[i][j] != 1)
return;
matrix[i][j] = -1;
dfsVisit(matrix, i + 1, j);
dfsVisit(matrix, i, j + 1);
dfsVisit(matrix, i - 1, j);
dfsVisit(matrix, i, j - 1);
}

private static void printMatrix(int[][] matrix) {
// TODO Auto-generated method stub
System.out.println();
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
System.out.printf("%4d", matrix[i][j]);
}
System.out.println();
}

}

}
</pre><pre title="CodeMonkey35139" input="yes">
</pre>

- Anonymous November 04, 2011 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

<pre lang="" line="1" title="CodeMonkey83180" class="run-this">class Cluster {

/**
* @param args
*/
public static void main(String[] args) {
// TODO Auto-generated method stub
int[][] matrix = { { 1, 0, 1 }, { 1, 0, 1 }, { 0, 1, 1 } };
int cluster = dfs(matrix);
System.out.print("Cluster:" + cluster);
}

private static int dfs(int[][] matrix) {
// TODO Auto-generated method stub
int cluster = 0;
printMatrix(matrix);
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
if (matrix[i][j] == 1) {
dfsVisit(matrix, i, j);
cluster++;
}
}
}
printMatrix(matrix);
return cluster;
}

private static void dfsVisit(int[][] matrix, int i, int j) {
// TODO Auto-generated method stub
if (i < 0 || i >= matrix.length)
return;
if (j < 0 || j >= matrix[0].length)
return;
if (matrix[i][j] != 1)
return;
matrix[i][j] = -1;
dfsVisit(matrix, i + 1, j);
dfsVisit(matrix, i, j + 1);
dfsVisit(matrix, i - 1, j);
dfsVisit(matrix, i, j - 1);
}

private static void printMatrix(int[][] matrix) {
// TODO Auto-generated method stub
System.out.println();
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
System.out.printf("%4d", matrix[i][j]);
}
System.out.println();
}

}

}
</pre><pre title="CodeMonkey83180" input="yes">
</pre>

- Anonymous November 04, 2011 | Flag Reply
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0
of 0 vote

Look at connected component labeling problem
en.wikipedia.org/wiki/Connected-component_labeling

- chennavarri November 04, 2011 | Flag Reply
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0
of 0 vote

int FindAdjacentOns(int** matrix, int rows, int cols)
    {
	int groupCount = 0;

        for (int i=0; i< rows ; i++)
        {
            for(int j=0; j< cols ; j++)
            {
                if(matrix[i][j] == 1)
                {
	                groupCount ++;
	                CheckAdjacent( matrix,  i, j)
                }
            }
        }
        return groupCount;
    }

    void CheckAdjacent(int** matrix, int i, int j, int rows, int cols)
    {
        if (matrix != null && matrix[i][j] == 0)
        {
            return;
        }
        matrix[i][j] = 0;

        if (i > 0)
        {
            CheckAdjacent(matrix, i - 1, j, rows, cols);
        }
        if (j > 0)
        {
            CheckAdjacent(matrix, i - 1, j, rows, cols);
        }
        if (i < rows - 1)
        {
            CheckAdjacent(matrix, i + 1, j, rows, cols);
        }
        if (j < cols - 1)
        {
            CheckAdjacent(matrix, i, j + 1, rows, cols);
        }
    }

- satishl November 05, 2011 | Flag Reply
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0
of 0 vote

I don't understand the question completely?
Do i need to find out group of 1 or 0 in rows?
Please explain?

- sathishperias November 07, 2011 | Flag Reply
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0
of 0 vote

is it calculation of group of 1's sequentially ? or group 1's in all directions?

- Krishna November 08, 2011 | Flag Reply
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0
of 0 votes

Need to find the closet 1s. While finding the 1s we can move up/down.left and right. There is no diagonal move for finding group of 1s

- nmarasu November 08, 2011 | Flag
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0
of 0 vote

If some one can provide 0(n) complexity solution for this?

- ashu November 09, 2011 | Flag Reply
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0
of 0 vote

int  count  = 0;
for(int row = 1; row < m; row++) {
    for(int col = 0; col < n; col++) {
        if((A[row-1][col] == 1) && (A[row][col] == 0)) {
            count++; // This is a move from 1 to 0.
        }
    }
}
return 0;

So whenever we transition from 1 to 0 we increment the count.

- Anonymous November 20, 2011 | Flag Reply
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0
of 0 vote

Running C++ code:

#include<iostream>
using namespace std;
void numGroups(int a[][5],int i,int j,int n,int curr )
{

if(i < 0 || j < 0 )
return;
if((i == n) || (j == n))
return;


if(a[i][j]==1)
{

a[i][j] = curr;

numGroups(a,i-1,j,n,curr);
numGroups(a,i+1,j,n,curr);

numGroups(a,i,j-1,n,curr);
numGroups(a,i,j+1,n,curr);

numGroups(a,i+1,j+1,n,curr);
numGroups(a,i-1,j+1,n,curr);

numGroups(a,i+1,j-1,n,curr);
numGroups(a,i-1,j-1,n,curr);

}
}

bool discoverAll(int arr[][5],int n,int &next_i,int &next_j)
{
for(int i =0;i<n;i++)
for(int j=0;j<n;j++)
if(arr[i][j]==1)
{
next_i = i;
next_j = j;
return false;
}
return true;
}

int numGroupsTrigger(int arr[][5],int n)
{
int groupCount = 2;
int start_i=0,start_j=0;
while(!discoverAll(arr,n,start_i,start_j))
{

numGroups(arr,start_i,start_j,n,groupCount);
groupCount++;

}

return groupCount-2;
}
void showArr(int arr[][5],int n)
{
for(int i =0;i<n;i++)
{

for(int j=0;j<n;j++)
cout<<arr[i][j]<<" ";

cout<<endl;
}
}
int main()
{
int array2D[5][5] = { {1,0,1,0,1},
{0,1,1,0,0},
{1,1,1,1,1},
{0,0,1,1,0},
{1,0,1,0,1}
};
int n = 5;

cout<<"# 1s: "<<numGroupsTrigger(array2D,5);
cout<<endl;
showArr(array2D,n);
cout<<endl;
char c;
cin>>c;
return 0;
}

- Ratna Kumar November 22, 2011 | Flag Reply
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0
of 0 vote

Assuming that I am writing code for a 4 by 4 matrix.

static void GroupCount(ref int[,] Arr)
        {
            int groupCount = 1;
            for (int i = 0; i < 4; i++)
            {
                for (int j = 0; j < 4; j++)
                {
                    if (Arr[i, j] == 1)
                    {
                        groupCount += 1;
                        Arr[i, j] = groupCount;
                        if (i + 1 < 4)
                        {
                            if (Arr[i + 1, j] == 1)
                            {
                                Arr[i + 1, j] = Arr[i, j];
                                CombineGroup(ref Arr, i + 1, j);
                            }
                        }
                        if (j + 1 < 4)
                        {
                            if (Arr[i, j + 1] == 1)
                            {
                                Arr[i, j + 1] = Arr[i, j];
                                CombineGroup(ref Arr, i, j + 1);
                            }
                        }
                    }
                }
            }

            Console.WriteLine("GroupCount is {0}", groupCount-1);
            Console.ReadLine();
        }

        private static void CombineGroup(ref int[,] Arr, int i, int j)
        {
            if (i + 1 < 4)
            {
                if (Arr[i + 1, j] == 1)
                {
                    Arr[i + 1, j] = Arr[i, j];
                    CombineGroup(ref Arr, i + 1, j);
                }
            }
            if (j + 1 < 4)
            {
                if (Arr[i, j + 1] == 1)
                {
                    Arr[i, j + 1] = Arr[i, j];
                    CombineGroup(ref Arr, i, j + 1);
                }
            }
        }

- anonymous July 24, 2015 | Flag Reply
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0
of 0 vote

Assuming that I am working on a 4by4 matrix

static void GroupCount(ref int[,] Arr)
        {
            int groupCount = 1;
            for (int i = 0; i < 4; i++)
            {
                for (int j = 0; j < 4; j++)
                {
                    if (Arr[i, j] == 1)
                    {
                        groupCount += 1;
                        Arr[i, j] = groupCount;
                        if (i + 1 < 4)
                        {
                            if (Arr[i + 1, j] == 1)
                            {
                                Arr[i + 1, j] = Arr[i, j];
                                CombineGroup(ref Arr, i + 1, j);
                            }
                        }
                        if (j + 1 < 4)
                        {
                            if (Arr[i, j + 1] == 1)
                            {
                                Arr[i, j + 1] = Arr[i, j];
                                CombineGroup(ref Arr, i, j + 1);
                            }
                        }
                    }
                }
            }

            Console.WriteLine("GroupCount is {0}", groupCount-1);
            Console.ReadLine();
        }

        private static void CombineGroup(ref int[,] Arr, int i, int j)
        {
            if (i + 1 < 4)
            {
                if (Arr[i + 1, j] == 1)
                {
                    Arr[i + 1, j] = Arr[i, j];
                    CombineGroup(ref Arr, i + 1, j);
                }
            }
            if (j + 1 < 4)
            {
                if (Arr[i, j + 1] == 1)
                {
                    Arr[i, j + 1] = Arr[i, j];
                    CombineGroup(ref Arr, i, j + 1);
                }
            }
        }

- anonymous July 24, 2015 | Flag Reply
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0
of 0 vote

this solution assumes the input matrix order is 4by4.

static void GroupCount(ref int[,] Arr)
        {
            int groupCount = 1;
            for (int i = 0; i < 4; i++)
            {
                for (int j = 0; j < 4; j++)
                {
                    if (Arr[i, j] == 1)
                    {
                        groupCount += 1;
                        Arr[i, j] = groupCount;
                        if (i + 1 < 4)
                        {
                            if (Arr[i + 1, j] == 1)
                            {
                                Arr[i + 1, j] = Arr[i, j];
                                CombineGroup(ref Arr, i + 1, j);
                            }
                        }
                        if (j + 1 < 4)
                        {
                            if (Arr[i, j + 1] == 1)
                            {
                                Arr[i, j + 1] = Arr[i, j];
                                CombineGroup(ref Arr, i, j + 1);
                            }
                        }
                    }
                }
            }

            Console.WriteLine("GroupCount is {0}", groupCount-1);
            Console.ReadLine();
        }

        private static void CombineGroup(ref int[,] Arr, int i, int j)
        {
            if (i + 1 < 4)
            {
                if (Arr[i + 1, j] == 1)
                {
                    Arr[i + 1, j] = Arr[i, j];
                    CombineGroup(ref Arr, i + 1, j);
                }
            }
            if (j + 1 < 4)
            {
                if (Arr[i, j + 1] == 1)
                {
                    Arr[i, j + 1] = Arr[i, j];
                    CombineGroup(ref Arr, i, j + 1);
                }
            }
        }

- varunpvk July 24, 2015 | Flag Reply


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