Programming Interview Questions

Nascent

Questions (68)

Comments (18)

Reputation 1630

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Comment hidden because of low score. Click to expand.

of 4 vote

Convert the tree to sorted doubly linked list. Then follow the same procedure as in case of array.

- Nascent February 12, 2014 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

initialize i =1 and j=N-1

put loop condition while(i<j)
{
if(i^3 + j^3 >N)
--j;
elseif(i^3 + j^3 <N)
++i;
else
cout<<i<<j;
++i;--j;

thats all
O(N) time.

- Nascent January 31, 2014 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Same qns was asked to me. I sorted the array then started from the end comparing with a[0]. If a[0]x2 >= a[end] then fine else keep on decrementing. Answer is the number of decrements made. It passed 3 out of 4 test cases.

- Nascent January 27, 2014 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Just calculate the number of set bits. These need to be flipped. I guess this would be a respectable answer.

- Nascent January 26, 2014 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Using this method we can find the absolute minimum difference:

Assume numbers are array a[ ].

for(int i=1;i<(sizeof(a)/sizeof(int));i++)
    {
         min=a[1]-a[0];
         max=a[0];
         if((a[i]-min)<min)
         {
            min=a[i]-max;
            start=a[i];
            end=max;
         }
         if(a[i]>max)
         max=a[i];

}

after finding absolute min we can get the elements using start and end.

but since we need to find all. After getting the absolute minimum we will traverse the liist and find out two numbers having difference equal to absolute min.

- Nascent November 30, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

-2

of 2 vote

void mirror(tree_t *t)
{
if(t==NULL)
return;
else
{
tree_t *temp;
mirror(t->left->left);
mirror(t->right->right);
temp=t->left;
t->left=t->right;
t->right=temp;
}
}

- Nascent July 15, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

-1

of 1 vote

#include<iostream>
using namespace std;
int main()
{
int a[]={3,3,2,1,1};
cout<<"repeated:"<<endl;
for(int i=0;i<5;i++)
{
if((a[abs(a[i])-1]) > 0)
{
a[abs(a[i])-1]=- a[abs(a[i])-1];

}
else
{
cout<<a[i]<<endl;
}
}
cout<<"missing:"<<endl;
for(int i=0;i<5;i++)
{
if(a[i]>0)
cout<<i+1<<endl;
}
getchar();
return 0;
}

- Nascent July 15, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

I forgot to mention with the question, its for C++ and not java. It was asked to code single singleton in C++ and then suggest possible measures to make it thread safe. The stress was on proper use of MUTEX and semaphors.

- Nascent July 03, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Its not consecutive. Its an increasing sequence. This is the tricky part else the problem is simple.

- Nascent July 03, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

@Dumbo, Can u elaborate more about how u will construct the tree also how tag wise check will be applied while adding to the queue.

- Nascent July 02, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Assume the file to be a static file with well formed XML tags. Suppose a tag is <Colour>Red</Colour>, then corresponding to this tage we should get all the values.

- Nascent July 02, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 6 vote

How can they ask such question in interviews. Its something very weird and i doubt it to be an interview question.

- Nascent June 15, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

Anything mentioned about the desired time and space complexities?

- Nascent April 15, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 1 vote

I think this question is similar to the closest pair in a given set of point. Instead of taking the mid point we consider the given point p(x,y) and compute the points based on x co-ordinate which are closest. Repeat for the closest point with the y - co ordinate. I think this will give the answer.

- Nascent March 10, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 2 vote

Approach:
Let D be the distance array.
Let P be the petrol array.
int [] D = {4, 7, 4, 8, 4, 1};
int [] P = {3, 5, 3, 8, 3, 6};

Now start with i=0;
D[0] = 4 > P[0] = 3; i=0 can not be starting point so we move to i=1;
D[1] = 7 > P[1] = 5; i=1 can not be starting point So we move to i=2;
D[2] = 4 > P[2] = 3; i=2 can not be starting pointso we move to i=3;
D[3] = 8 <= P[3] = 8; i=3 can be starting point, now keep on summing the D[i] and P[i] and P[i] >= D[i]
D[4] = 4 > P[4] = 3; here the P[i] is not greater or equal to D[i] so we break; and start from next i=5;

D[5]=1 < P[5] = 6; and carry over petrol is 5
now,
D[0] = 4 < P[0] + carry = 3 + 5;
D[1] = 7 < P[1] + carry = 5 + 4;
D[2] = 4 < P[2] + carry = 3 + 2;
D[3] = 8 < P[3] + carry = 8 + 1;
D[4] = 4 <= P[4] + carry = 3 + 1; // now next i = 5 is the same as the starting point ie index 5.
So the starting point is index=5;

Total time complexity: O(n)

- Nascent January 26, 2013 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

#include<stdio.h>

/* A utility function that prints a given arr[] of length size*/
void printArray(int arr[], int size)
{
for(int i = 0; i < size; i++)
printf("%d ", arr[i]);
printf("\n");
return;
}

/* The core function that recursively generates and prints all sequences of
length k */
void printSequencesRecur(int a[],int arr[], int n, int k, int index)
{
int i;
if (k == 0)
{
printArray(arr, index);
}
if (k > 0)
{
for(i = 0; i<n; ++i)
{
arr[index] = a[i];
printSequencesRecur(a,arr, n, k-1, index+1);
}
}
}

/* A function that uses printSequencesRecur() to prints all sequences
from 1, 1, ..1 to n, n, ..n */
void printSequences(int a[],int n, int k)
{
int *arr = new int[k];
printSequencesRecur(a,arr, n, k, 0);

delete(arr); // free dynamically allocated array
return;
}

/* Driver Program to test above functions */
int main()
{
int a[]={2,8,9,16};
int n = 3;
int k = 2;
printSequences(a,4,k);
getchar();
return 0;
}

- Nascent October 21, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

We can use a doublylinked list in association with hash map. The map will have key as name i.e "abc","def" etc and the value field will be pointer to the doubly link list node which contains the numerical value. Whenever we delete the the entry will be removed both from the map as well as the doubly linked list. Becoz of the doubly linked list the order will be maintained. All operation can be done in O(1).

- Nascent October 21, 2012 | Flag Reply

Comment hidden because of low score. Click to expand.

of 0 vote

After seeing so many comments , I felt that I need to give more details. Consider the array below:

a[ ]={2,8,1,9,10,5,7,11,13}

output={1,2,9,8,5,10,7,11,13}. Here the order of odd and even numbers are same. The one which is more in number will be assembled at last. I hope the qns is now clear.

- Nascent September 17, 2012 | Flag Reply

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