reci January 20, 2018 at 11:23

Issue # 7: IT training - current issues and challenges from leading companies

We have prepared a new issue of IT training with questions and tasks from leading IT companies.

KDPV

The selection includes questions encountered during interviews in Adobe (yes, the question about color is included in the selection :). Tasks of various difficulty levels, but all are solvable. Especially if you have already answered questions from past issues.

We hope that the above tasks will help you to prepare for the upcoming interviews.

Questions

8 marbles
Let's say you have 8 marbles and a two-pan balance.
All of the marbles look the same. Each marble weighs 2.0 grams except for one, which is slightly heavier at 2.05 grams.
How would you find the heaviest marble if you are only allowed to weigh the marbles 2 times using the balance scale?

Transfer
Suppose you have 8 glass balls and a cup scale. All balls look the same. Each weighs 2 grams, with the exception of one that is slightly heavier - 2.05 grams.
How to find the heaviest ball, if only 2 weighings are allowed?
Falling bear
A bear fell from a height of 10m on the ground in √2 seconds. But somehow, it didn't get hurt. What is the color of bear?

Transfer
Bear, falling from a height of 10m, reaches the ground in √2 seconds and, for some reason, remains without damage. What color is the bear?

Note: I thought for a while, but still looked back. The question is a little trick, but you can answer. I decided to bring it here, if this is found in interviews.

Tasks

Product of max and min in 2 arrays
Given two arrays of integers, the task is to calculate the product of max element of first array and min element of second array.

Ex:
Input: arr1 [] = {5, 7, 9, 3, 6, 2},
arr2 [] = {1, 2, 6, -1, 0, 9}
Output: max element in first array
is 9 and min element in second array
is -1. The product of these two is -9.

Input: arr1 [] = {1, 4, 2, 3, 10, 2},
arr2 [] = {4, 2, 6, 5, 2, 9}
Output: 20.

Transfer
Given 2 arrays of integers. The task is to calculate the product of the maximum element from the first array and the minimum from the second array.

Examples:
Given: arr1 [] = {5, 7, 9, 3, 6, 2}, arr2 [] = {1, 2, 6, -1, 0, 9}
Answer: the maximum element of the first array is 9, the minimum element of the second is -1. Product -9.

Given: arr1 [] = {1, 4, 2, 3, 10, 2},
arr2 [] = {4, 2, 6, 5, 2, 9}
Answer: 20.
Maximum chocolates
You have $ 15 with you. You go to a shop and shopkeeper tells you price as $ 1 per chocolate. He also tells you that you can get a chocolate in return of 3 wrappers. How many maximum chocolates you can eat?

Write a program to find solution for variable inputs. Given the following three values, the task is to find the total number of maximum chocolates you can eat.

money: Money you have to buy chocolates
price: Price of a chocolate
wrap: Number of wrappers to be returned for getting one extra chocolate.

It may be assumed that all given values are positive integers and greater than 1.

Transfer
You have $ 15. In the store you will find out the price for a chocolate bar from the seller - $ 1. The seller also reports that for 3 wrappers will give you another chocolate bar. What maximum chocolates can you count on?

Write a program for variable input, the task of which is to find the maximum of chocolates:

money: Available money
price: Price for a chocolate
bar wrap: The number of wrappers you need to get another chocolate bar.

We can assume that all input data are integers and greater than 1.
Sort array larger than RAM
Given 2 machines, each having 64 GB RAM, containing all integers (8 byte). You need to sort the entire 128 GB data. You may assume a small amount of additional RAM.

Transfer
There are 2 computers, each with 64 GB of RAM, occupied by integers (8 bytes). You need to sort all 128 GB of data. You can use a small amount of additional RAM memory.

Answers, as usual, will be given over the next week - have time to decide. Good luck

Solutions

Question 1
In the comments, they found the correct answer:
1. take six balls of 3 for each bowl
2. if the bowls are equal, then the desired ball in the remaining two
3. if they are not equal, then we throw out the three lightest
4. weighed 2 of the remaining three - the problem is solved: )
Question 2
In the original, the color of the bear is white, because it fell with an acceleration of 10 m / s2, which can be at the poles. The correct solution was found, and several alternative ones were suggested :)

Task 1

An array sorting solution is not optimal. A simple pass through arrays in search of min / max is better:

// C++ program to find the to calculate
// the product of max element of first 
// array and min element of second array
#include 
using namespace std;
// Function to calculate the product
int minMaxProduct(int arr1[], int arr2[], 
                         int n1, int n2)
{
    // Initialize max of first array
    int max = arr1[0];
    // initialize min of second array
    int min = arr2[0];
    int i;
    for (i = 1; i < n1 && i < n2; ++i) {
        // To find the maximum element 
        // in first array
        if (arr1[i] > max)
            max = arr1[i];
        // To find the minimum element
        // in second array
        if (arr2[i] < min)
            min = arr2[i];
    }
    // Process remaining elements
    while (i < n1)
    {
        if (arr1[i] > max)
           max = arr1[i];  
        i++;
    }
    while (i < n2)
    {
        if (arr2[i] < min)
           min = arr2[i];  
        i++;
    }
    return max * min;
}
// Driven code
int main()
{
    int arr1[] = { 10, 2, 3, 6, 4, 1 };
    int arr2[] = { 5, 1, 4, 2, 6, 9 };
    int n1 = sizeof(arr1) / sizeof(arr1[0]);
    int n2 = sizeof(arr1) / sizeof(arr1[0]);
    cout << minMaxProduct(arr1, arr2, n1, n2) << endl;
    return 0;
}

Task 2

A recursion solution is not the best one in this case. In the original solution, by observation, the formula totalChoc = (choc-1) / (wrap-1) was derived and the program in this case took the form:

// Efficient C++ program to find maximum
// number of chocolates
#include 
using namespace std;
// Returns maximum number of chocolates we can eat
// with given money, price of chocolate and number
// of wrapprices required to get a chocolate.
int countMaxChoco(int money, int price, int wrap)
{
    // Corner case
    if (money < price)
       return 0;
    // First find number of chocolates that
    // can be purchased with the given amount
    int choc = money / price;
    // Now just add number of chocolates with the
    // chocolates gained by wrapprices
    choc = choc + (choc - 1) / (wrap - 1);
    return choc;
}
// Driver code
int main()
{
    int money = 15 ; // total money
    int price = 1; // cost of each candy
    int wrap = 3 ; // no of wrappers needs to be
    // exchanged for one chocolate.
    cout << countMaxChoco(money, price, wrap);
    return 0;
}

Task 3

As a solution to the problem, External merge sort is proposed :

// C++ program to implement external sorting using 
// merge sort
#include 
using namespace std;
struct MinHeapNode
{
    // The element to be stored
    int element;
    // index of the array from which the element is taken
    int i;
};
// Prototype of a utility function to swap two min heap nodes
void swap(MinHeapNode* x, MinHeapNode* y);
// A class for Min Heap
class MinHeap
{
    MinHeapNode* harr; // pointer to array of elements in heap
    int heap_size;     // size of min heap
public:
    // Constructor: creates a min heap of given size
    MinHeap(MinHeapNode a[], int size);
    // to heapify a subtree with root at given index
    void MinHeapify(int);
    // to get index of left child of node at index i
    int left(int i) { return (2 * i + 1); }
    // to get index of right child of node at index i
    int right(int i) { return (2 * i + 2); }
    // to get the root
    MinHeapNode getMin() {  return harr[0]; }
    // to replace root with new node x and heapify()
    // new root
    void replaceMin(MinHeapNode x)
    {
        harr[0] = x;
        MinHeapify(0);
    }
};
// Constructor: Builds a heap from a given array a[]
// of given size
MinHeap::MinHeap(MinHeapNode a[], int size)
{
    heap_size = size;
    harr = a; // store address of array
    int i = (heap_size - 1) / 2;
    while (i >= 0)
    {
        MinHeapify(i);
        i--;
    }
}
// A recursive method to heapify a subtree with root
// at given index. This method assumes that the
// subtrees are already heapified
void MinHeap::MinHeapify(int i)
{
    int l = left(i);
    int r = right(i);
    int smallest = i;
    if (l < heap_size && harr[l].element < harr[i].element)
        smallest = l;
    if (r < heap_size && harr[r].element < harr[smallest].element)
        smallest = r;
    if (smallest != i)
    {
        swap(&harr[i], &harr[smallest]);
        MinHeapify(smallest);
    }
}
// A utility function to swap two elements
void swap(MinHeapNode* x, MinHeapNode* y)
{
    MinHeapNode temp = *x;
    *x = *y;
    *y = temp;
}
// Merges two subarrays of arr[].
// First subarray is arr[l..m]
// Second subarray is arr[m+1..r]
void merge(int arr[], int l, int m, int r)
{
    int i, j, k;
    int n1 = m - l + 1;
    int n2 = r - m;
    /* create temp arrays */
    int L[n1], R[n2];
    /* Copy data to temp arrays L[] and R[] */
    for(i = 0; i < n1; i++)
        L[i] = arr[l + i];
    for(j = 0; j < n2; j++)
        R[j] = arr[m + 1 + j];
    /* Merge the temp arrays back into arr[l..r]*/
    i = 0; // Initial index of first subarray
    j = 0; // Initial index of second subarray
    k = l; // Initial index of merged subarray
    while (i < n1 && j < n2)
    {
        if (L[i] <= R[j])
            arr[k++] = L[i++];
        else
            arr[k++] = R[j++];
    }
    /* Copy the remaining elements of L[], if there
       are any */
    while (i < n1)
        arr[k++] = L[i++];
    /* Copy the remaining elements of R[], if there
       are any */
    while(j < n2)
        arr[k++] = R[j++];
}
/* l is for left index and r is right index of the
   sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
    if (l < r)
    {
        // Same as (l+r)/2, but avoids overflow for
        // large l and h
        int m = l + (r - l) / 2;
        // Sort first and second halves
        mergeSort(arr, l, m);
        mergeSort(arr, m + 1, r);
        merge(arr, l, m, r);
    }
}
FILE* openFile(char* fileName, char* mode)
{
    FILE* fp = fopen(fileName, mode);
    if (fp == NULL)
    {
        perror("Error while opening the file.\n");
        exit(EXIT_FAILURE);
    }
    return fp;
}
// Merges k sorted files.  Names of files are assumed
// to be 1, 2, 3, ... k
void mergeFiles(char *output_file, int n, int k)
{
    FILE* in[k];
    for (int i = 0; i < k; i++)
    {
        char fileName[2];
        // convert i to string
        snprintf(fileName, sizeof(fileName), "%d", i);
        // Open output files in read mode.
        in[i] = openFile(fileName, "r");
    }
    // FINAL OUTPUT FILE
    FILE *out = openFile(output_file, "w");
    // Create a min heap with k heap nodes.  Every heap node
    // has first element of scratch output file
    MinHeapNode* harr = new MinHeapNode[k];
    int i;
    for (i = 0; i < k; i++)
    {
        // break if no output file is empty and
        // index i will be no. of input files
        if (fscanf(in[i], "%d ", &harr[i].element) != 1)
            break;
        harr[i].i = i; // Index of scratch output file
    }
    MinHeap hp(harr, i); // Create the heap
    int count = 0;
    // Now one by one get the minimum element from min
    // heap and replace it with next element.
    // run till all filled input files reach EOF
    while (count != i)
    {
        // Get the minimum element and store it in output file
        MinHeapNode root = hp.getMin();
        fprintf(out, "%d ", root.element);
        // Find the next element that will replace current
        // root of heap. The next element belongs to same
        // input file as the current min element.
        if (fscanf(in[root.i], "%d ", &root.element) != 1 )
        {
            root.element = INT_MAX;
            count++;
        }
        // Replace root with next element of input file
        hp.replaceMin(root);
    }
    // close input and output files
    for (int i = 0; i < k; i++)
        fclose(in[i]);
    fclose(out);
}
// Using a merge-sort algorithm, create the initial runs
// and divide them evenly among the output files
void createInitialRuns(char *input_file, int run_size,
                       int num_ways)
{
    // For big input file
    FILE *in = openFile(input_file, "r");
    // output scratch files
    FILE* out[num_ways];
    char fileName[2];
    for (int i = 0; i < num_ways; i++)
    {
        // convert i to string
        snprintf(fileName, sizeof(fileName), "%d", i);
        // Open output files in write mode.
        out[i] = openFile(fileName, "w");
    }
    // allocate a dynamic array large enough
    // to accommodate runs of size run_size
    int* arr = (int*)malloc(run_size * sizeof(int));
    bool more_input = true;
    int next_output_file = 0;
    int i;
    while (more_input)
    {
        // write run_size elements into arr from input file
        for (i = 0; i < run_size; i++)
        {
            if (fscanf(in, "%d ", &arr[i]) != 1)
            {
                more_input = false;
                break;
            }
        }
        // sort array using merge sort
        mergeSort(arr, 0, i - 1);
        // write the records to the appropriate scratch output file
        // can't assume that the loop runs to run_size
        // since the last run's length may be less than run_size
        for (int j = 0; j < i; j++)
            fprintf(out[next_output_file], "%d ", arr[j]);
        next_output_file++;
    }
    // close input and output files
    for (int i = 0; i < num_ways; i++)
        fclose(out[i]);
    fclose(in);
}
// For sorting data stored on disk
void externalSort(char* input_file,  char *output_file,
                  int num_ways, int run_size)
{
    // read the input file, create the initial runs,
    // and assign the runs to the scratch output files
    createInitialRuns(input_file, run_size, num_ways);
    // Merge the runs using the K-way merging
    mergeFiles(output_file, run_size, num_ways);
}
// Driver program to test above
int main()
{
    // No. of Partitions of input file.
    int num_ways = 10;
    // The size of each partition
    int run_size = 1000;
    char input_file[] = "input.txt";
    char output_file[] = "output.txt";
    FILE* in = openFile(input_file, "w");
    srand(time(NULL));
    // generate input
    for (int i = 0; i < num_ways * run_size; i++)
        fprintf(in, "%d ", rand());
    fclose(in);
    externalSort(input_file, output_file, num_ways,
                run_size);
    return 0;
}

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Issue # 7: IT training - current issues and challenges from leading companies

Questions

Tasks

Solutions

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