# 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.

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.

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

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; }