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Anteshkumar Sharma
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Merge pull request #5 from MissionJava/Alten_codility_test
Alten codility test
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‎README.md

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# data-structure-and-algorithm-using-java
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This repository includes data structure and algorithm solutions, and solutions of the live coding interview questions asked at different companies over a video call.

‎src/main/java/com/antesh/agoda/App.java

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‎src/main/java/com/antesh/interview/agoda/MaximumRevenueFromRodCuttingProblem.java

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package com.antesh.interview.agoda;
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/*
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* Suppose we
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have a rod of length 5, which can be cut into smaller pieces of integer lengths 1, 2, 3, or 4. These smaller rods
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can then be further cut into smaller pieces. We stress that the rod pieces can only be cut into pieces of integer
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length; so pieces of size 1/2 or π are not considered in this problem. Now suppose that we can sell rods of
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length 1 for $1, which we denote p1 = 1. Similarly, suppose that the prices for rods of length 2, . . . , 5 are given
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by p2 = 4, p3 = 7, p4 = 8, and p5 = 9 respectively. We make two key assumptions: we will sell all the smaller
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rods, regardless of the cuts; and that each cut is free. Under these assumptions, how should the rod be cut to
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maximize the profit? We note the following cuts and the corresponding profits.
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- If the rod is cut into five pieces of length 1, we stand to make 5 · p1 = $5.
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- If the rod is cut into one piece of length 2 and one piece of length 3, we stand to make p2+p3 = 4+7 = $11.
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-If the rod is cut into two pieces of length 2 and one piece of length 1, we stand to make 2 · p2 + p1 =
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8 + 1 = $9.
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Out of the above options, cutting the rod into one piece of length 2 and one rod of length 3 is the most
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profitable. Of course, there are other possible cuts not listed above, such as cutting the rod into one piece
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of length 1 and one piece of length 4. The goal is to determine the most profitable cut. The Rod-Cutting
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Problem is formalized as follows.
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Definition 1 (Rod-Cutting Problem).
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- Instance: Let n ∈ N be the length of the rod, and let p1, p2, . . . , pn be non-negative real numbers. Here,
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pi
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is the price of a length i rod.
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- Solution: The maximum revenue, which we denote rn, obtained by cutting the rod into smaller pieces
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of integer lengths and selling the smaller rods
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* Example:
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* For input N = 4, v[i] = {0, 2, 4, 7, 7}
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* Output: 2 (cuts of length 1 and 3 gives revenue of 2 + 7 = 9)
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* Explanation:
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In below price by rod length array,
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price 0 at index i=0 is for if rod length: 0,
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price 2 at index i = 1 is for rod length 1
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price 4 at index i = 2 is for rod length 2
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price 7 at index i = 3 is for rod length 3
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price 7 at index i = 4 is for rod length 4
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so on can be extended...
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Input:
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int[] prices = { 2, 4, 7, 7};
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int rod_size = 4;
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Output:
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9
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* */
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public class MaximumRevenueFromRodCuttingProblem {
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public static void main(String[] args) {
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System.out.println(solution(4, new int[]{2, 4, 7, 7}));
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System.out.println(solution(8, new int[]{1, 5, 8, 9, 10, 17, 17, 20}));
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System.out.println(solution(8, new int[]{3, 5, 8, 9, 10, 17, 17, 20}));
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}
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private static int solution(int rod_size, int[] prices) {
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int max_profit = Integer.MIN_VALUE;
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int profit = 0;
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if (rod_size == 1) {
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return prices[0];
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}
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for (int i = rod_size - 1; i >= 0; i--) {
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profit = profit + prices[i];
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int remain = rod_size - i - 1;
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if (remain > 0) {
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profit = profit + solution(remain, prices);
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}
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max_profit = Math.max(max_profit, profit);
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profit = 0;
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}
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return max_profit;
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}
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}

‎src/main/java/com/antesh/alten/codility/SolutionIter.java renamed to ‎src/main/java/com/antesh/interview/alten/SolutionIter.java

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package com.antesh.alten.codility;
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package com.antesh.interview.alten;
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import java.io.BufferedReader;
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import java.io.FileReader;

‎src/main/java/com/antesh/interview/nvidia/CardDeckProblem.java

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/*
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Coding Round 3 - at Nvidia
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* // 1. check all 52 cards available or not
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// 2. check which card is missing
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// 3. check if we have a duplicate card
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String[] cardsDeck = new String[]
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{ "AH", "KH", "10D", "5H", "AC", "10S", "AS", "8H", "9D", "5S", "3D", "7S", "2D", "JH",
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"QS", "2H", "QD", "6S", "5D", "8D", "2C", "JC", "KS", "KD", "4H", "3H", "3C", "6D", "QH", "9H", "JD", "7D",
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"AD", "2S", "3S", "4S", "9S", "9C", "5C", "7C", "QC", "10H", "10C", "8S", "JS", "4D", "6H", "4C", "8C",
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"7H", "6C", "KC" };
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Cards: A J Q K, 2 - 10
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Suit
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D - Diamond
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S - Spade
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H - Heart
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C - Club
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*
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*
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*
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*
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* */
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import java.util.*;
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import java.util.stream.Collectors;
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public class CardDeckProblem {
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public static void main(String[] args) {
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Set<String> prePopulatedCardDeck = prepopulateUniqueCards();
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testNullCardDeck();
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test52CardsWithEmptyCardDeck();
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testDuplicateAndMissingCards(prePopulatedCardDeck);
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testMissingCards(prePopulatedCardDeck);
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testNoMissingCards(prePopulatedCardDeck);
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}
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public static void testNullCardDeck() {
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validateCardDeck(null);
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}
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public static void test52CardsWithEmptyCardDeck() {
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String[] cardsDeck = new String[]{"AH"};
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validateCardDeck(cardsDeck);
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}
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public static void testDuplicateAndMissingCards(Set<String> prePopulatedCardDeck) {
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//duplicate AH
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String[] cardsDeck = new String[]
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{"AH", "AH", "10D", "5H", "AC", "10S", "AS", "8H", "9D", "5S", "3D", "7S", "2D", "JH",
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"QS", "2H", "QD", "6S", "5D", "8D", "2C", "JC", "KS", "KD", "4H", "3H", "3C", "6D", "QH", "9H", "JD", "7D",
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"AD", "2S", "3S", "4S", "9S", "9C", "5C", "7C", "QC", "10H", "10C", "8S", "JS", "4D", "6H", "4C", "8C",
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"7H", "6C", "KC"};
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Set<String> uniqueCardDeck = getUniqueCardDeck(cardsDeck);
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findMissingCards(prePopulatedCardDeck, uniqueCardDeck);
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}
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public static void testNoMissingCards(Set<String> prePopulatedCardDeck) {
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//All valid 52 cards
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String[] cardsDeck = new String[]
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{"AH", "KH", "10D", "5H", "AC", "10S", "AS", "8H", "9D", "5S", "3D", "7S", "2D", "JH",
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"QS", "2H", "QD", "6S", "5D", "8D", "2C", "JC", "KS", "KD", "4H", "3H", "3C", "6D", "QH", "9H", "JD", "7D",
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"AD", "2S", "3S", "4S", "9S", "9C", "5C", "7C", "QC", "10H", "10C", "8S", "JS", "4D", "6H", "4C", "8C",
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"7H", "6C", "KC"};
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Set<String> uniqueCardDeck = getUniqueCardDeck(cardsDeck);
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findMissingCards(prePopulatedCardDeck, uniqueCardDeck);
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}
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public static void testMissingCards(Set<String> prePopulatedCardDeck) {
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//missing AK, AH
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String[] cardsDeck = new String[]
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{"ZK", "ZK", "10D", "5H", "AC", "10S", "AS", "8H", "9D", "5S", "3D", "7S", "2D", "JH",
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"QS", "2H", "QD", "6S", "5D", "8D", "2C", "JC", "KS", "KD", "4H", "3H", "3C", "6D", "QH", "9H", "JD", "7D",
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"AD", "2S", "3S", "4S", "9S", "9C", "5C", "7C", "QC", "10H", "10C", "8S", "JS", "4D", "6H", "4C", "8C",
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"7H", "6C", "KC"};
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Set<String> uniqueCardDeck = getUniqueCardDeck(cardsDeck);
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findMissingCards(prePopulatedCardDeck, uniqueCardDeck);
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}
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private static void validateCardDeck(String[] cardsDeck) {
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if (cardsDeck == null || cardsDeck.length == 0) {
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System.out.println("Invalid input or empty input");
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return;
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}
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if (cardsDeck.length < 52) {
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System.out.println("Number of cards in the deck are less than 52");
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}
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}
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private static Set<String> getUniqueCardDeck(String[] cardsDeck) {
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Set<String> uniqueCardDeck = new TreeSet<>();
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StringBuilder duplicate = new StringBuilder();
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for (String card : cardsDeck) {
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boolean result = uniqueCardDeck.add(card);
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if (!result) {
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duplicate.append(card + " ");
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}
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}
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String duplicateMessage = duplicate.toString().length() > 0 ? "Found duplicate card: " + duplicate.toString() : "No duplicate card found";
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System.out.println(duplicateMessage);
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return uniqueCardDeck;
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}
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private static void findMissingCards(Set<String> prePopulatedCardDeck, Set<String> uniqueCardDeck) {
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Set<String> missingCards = prePopulatedCardDeck.stream().filter(prePopulatedCard -> {
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return uniqueCardDeck.stream().noneMatch(givenCard -> {
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return prePopulatedCard.equals(givenCard);
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});
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}).collect(Collectors.toSet());
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String missingCardMessage = missingCards.size() > 0 ? "Missing cards: " + missingCards : "No missing card";
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System.out.println(missingCardMessage);
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}
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private static Set<String> prepopulateUniqueCards() {
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Set<String> uniqueCardDeck = new TreeSet<>();
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for (int i = 0; i < 4; i++) {
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for (int j = 1; j <= 13; j++) {
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StringBuilder sb = new StringBuilder();
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if (j == 1) {
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sb.append("A");
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} else if (j == 11) {
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sb.append("J");
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} else if (j == 12) {
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sb.append("Q");
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} else if (j == 13) {
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sb.append("K");
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} else {
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sb.append(j);
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}
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if (i == 0) {
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sb.append("D");
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} else if (i == 1) {
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sb.append("S");
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} else if (i == 2) {
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sb.append("H");
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} else if (i == 3) {
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sb.append("C");
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}
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uniqueCardDeck.add(sb.toString());
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}
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}
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return uniqueCardDeck;
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}
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}

‎src/main/java/com/antesh/interview/nvidia/ReadMe.md

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#HackerRank coding round interview asked at Nvidia
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##Football Scores
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The number of goals achieved by two football teams in matches in a league is given in the form of two
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lists. Consider:
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Football team A, has played three matches, and has scored teamA = [1, 2, 3] goals in each match respectively.
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Football team B, has played two matches, and has scored teamB = [2, 4] goals in each match respectively.
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For each match of team B, compute the total number of matches of team A where team A has scored less than or equal to
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the number of goals scored by team B in that match.
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###In the above case:
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For 2 goals scored by team B in its first match, team A has 2 matches with scores 1 and 2. For 4 goals scored by team B
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in its second match, team A has 3 matches with scores 1, 2 and 3. Hence, the answer: [2, 3].
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###Function Description
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Complete the function counts in the editor below. The function must return an array of m positive integers, one for each
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teamB[i] representing the total number of elements teamA[j] satisfying teamA[j] = teamB[i] where 0 = j < n
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and 0 = i < m, in the given order.
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###counts has the following parameter(s):
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teamA[teamA[0],...teamA[n-1]]: first array of positive integers
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teamB[teamB[0],...teamB[n-1]]: second array of positive integers
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##Constraints
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2 = n, m = 105 1 = teamA[j] = 109, where 0 = j < n. 1 = teamB[i] = 109, where 0 = i < m.
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Input Format For Custom Testing Input from stdin will be processed as follows and passed to the function.
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The first line contains an integer n, the number of elements in teamA.
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The next n lines each contain an integer describing teamA[j] where 0 = j < n.
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The next line contains an integer m, the number of elements in teamB.
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The next m lines each contain an integer describing teamB[i] where 0 = i < m.
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##Sample Case 0
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###Sample Input 0
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4 1 4 2 4 2 3 5
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###Sample Output 0
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2 4
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###Explanation 0
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Given values are n = 4, teamA = [1, 4, 2, 4], m = 2, and teamB = [3, 5].
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For teamB[0] = 3, we have 2 elements in teamA (teamA[0] = 1 and teamA[2] = 2) that are = teamB[0]. For teamB[1] = 5,
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we have 4 elements in teamA (teamA[0] = 1, teamA[1] = 4, teamA[2] = 2, and teamA[3] = 4) that are = teamB[1].
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Thus, the function returns the array [2, 4] as the answer.
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##Sample Case 1
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###Sample Input 1
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5 2 10 5 4 8 4 3 1 7 8
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###Sample Output 1
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1 0 3 4
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###Explanation 1
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Given values are n = 5, teamA = [2, 10, 5, 4, 8], m = 4, and teamB = [3, 1, 7, 8].
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For teamB[0] = 3, we have 1 element in teamA (teamA[0] = 2) that is = teamB[0]. For teamB[1] = 1, there are 0 elements
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in teamA that are = teamB[1]. For teamB[2] = 7, we have 3 elements in teamA (teamA[0] = 2, teamA[2] = 5,
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and teamA[3] = 4) that are = teamB[2]. For teamB[3] = 8, we have 4 elements in teamA (teamA[0] = 2, teamA[2] = 5,
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teamA[3] = 4, and teamA[4] = 8) that are = teamB[3]. Thus, the function returns the array [1, 0, 3, 4] as the answer.

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