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public class BestGrade { public static void main(String[] args) { String student[][] = {{"jerry","65"}, {"bob","1"}, {"jerry","23"},{"jerry","23"}, {"jerry","100"},{"Eric","83"}}; //Map counter = new HashMap(); List score = new ArrayList(); Map> map = new HashMap(); for(int i = 0;i(); score.add(Integer.parseInt(student[i][1])); map.put(student[i][0], score); } } List grade = new ArrayList(); for(String key:map.keySet()) { score = map.get(key); int sum = 0; for(int num : score) { sum+=num; } int average = sum/score.size(); grade.add(average); } Collections.sort(grade); System.out.println(grade.get(grade.size()-1)); } } Minder

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import java.util.*; import java.util.stream.Collectors; class Student{ private String name; private Float averageMark = 0.0f; private int numberOfSubjects = 0; public Student(String name){ this.name = name; } public void addMark(Float mark ){ averageMark += mark; numberOfSubjects += 1; averageMark = averageMark/numberOfSubjects; } public Float getAverageMark(){ return averageMark; } public String getName(){ return name; } public Student(Float mark, String n){ name = n; averageMark = mark; } } public class MaxScore { public static String s[][] = {{"jerry","65"}, {"bob","91"}, {"jerry","23"}, {"Eric","83"},{"Eric","99"}}; public static void main(String args[]){ Map studentMap = new HashMap(); for(int i=0; i studentList = studentMap.values().stream().collect(Collectors.toList()); Collections.sort(studentList, Comparator.comparing(Student::getAverageMark).reversed()); System.out.println(studentList.get(0).getName()); } } Minder

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//if vec[i][j] == 0 then i is not an influence //if vec[i][j] == 1 then j is not an influence //so time complexity is O(n) bool find_influences(vector > &vec) { int n = vec.size(); vector not_influence(n); for (int i = 0; i = 0; --j) { if (!vec[i][j]) { break; } not_influence[j] = 1; } if (j < 0) { return true; } } not_influence[i] = 1; } return false; } Minder

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X should be equal to Y, right?

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I think that you need to take into account that if you want to toggle 2 bits, you can only do if you flip bits from position 0..30. Toggling bit 31 is only going to toggle this bit no matter what. Therefore, you need to multiply (33-N)/32 to your proposed result, to keep this into account. Minder

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@Mythreya's analysis is correct but incomplete. To get the expected value, you have to multiply the number of bits by their probability. Answer is Sigma{k/(2^k)} for k = 1 to 32. Minder

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small mistake function power(x, n) { if n == 1 return x; // Even numbers else if (n%2 == 0) return square( power (x, n/2)); // Odd numbers else return power(x, n-1) * x; } Minder

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double Power(int x, int y) { double ret = 1; double power = x; while (y > 0) { if (y & 1) { ret *= power; } power *= power; y >>= 1; } return ret; } Minder

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sorry, typo while(top>p)

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I was asked of this question as well. I not sure how to implment a second stack in a way that the max number is always on top. Minder

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If you're writing it in ruby def find_repeat(numbers) numbers.length.times{|n| return numbers[n] if numbers[n] != n } end Minder

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Mat, try 1,2,2,3: 1+2+2+3= 8 1^2^2^3= 2 (8-2)/2=3?2

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void traverse(Node node){ if(node == null) return null; System.out.println(node.data); traverse(node.left); traverse(node.right); } Minder

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It can't be that this candidate really serves on a language committee. I find it difficult to believe that someone who serves on a language standards committee doesn't care about the difference between an O(N^2) and an O(logN) solution, for that would be horrifying indeed. And in defense of the employee conducting the interview, if I did see a candidate that didn't care about the difference between the two, I really wouldn't care what they have on their resume. They could be a Turing award winner for all I would care. There's a valid point about spatial locality, etc, but for something like N = 10^6 (for example), I'll take even O(logN) disk reads (*gasp*), over N^2 quick register operations any day.. Minder

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The solution was something like: Look for the candidates that have a bigger difference in cost first (in this case B and D since the difference for the price to each office are 400), choose the cheaper option for those. After one of the offices is full then just choose whatever spot is left. Minder

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I would use a Min Heap for SF of size = no. of candidates / 2 . k = no. of candidates / 2 ; for(int i : SFcosts){ minHeap.offer(i); if(minHeap.size() > k){ minHeap.poll(); } } // so at the end we have half candiates with minimum cost for SF then remaining will go ti NY Minder

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Nobody need to actually put golf balls in a school buss , nobody care if the number is 61,000,000 or whatever .Nobody in Google cares how good are you in giving estimates to the size of a buss. The answer is bussWidth*bussHight*bussLength / (ball/Diameter ^ 3 * sin (60) * sin(60) ^2 ) . This answer depends on the size of the school buss, different busses yield different answer. The sin(60) are since its gold balls and not cubics. Minder

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As Jehova put it. This tests your problem-solving stance being seeing that you leave the right things as variables. The question is really about trigonometry and understanding how tightly you can pack spheres. In other words, you need to figure the effective volume of each golf ball. To do so, first consider a single plane of balls, and a single row from it. If the ball radius is r, the width required for each ball is 2r. Now consider the depth required by each row. Since the rows will be "staved off" from each other, each row requires somewhat less than 2r for its depth. In fact, each requires r * tan 60, or about 1.73 * r. To see why, notice that each adjacent group of three spheres forms a triangle, and, because of the symmetry of the situation, their centers form an equilateral triangle. Thus each of its angles it 60 degrees. Since tan(theta) = opposite / adjacent, we have that tan(60) = depth / radius, or depth = radius * tan(60). Finally, consider the height require by each plane of balls. Take the three adjacent balls from the first plane and imagine adding a forth ball above them in the center. The centers of the four balls now form a tetrahedron (just a pyramid with a triangular base). We need to find the height from the center of the base of the tetrahedron to its top vertex. Looking at a diagram, one can see that the distance to the center of the base from one of the base vertices is found by cos(theta) = radius / hypotenuse, or cos(30) = r / distance, meaning distance = r / cos(30), or about 1.15 * r. If you create a line going from the base vertex to the base center, it then becomes clear that this line forms a vertical isosceles right triangle with the height line. In other words, the distance we just found is equal to the height of the tetrahedron. Bring it all together, we have that the effective volume of each ball is: effectiveWidth * effectiveDepth * effectiveHeight = 2r * r tan 60 * r / cos 30. So finally, the number of balls is the bus volume divided by each ball's effective volume. Minder