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Bloomberg Interview Questions

Review this list of Bloomberg interview questions and answers verified by hiring managers and candidates.
  • Bloomberg logoAsked at Bloomberg 

    " max Min 4, 3, 1 , 6, 7, 8 1 3 4 6 7 8 9 0 1 2 3 0 1 1 2 3 6 7 8 9 class MedianFinder{ std::priority_queue minHeap; std::priority_queue, greater> maxHeap; int numEleMaxheap = 0, numEleMinHeap = 0; public: void addNum( int n) { if(numEleMaxheap == numEleMinHeap ) { maxHeap.push(n); int maxofmaxheap = maxHeap.top(); maxHeap.pop(); "

    Ankush G. - " max Min 4, 3, 1 , 6, 7, 8 1 3 4 6 7 8 9 0 1 2 3 0 1 1 2 3 6 7 8 9 class MedianFinder{ std::priority_queue minHeap; std::priority_queue, greater> maxHeap; int numEleMaxheap = 0, numEleMinHeap = 0; public: void addNum( int n) { if(numEleMaxheap == numEleMinHeap ) { maxHeap.push(n); int maxofmaxheap = maxHeap.top(); maxHeap.pop(); "See full answer

    Software Engineer
    Data Structures & Algorithms
    +1 more
  • Bloomberg logoAsked at Bloomberg 
    Software Engineer
    Behavioral
  • "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."

    Techzen I. - "Implemented the Java code to find the largest island. It is similar to count the island. But in this we need to keep track of max island and compute its perimeter."See full answer

    Machine Learning Engineer
    Data Structures & Algorithms
    +2 more
  • "This problem can be solved with two approaches Iterative approach Recursive approach Quite easy to think about the iterative approach, you can make use of a while loop in that case. But what if you want to make use of previously computed values? That case going for the recursive solution is quite useful. class Collatz: def init(self) -> None: self.cache = {} self.steps = 0 def steps_from(self, n) -> int: # base case if n == 1: "

    Frederick A. - "This problem can be solved with two approaches Iterative approach Recursive approach Quite easy to think about the iterative approach, you can make use of a while loop in that case. But what if you want to make use of previously computed values? That case going for the recursive solution is quite useful. class Collatz: def init(self) -> None: self.cache = {} self.steps = 0 def steps_from(self, n) -> int: # base case if n == 1: "See full answer

    Software Engineer
    Coding
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