Learn and practice the Top K Elements and Two Heaps patterns to solve coding interview problems.

Speedrun the Oracle Coding Interview: Top Problems in Python

This module will teach you the underlying concepts and equip you with the techniques needed to use heaps to efficiently solve a diverse range of problems. In many problems, we need to find the k most/least frequent or k largest/smallest elements in a given set of elements. Such problems, common in data analysis, natural language processing, online algorithms, as well as in recommendation systems, may be solved using the Top K Elements pattern. The fast insertion and deletion operations possible with heaps make them ideal to implement efficient algorithms for such problems. Further, we can use two heaps when we need to simultaneously keep track of the k largest elements in a set, as well as the k smallest elements in the same set. Extending this idea, each heap may be based on a separate dataset. These techniques are used to solve problems categorized under the Two Heaps pattern.

Heaps

# Statement

Given an array of integers, `arr`, and an integer, `k`, return the $k$ most frequent elements.

> **Note:** You can return the answer in any order.

**Constraints:**
* $1$ $\leq$ `arr.length` $\leq$ $10^{3}$
* $10^{-4}$ $\leq$ `arr[i]` $\leq$ $10^{4}$
* 1 $\leq$ `k` $\leq$ number of unique elements in an array.
* It is guaranteed that the answer is unique.

# Examples


# Understand the problem

Let’s take a moment to make sure you’ve correctly understood the problem. The quiz below helps you check if you’re solving the correct problem:


What is the output if the following inputs are given?

arr = [1, 1, 2, 4, 5, 5]

k = 2

What is the output if the following inputs are given?
arr = [1, 1, 2, 4, 5, 5]
k = 2

What is the output if the following inputs are given?

arr = [1, 1, 1, 2, 2, 4, 5, 5]

k = 3

What is the output if the following inputs are given?
arr = [1, 1, 1, 2, 2, 4, 5, 5]
k = 3

Top K Frequent Elements

Top K Frequent Elements

# Figure it out!


We have a game for you to play. Rearrange the logical building blocks to develop a clearer understanding of how to solve this problem.




# Try it yourself

Implement your solution in the following coding playground:





Help me debug only the function Top K Frequent Elements . . I don't need the corrected code.I don't want explanations or code review. Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Please help me understand why this code failed for the last failed test case. Don't give me the corrected code. . If yes, then Just provide me the exact and to-the-point error that helps me run the code successfully on the test cases provided.

Your current task is to review only the top_k_frequent function and the other function(s) invoked in it/them. Do not review extra/helper files or functions not invoked in the solution function top_k_frequent. Do not review the code of data structures objects instantiated like a binary tree, linked list, trie, union find, etc., in the solution functions. Just review the solution/algorithm function of the problem. Also, Answer the following questions: Is the Top K Elements pattern used? Take your time and review the code.

Please provide a brief time complexity analysis of the function top_k_frequent. While calculating time complexity, include the time complexity of built-in Python functions invoked by the code. The Answer should be to the point and in a few lines.

Please provide a brief Space complexity analysis of the function top_k_frequent. While calculating space complexity, include the space complexity of built-in Python functions invoked by the code. Do not include the space used for the input parameters and the space used for just returning an output(not used as a scratch space). The Answer should be to the point and in a few lines

def evaluate_results(self) -> object:
        """
        self.__inputs is the list of inputs that are passed as input to function
        self.__actual_output is a list of outputs that is produced by the learner's code
        return : self.__edu__result object
        """
        from heapq import heappush, heappop


        def top_k_frequent_test(arr, k):

            # find the frequency of each number
            num_frequency_map = {}
            for num in arr:
                num_frequency_map[num] = num_frequency_map.get(num, 0) + 1
            top_k_elements = []

            # go through all numbers of the num_frequency_map
            # and push them in the top_k_elements, which will have
            # top k frequent numbers. If the heap size is more than k,
            # we remove the smallest(top) number
            for num, frequency in num_frequency_map.items():
                heappush(top_k_elements, (frequency, num))
                if len(top_k_elements) > k:
                    heappop(top_k_elements)

            # create a list of top k numbers
            top_numbers = []
            while top_k_elements:
                top_numbers.append(heappop(top_k_elements)[1])

            return top_numbers

        expected = top_k_frequent_test(self.__inputs[0], self.__inputs[1])
        actual = self.__actual_outputs[0]
        # multiple outputs will be added at each index respectively for a single test_case
        expected_outputs = [] 
        expected_outputs.append(expected)

        if sorted(actual) == sorted(expected):
            self.__edu_result.store_result(expected_outputs, TestCaseResult.PASS)
        else: 
            self.__edu_result.store_result(expected_outputs, TestCaseResult.FAIL)   

        return self.__edu_result

from heapq import heappush, heappop

def top_k_frequent(arr, k):

    # Replace this placeholder return statement with your code
    return []

top_k_frequent_elements.yaml

testcases:
  - name: "testcase 1"
    inputs:
    - 1: [3,2,3,1,2,4,5,5,6,7,7,8,2,3,1,1,1,10,11,5,6,2,4,7,8,5,6]
    - 2: 10
    output:
    - 1: [1,2,5,3,6,7,4,8,10,11]
  - name: "testcase 2"
    inputs:
    - 1: [5,1,-1,-8,-7,8,-5,0,1,10,8,0,-4,3,-1,-1,4,-5,4,-3,0,2,2,2,4,-2,-4,8,-7,-7,2,-8,0,-8,10,8,-8,-2,-9,4,-7,6,6,-1,4,2,8,-3,5,-9,-3,6,-8,-5,5,10,2,-5,-1,-5,1,-3,7,0,8,-2,-3,-1,-5,4,7,-9,0,2,10,4,4,-4,-1,-1,6,-8,-9,-1,9,-9,3,5,1,6,-1,-2,4,2,4,-6,4,4,5,-5]
    - 2: 7
    output:
    - 1: [4,-1,2,-5,-8,0,8]
  - name: "testcase 3"
    inputs:
    - 1: [1, 1, 1, 1, 1, 1]
    - 2: 1
    output:
    - 1: [1]
  - name: "testcase 4"
    inputs:
    - 1: [6,0,1,4,9,7,-3,1,-4,-8,4,-7,-3,3,2,-3,9,5,-4,0]
    - 2: 6
    output:
    - 1: [-3,-4,0,1,4,9]
  - name: "testcase 5"
    inputs:
    - 1 : [1]
    - 2 : 1
    output: 
    - 1 : [1]

def sum(input_1, input_2):
    return input_1+input_2

def validate_inputs(self) -> None:
 """
 self.__inputs is the list of inputs that are passed as input to function
 store the result using the self.store_result(error_msg, status)
 return : void
 """
 from collections import Counter

 def is_unique_answer(nums, k):
 # Count the frequency of each element in nums
 frequency = Counter(nums)
 
 # Get the most common elements and their frequencies
 most_common = frequency.most_common()
 
 # If there are fewer than k unique elements, the answer cannot be unique
 if len(most_common) < k:
 return False
 
 # Get the frequencies of the top k most common elements
 top_k_frequencies = [count for _, count in most_common[:k]]
 
 # Check if there is a tie for the k-th most frequent element
 if len(most_common) > k:
 # If the k-th frequency is the same as the (k+1)-th frequency, it's not unique
 if most_common[k-1][1] == most_common[k][1]:
 return False
 
 return True
 
 def is_valid_1d_array(arr):
 if not isinstance(arr, list):
 return False # Not a valid 1D array if it's not a list
 
 return all(isinstance(elem, int) and not isinstance(elem, bool) for elem in arr)


 arr = self.__inputs[0]
 k = self.__inputs[1]
 
 if not is_valid_1d_array(arr):
 self.store_result ("'arr' should be a valid 1D array of integers", ValidationFunction.FAIL)
 elif not (1 <= len(arr) <= 1000):
 self.store_result ("'arr' length should be in the range[1,1000]", ValidationFunction.FAIL)
 elif not (1 <= k <= len(list(set(arr)))):
 self.store_result ("'k' should be in the range [1, number of unique elements in 'arr']", ValidationFunction.FAIL)
 elif not all(-10000 <= num <= 10000 for num in arr):
 self.store_result ("'arr[i]' should be in the range [-10000,10000]", ValidationFunction.FAIL)
 elif not is_unique_answer(arr, k):
 self.store_result ("The answer is not unique", ValidationFunction.FAIL)
 else:
 self.store_result("", ValidationFunction.PASS)

Try to solve the Top K Frequent Elements problem.


Top K Elements

Two Heaps

Conclusion

Top K Frequent Elements

Statement

Examples

Understand the problem

Figure it out!

Try it yourself