LeetCode 1086. High Five

Description

https://leetcode.com/problems/high-five/

Given a list of the scores of different students, items, where items[i] = [IDi, scorei] represents one score from a student with IDi, calculate each student’s top five average.

Return the answer as an array of pairs result, where result[j] = [IDj, topFiveAveragej] represents the student with IDj and their top five average. Sort result by IDj in increasing order.

A student’s top five average is calculated by taking the sum of their top five scores and dividing it by 5 using integer division.

Example 1:

Input: items = [[1,91],[1,92],[2,93],[2,97],[1,60],[2,77],[1,65],[1,87],[1,100],[2,100],[2,76]]
Output: [[1,87],[2,88]]
Explanation: 
The student with ID = 1 got scores 91, 92, 60, 65, 87, and 100. Their top five average is (100 + 92 + 91 + 87 + 65) / 5 = 87.
The student with ID = 2 got scores 93, 97, 77, 100, and 76. Their top five average is (100 + 97 + 93 + 77 + 76) / 5 = 88.6, but with integer division their average converts to 88.

Example 2:

Input: items = [[1,100],[7,100],[1,100],[7,100],[1,100],[7,100],[1,100],[7,100],[1,100],[7,100]]
Output: [[1,100],[7,100]]

Constraints:

• 1 <= items.length <= 1000
• items[i].length == 2
• 1 <= IDi <= 1000
• 0 <= scorei <= 100
• For each IDi, there will be at least five scores.

Explanation

Use a heap to maintain the scores in order.

Python Solution

class Solution:
    def highFive(self, items: List[List[int]]) -> List[List[int]]:
        
        scores = defaultdict(list)
        
        for item in items:            
            heapq.heappush(scores[item[0]], -item[1])
        
        results = []        
        for key, value in scores.items():
            sum = 0            
            for i in range(5):
                sum += -heappop(value)
            
            average = sum // 5
            
            heapq.heappush(results, [key, average])
            
        return results

• Time Complexity: O(N logN). Similarly pushing an item in the max heap also takes O(log N). Hence to insert all the N elements, the total time taken is O(N \log N). Iterating over the map takes O(N) time and extracting the top 5 elements is a constant time operation. Hence the overall time taken is O(N log N).
• Space Complexity: O(N).