LeetCode 1086. High Five Solution in Java, C++, Python & More | Explanation + Code

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1086. High Five

Description

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.

Solutions

Solution 1

PythonJavaC++GoTypeScript
class Solution: def highFive(self, items: List[List[int]]) -> List[List[int]]: d = defaultdict(list) m = 0 for i, x in items: d[i].append(x) m = max(m, i) ans = [] for i in range(1, m + 1): if xs := d[i]: avg = sum(nlargest(5, xs)) // 5 ans.append([i, avg]) return ans(code-box)

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