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

Python3JavaC++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

class Solution {
    public int[][] highFive(int[][] items) {
        int size = 0;
        PriorityQueue[] s = new PriorityQueue[101];
        int n = 5;
        for (int[] item : items) {
            int i = item[0], score = item[1];
            if (s[i] == null) {
                ++size;
                s[i] = new PriorityQueue<>(n);
            }
            s[i].offer(score);
            if (s[i].size() > n) {
                s[i].poll();
            }
        }
        int[][] res = new int[size][2];
        int j = 0;
        for (int i = 0; i < 101; ++i) {
            if (s[i] == null) {
                continue;
            }
            int avg = sum(s[i]) / n;
            res[j][0] = i;
            res[j++][1] = avg;
        }
        return res;
    }

    private int sum(PriorityQueue<Integer> q) {
        int s = 0;
        while (!q.isEmpty()) {
            s += q.poll();
        }
        return s;
    }
}

class Solution {
public:
    vector<vector<int>> highFive(vector<vector<int>>& items) {
        vector<int> d[1001];
        for (auto& item : items) {
            int i = item[0], x = item[1];
            d[i].push_back(x);
        }
        vector<vector<int>> ans;
        for (int i = 1; i <= 1000; ++i) {
            if (!d[i].empty()) {
                sort(d[i].begin(), d[i].end(), greater<int>());
                int s = 0;
                for (int j = 0; j < 5; ++j) {
                    s += d[i][j];
                }
                ans.push_back({i, s / 5});
            }
        }
        return ans;
    }
};

func highFive(items [][]int) (ans [][]int) {
    d := make([][]int, 1001)
    for _, item := range items {
        i, x := item[0], item[1]
        d[i] = append(d[i], x)
    }
    for i := 1; i <= 1000; i++ {
        if len(d[i]) > 0 {
            sort.Ints(d[i])
            s := 0
            for j := len(d[i]) - 1; j >= len(d[i])-5; j-- {
                s += d[i][j]
            }
            ans = append(ans, []int{i, s / 5})
        }
    }
    return ans
}

function highFive(items: number[][]): number[][] {
    const d: number[][] = Array(1001)
        .fill(0)
        .map(() => Array(0));
    for (const [i, x] of items) {
        d[i].push(x);
    }
    const ans: number[][] = [];
    for (let i = 1; i <= 1000; ++i) {
        if (d[i].length > 0) {
            d[i].sort((a, b) => b - a);
            const s = d[i].slice(0, 5).reduce((a, b) => a + b);
            ans.push([i, Math.floor(s / 5)]);
        }
    }
    return ans;
}

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