2512. Reward Top K Students

Description

You are given two string arrays positive_feedback and negative_feedback, containing the words denoting positive and negative feedback, respectively. Note that no word is both positive and negative.

Initially every student has 0 points. Each positive word in a feedback report increases the points of a student by 3, whereas each negative word decreases the points by 1.

You are given n feedback reports, represented by a 0-indexed string array report and a 0-indexed integer array student_id, where student_id[i] represents the ID of the student who has received the feedback report report[i]. The ID of each student is unique.

Given an integer k, return the top k students after ranking them in non-increasing order by their points. In case more than one student has the same points, the one with the lower ID ranks higher.

Example 1:

Input: positive_feedback = ["smart","brilliant","studious"], negative_feedback = ["not"], report = ["this student is studious","the student is smart"], student_id = [1,2], k = 2
Output: [1,2]
Explanation: 
Both the students have 1 positive feedback and 3 points but since student 1 has a lower ID he ranks higher.

Example 2:

Input: positive_feedback = ["smart","brilliant","studious"], negative_feedback = ["not"], report = ["this student is not studious","the student is smart"], student_id = [1,2], k = 2
Output: [2,1]
Explanation: 
- The student with ID 1 has 1 positive feedback and 1 negative feedback, so he has 3-1=2 points. 
- The student with ID 2 has 1 positive feedback, so he has 3 points. 
Since student 2 has more points, [2,1] is returned.

Constraints:

1 <= positive_feedback.length, negative_feedback.length <= 10⁴
1 <= positive_feedback[i].length, negative_feedback[j].length <= 100
Both positive_feedback[i] and negative_feedback[j] consists of lowercase English letters.
No word is present in both positive_feedback and negative_feedback.
n == report.length == student_id.length
1 <= n <= 10⁴
report[i] consists of lowercase English letters and spaces ' '.
There is a single space between consecutive words of report[i].
1 <= report[i].length <= 100
1 <= student_id[i] <= 10⁹
All the values of student_id[i] are unique.
1 <= k <= n

Solutions

Solution 1: Hash Table + Sorting

We can store the positive words in a hash table $ps$ and the negative words in a hash table $ns$.

Then, we traverse the $report$ and for each student, we store their score in an array $arr$, where each element is a tuple $(t, sid)$, where $t$ represents the student’s score and $sid$ represents the student’s ID.

Finally, we sort the array $arr$ in descending order by score, and if the scores are the same, we sort by ID in ascending order. Then, we take the IDs of the top $k$ students.

The time complexity is $O(n \times \log n + (

+ n) \times

)$, and the space complexity is $O((

) \times

+ n)$. Here, $n$ is the number of students, $

$ and $

$ are the number of positive and negative words, respectively, and $

$ is the average length of a word.

class Solution {
    public List<Integer> topStudents(String[] positive_feedback, String[] negative_feedback,
        String[] report, int[] student_id, int k) {
        Set<String> ps = new HashSet<>();
        Set<String> ns = new HashSet<>();
        for (var s : positive_feedback) {
            ps.add(s);
        }
        for (var s : negative_feedback) {
            ns.add(s);
        }
        int n = report.length;
        int[][] arr = new int[n][0];
        for (int i = 0; i < n; ++i) {
            int sid = student_id[i];
            int t = 0;
            for (var s : report[i].split(" ")) {
                if (ps.contains(s)) {
                    t += 3;
                } else if (ns.contains(s)) {
                    t -= 1;
                }
            }
            arr[i] = new int[] {t, sid};
        }
        Arrays.sort(arr, (a, b) -> a[0] == b[0] ? a[1] - b[1] : b[0] - a[0]);
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < k; ++i) {
            ans.add(arr[i][1]);
        }
        return ans;
    }
}

class Solution {
public:
    vector<int> topStudents(vector<string>& positive_feedback, vector<string>& negative_feedback, vector<string>& report, vector<int>& student_id, int k) {
        unordered_set<string> ps(positive_feedback.begin(), positive_feedback.end());
        unordered_set<string> ns(negative_feedback.begin(), negative_feedback.end());
        vector<pair<int, int>> arr;
        int n = report.size();
        for (int i = 0; i < n; ++i) {
            int sid = student_id[i];
            vector<string> ws = split(report[i], ' ');
            int t = 0;
            for (auto& w : ws) {
                if (ps.count(w)) {
                    t += 3;
                } else if (ns.count(w)) {
                    t -= 1;
                }
            }
            arr.push_back({-t, sid});
        }
        sort(arr.begin(), arr.end());
        vector<int> ans;
        for (int i = 0; i < k; ++i) {
            ans.emplace_back(arr[i].second);
        }
        return ans;
    }

    vector<string> split(string& s, char delim) {
        stringstream ss(s);
        string item;
        vector<string> res;
        while (getline(ss, item, delim)) {
            res.emplace_back(item);
        }
        return res;
    }
};

class Solution:
    def topStudents(
        self,
        positive_feedback: List[str],
        negative_feedback: List[str],
        report: List[str],
        student_id: List[int],
        k: int,
    ) -> List[int]:
        ps = set(positive_feedback)
        ns = set(negative_feedback)
        arr = []
        for sid, r in zip(student_id, report):
            t = 0
            for w in r.split():
                if w in ps:
                    t += 3
                elif w in ns:
                    t -= 1
            arr.append((t, sid))
        arr.sort(key=lambda x: (-x[0], x[1]))
        return [v[1] for v in arr[:k]]

func topStudents(positive_feedback []string, negative_feedback []string, report []string, student_id []int, k int) (ans []int) {
	ps := map[string]bool{}
	ns := map[string]bool{}
	for _, s := range positive_feedback {
		ps[s] = true
	}
	for _, s := range negative_feedback {
		ns[s] = true
	}
	arr := [][2]int{}
	for i, sid := range student_id {
		t := 0
		for _, w := range strings.Split(report[i], " ") {
			if ps[w] {
				t += 3
			} else if ns[w] {
				t -= 1
			}
		}
		arr = append(arr, [2]int{t, sid})
	}
	sort.Slice(arr, func(i, j int) bool { return arr[i][0] > arr[j][0] || (arr[i][0] == arr[j][0] && arr[i][1] < arr[j][1]) })
	for _, v := range arr[:k] {
		ans = append(ans, v[1])
	}
	return
}

function topStudents(
    positive_feedback: string[],
    negative_feedback: string[],
    report: string[],
    student_id: number[],
    k: number,
): number[] {
    const n = student_id.length;
    const map = new Map<number, number>();
    const ps = new Set(positive_feedback);
    const ns = new Set(negative_feedback);
    for (let i = 0; i < n; i++) {
        map.set(
            student_id[i],
            report[i].split(' ').reduce((r, s) => {
                if (ps.has(s)) {
                    return r + 3;
                }
                if (ns.has(s)) {
                    return r - 1;
                }
                return r;
            }, 0),
        );
    }
    return [...map.entries()]
        .sort((a, b) => {
            if (a[1] === b[1]) {
                return a[0] - b[0];
            }
            return b[1] - a[1];
        })
        .map(v => v[0])
        .slice(0, k);
}

use std::collections::{ HashMap, HashSet };
impl Solution {
    pub fn top_students(
        positive_feedback: Vec<String>,
        negative_feedback: Vec<String>,
        report: Vec<String>,
        student_id: Vec<i32>,
        k: i32
    ) -> Vec<i32> {
        let n = student_id.len();
        let ps = positive_feedback.iter().collect::<HashSet<&String>>();
        let ns = negative_feedback.iter().collect::<HashSet<&String>>();
        let mut map = HashMap::new();
        for i in 0..n {
            let id = student_id[i];
            let mut count = 0;
            for s in report[i].split(' ') {
                let s = &s.to_string();
                if ps.contains(s) {
                    count += 3;
                } else if ns.contains(s) {
                    count -= 1;
                }
            }
            map.insert(id, count);
        }
        let mut t = map.into_iter().collect::<Vec<(i32, i32)>>();
        t.sort_by(|a, b| {
            if a.1 == b.1 {
                return a.0.cmp(&b.0);
            }
            b.1.cmp(&a.1)
        });
        t.iter()
            .map(|v| v.0)
            .collect::<Vec<i32>>()
            [0..k as usize].to_vec()
    }
}

2512 - Reward Top K Students

2512. Reward Top K Students

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2512. Reward Top K Students

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