Question

2512. Reward Top K Students

中文文档

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 $O(n \times \log n + (|ps| + |ns| + n) \times |s|)$, and the space complexity is $O((|ps|+|ns|) \times |s| + n)$. Here, $n$ is the number of students, $|ps|$ and $|ns|$ are the number of positive and negative words, respectively, and $|s|$ is the average length of a word.

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]]

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;
  }
};

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()
  }
}