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发布于 2024-06-17 01:03:33 字数 6131 浏览 0 评论 0 收藏 0

851. Loud and Rich

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Description

There is a group of n people labeled from 0 to n - 1 where each person has a different amount of money and a different level of quietness.

You are given an array richer where richer[i] = [ai, bi] indicates that ai has more money than bi and an integer array quiet where quiet[i] is the quietness of the ith person. All the given data in richer are logically correct (i.e., the data will not lead you to a situation where x is richer than y and y is richer than x at the same time).

Return _an integer array _answer_ where _answer[x] = y_ if _y_ is the least quiet person (that is, the person _y_ with the smallest value of _quiet[y]_) among all people who definitely have equal to or more money than the person _x.

 

Example 1:

Input: richer = [[1,0],[2,1],[3,1],[3,7],[4,3],[5,3],[6,3]], quiet = [3,2,5,4,6,1,7,0]
Output: [5,5,2,5,4,5,6,7]
Explanation: 
answer[0] = 5.
Person 5 has more money than 3, which has more money than 1, which has more money than 0.
The only person who is quieter (has lower quiet[x]) is person 7, but it is not clear if they have more money than person 0.
answer[7] = 7.
Among all people that definitely have equal to or more money than person 7 (which could be persons 3, 4, 5, 6, or 7), the person who is the quietest (has lower quiet[x]) is person 7.
The other answers can be filled out with similar reasoning.

Example 2:

Input: richer = [], quiet = [0]
Output: [0]

 

Constraints:

  • n == quiet.length
  • 1 <= n <= 500
  • 0 <= quiet[i] < n
  • All the values of quiet are unique.
  • 0 <= richer.length <= n * (n - 1) / 2
  • 0 <= ai, bi < n
  • ai != bi
  • All the pairs of richer are unique.
  • The observations in richer are all logically consistent.

Solutions

Solution 1

class Solution:
  def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
    def dfs(i: int):
      if ans[i] != -1:
        return
      ans[i] = i
      for j in g[i]:
        dfs(j)
        if quiet[ans[j]] < quiet[ans[i]]:
          ans[i] = ans[j]

    g = defaultdict(list)
    for a, b in richer:
      g[b].append(a)
    n = len(quiet)
    ans = [-1] * n
    for i in range(n):
      dfs(i)
    return ans
class Solution {
  private List<Integer>[] g;
  private int n;
  private int[] quiet;
  private int[] ans;

  public int[] loudAndRich(int[][] richer, int[] quiet) {
    n = quiet.length;
    this.quiet = quiet;
    g = new List[n];
    ans = new int[n];
    Arrays.fill(ans, -1);
    Arrays.setAll(g, k -> new ArrayList<>());
    for (var r : richer) {
      g[r[1]].add(r[0]);
    }
    for (int i = 0; i < n; ++i) {
      dfs(i);
    }
    return ans;
  }

  private void dfs(int i) {
    if (ans[i] != -1) {
      return;
    }
    ans[i] = i;
    for (int j : g[i]) {
      dfs(j);
      if (quiet[ans[j]] < quiet[ans[i]]) {
        ans[i] = ans[j];
      }
    }
  }
}
class Solution {
public:
  vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
    int n = quiet.size();
    vector<vector<int>> g(n);
    for (auto& r : richer) {
      g[r[1]].push_back(r[0]);
    }
    vector<int> ans(n, -1);
    function<void(int)> dfs = [&](int i) {
      if (ans[i] != -1) {
        return;
      }
      ans[i] = i;
      for (int j : g[i]) {
        dfs(j);
        if (quiet[ans[j]] < quiet[ans[i]]) {
          ans[i] = ans[j];
        }
      }
    };
    for (int i = 0; i < n; ++i) {
      dfs(i);
    }
    return ans;
  }
};
func loudAndRich(richer [][]int, quiet []int) []int {
  n := len(quiet)
  g := make([][]int, n)
  ans := make([]int, n)
  for i := range g {
    ans[i] = -1
  }
  for _, r := range richer {
    a, b := r[0], r[1]
    g[b] = append(g[b], a)
  }
  var dfs func(int)
  dfs = func(i int) {
    if ans[i] != -1 {
      return
    }
    ans[i] = i
    for _, j := range g[i] {
      dfs(j)
      if quiet[ans[j]] < quiet[ans[i]] {
        ans[i] = ans[j]
      }
    }
  }
  for i := range ans {
    dfs(i)
  }
  return ans
}
function loudAndRich(richer: number[][], quiet: number[]): number[] {
  const n = quiet.length;
  const g: number[][] = new Array(n).fill(0).map(() => []);
  for (const [a, b] of richer) {
    g[b].push(a);
  }
  const ans: number[] = new Array(n).fill(-1);
  const dfs = (i: number) => {
    if (ans[i] != -1) {
      return ans;
    }
    ans[i] = i;
    for (const j of g[i]) {
      dfs(j);
      if (quiet[ans[j]] < quiet[ans[i]]) {
        ans[i] = ans[j];
      }
    }
  };
  for (let i = 0; i < n; ++i) {
    dfs(i);
  }
  return ans;
}

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