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.length1 <= n <= 5000 <= quiet[i] < nquiet are unique.0 <= richer.length <= n * (n - 1) / 20 <= ai, bi < nai != biricher are unique.richer are all logically consistent.class Solution {
public int[] loudAndRich(int[][] richer, int[] quiet) {
int n = quiet.length;
int[] memo = new int[n], indegree = new int[n];
Arrays.fill(memo, -1);
List<Integer>[] adj = new List[n];
for (int i = 0; i < n; i++)
adj[i] = new ArrayList<>();
for (int[] edge : richer){
adj[edge[1]].add(edge[0]);
indegree[edge[0]]++;
}
for (int i = 0; i < n; i++)
if(indegree[i]==0)
dfs(adj, quiet, memo, i);
return memo;
}
private int dfs(List<Integer>[] adj, int[] quiet, int[] memo, int node) {
if(memo[node]!=-1)
return memo[node];
int quietest = node;
for (int nbr : adj[node]) {
int quietestNbr = dfs(adj, quiet, memo, nbr);
if (quiet[quietestNbr] < quiet[quietest])
quietest = quietestNbr;
}
return memo[node] = quietest;
}
}class Solution {
public int[] loudAndRich(int[][] richer, int[] quiet) {
int n = quiet.length;
int[] memo = new int[n];
Arrays.fill(memo, -1);
List<Integer>[] adj = new List[n];
for (int i = 0; i < n; i++)
adj[i] = new ArrayList<>();
for (int[] edge : richer)
adj[edge[1]].add(edge[0]);
for (int i = 0; i < n; i++)
dfs(adj, quiet, memo, i);
return memo;
}
private int dfs(List<Integer>[] adj, int[] quiet, int[] memo, int node) {
if(memo[node]!=-1)
return memo[node];
int quietest = node;
for (int nbr : adj[node]) {
int quietestNbr = dfs(adj, quiet, memo, nbr);
if (quiet[quietestNbr] < quiet[quietest])
quietest = quietestNbr;
}
return memo[node] = quietest;
}
}