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Loud and Rich

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.

Solution Explanation:

This problem can be modeled as a directed graph where each person is a node, and a directed edge from person b to person a indicates that a is richer than b. The goal is to find the quietest person reachable from each person in the graph. This can be efficiently solved using Depth-First Search (DFS).

Approach:

  1. Graph Construction: Create an adjacency list representation of the graph. The richer array provides the edges. If [a, b] is in richer, add a directed edge from b to a (because a is richer than b).

  2. Depth-First Search (DFS): Perform a DFS starting from each person. During the DFS:

    • Maintain an array ans to store the quietest person reachable from each person. Initially, all entries are -1.
    • For each person i, initially set ans[i] = i.
    • Recursively explore neighbors (richer people) using DFS.
    • If a neighbor j has a quieter person than the current quietest person (ans[i]), update ans[i] to the index of the quieter person.

  3. Return Result: The ans array will hold the solution. ans[x] is the index of the quietest person who has at least as much money as person x.

Time Complexity Analysis:

  • Graph construction takes O(R), where R is the length of the richer array.
  • DFS visits each node at most once. The total number of nodes is N (number of people).
  • The work done at each node is proportional to its out-degree (number of richer people). The sum of out-degrees across all nodes is at most R (the number of edges).
  • Therefore, the overall time complexity of the DFS is O(N + R).

Space Complexity Analysis:

  • The adjacency list g uses O(N + R) space in the worst case.
  • The ans array uses O(N) space.
  • The recursive call stack in DFS uses O(N) space in the worst case (deepest recursion path).
  • Therefore, the overall space complexity is O(N + R).

Code Explanation (Python):

from collections import defaultdict
 
class Solution:
    def loudAndRich(self, richer: List[List[int]], quiet: List[int]) -> List[int]:
        # Create adjacency list
        g = defaultdict(list) 
        for a, b in richer:
            g[b].append(a)  # Edge from b to a (a is richer than b)
 
        n = len(quiet)
        ans = [-1] * n  # Initialize answer array
 
        def dfs(i: int):
            if ans[i] != -1:  # Already processed
                return
            ans[i] = i  # Initially, quietest person is themselves
            for j in g[i]:  # Explore richer neighbors
                dfs(j)
                if quiet[ans[j]] < quiet[ans[i]]:
                    ans[i] = ans[j]  # Update if a quieter person is found
 
        for i in range(n):
            dfs(i)
        return ans

The other language solutions follow the same algorithmic approach; only the syntax and data structures differ slightly. For example, Java uses ArrayList for the adjacency list, while C++ uses vector<vector<int>>. The core logic of DFS and the update rule remains consistent across all implementations.