algoadvance

  1. 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 loudness. You are given an array richer where richer[i] = [ai, bi] indicates that person ai has more money than person bi and an integer array quiet where quiet[i] is the quietness of the i-th person. All the arrays are 0-indexed.

Return an integer array answer where answer[x] = y indicates that the person y that the person x would choose to be the quietest among all people who definitely have equal to or more money than person x.

Clarifying Questions

  1. Input Size:
    • What are the constraints on the number of people n and the size of the richer array?
  2. Edge Cases:
    • Should we consider situations where ricker array is empty?
    • Is it possible to have cycles in the richer relationships?
  3. Output:
    • Should the output always have the same length as the number of people n?
    • Can there be multiple valid outputs or is the result unique for given inputs?
  4. Performance:
    • What kind of time complexity is acceptable for this problem?

Strategy

  1. Graph Representation:
    • Represent the richer relationships as a directed graph where an edge from node a to node b indicates that a has more money than b.
  2. DFS for Traversal:
    • Use Depth-First Search (DFS) to determine the quietest person for each node in the graph.
  3. Memoization:
    • Use memoization to store the quietest person for each node to avoid redundant computations.
  4. Result Construction:
    • Construct the result array by iterating over each person and determining the quietest person who has equal to or more money than them.

Code

def loudAndRich(richer, quiet):
    n = len(quiet)
    
    # Build the adjacency list for the graph
    graph = [[] for _ in range(n)]
    for u, v in richer:
        graph[v].append(u)
    
    # Result array and memoization
    answer = [-1] * n
    
    def dfs(node):
        # If answer[node] has already been computed, return it.
        if answer[node] != -1:
            return answer[node]
        
        # Start with the assumption that the node itself is the quietest
        answer[node] = node
        for neighbor in graph[node]:
            candidate = dfs(neighbor)
            if quiet[candidate] < quiet[answer[node]]:
                answer[node] = candidate
        
        return answer[node]
    
    # Compute for each person
    for i in range(n):
        dfs(i)
    
    return answer

Time Complexity

This approach efficiently computes the required answers and leverages memoization to avoid redundant calculations, making it optimal for large inputs.

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