851. Loud and Rich

• Link: https://leetcode.com/problems/loud-and-rich/

Question

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|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.

Approach 1: Optimal

Intuition

• Edge u -> v = u richer than v.
• For each person i, we want the quietest person among all X that can reach i via -> (including i).
• Process people from richest to poorest (topo order).
• When a richer node knows its best quiet candidate, it pushes that candidate down to poorer nodes.

Algorithm

1. Build directed graph adj[richerPerson] -> poorerPerson.
2. Compute indegree for each node.
3. Initialize ans[i] = i (best candidate for each is themselves).
4. Push all indegree == 0 nodes (richest) into a queue.
5. While queue not empty:
   • Pop current.
   • For each neighbor in adj[current]:
     • If quiet[ans[current]] < quiet[ans[neighbor]], set ans[neighbor] = ans[current].
     • Decrement indegree[neighbor]; if it hits 0, push neighbor.
6. Return ans.

Code

class Solution {
public:
    vector<int> loudAndRich(vector<vector<int>>& richer, vector<int>& quiet) {
        int n = quiet.size();

        // Build directed graph: richerPerson -> poorerPerson
        vector<vector<int>> adj(n, vector<int>());
        for (auto& edge: richer) {
            int richerPerson = edge[0];
            int poorerPerson = edge[1];
            adj[richerPerson].push_back(poorerPerson);
        }

        // Compute indegree: number of richer people above each person
        vector<int> indegree(n, 0);
        for (int person = 0 ; person < n ; person++) {
            for (int neighbor: adj[person]) {
                indegree[neighbor]++;
            }
        }

        // ans[i] = quietest person among all people richer-or-equal to i
        // start by assuming each person is their own best candidate
        vector<int> ans(n, 0);
        for (int person = 0 ; person < n ; person++) {
            ans[person] = person;
        }

        // Topological BFS starting from richest people (indegree 0)
        queue<int> q;
        for (int person = 0 ; person < n ; person++) {
            if (indegree[person] == 0) {
                q.push(person);
            }
        }

        // Propagate quietest richer candidate along edges
        while (!q.empty()) {
            int current = q.front();
            q.pop();

            // current -> neighbor (neighbor is poorer)
            for (int neighbor: adj[current]) {

                // If the quietest richer-or-equal person for current
                // is quieter than the quietest known for neighbor,
                // update neighbor's best candidate
                if (quiet[ans[current]] < quiet[ans[neighbor]]) {
                    ans[neighbor] = ans[current];
                }

                // Maintain topo order
                indegree[neighbor]--;
                if (indegree[neighbor] == 0) {
                    q.push(neighbor);
                }
            }
        }

        return ans;
    }
};

Complexity Analysis

• Time Complexity: $O(N+M)$
• Space Complexity: $O(N+M)$

Approach 2: Better

Intuition

Algorithm

Code

Complexity Analysis

• Time Complexity: $O()$
• Space Complexity: $O()$

Approach 3: Optimal

Intuition

Algorithm

Code

Complexity Analysis

• Time Complexity: $O()$
• Space Complexity: $O()$