851. Loud and Rich

Problem

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 (Java)

class Solution {
    public int[] loudAndRich(int[][] richer, int[] quiet) {
        int[] result = new int[quiet.length];
        for (int i = 0; i < quiet.length; i++) {
            result[i] = i;
        }
        for (int k = 0; k < quiet.length; k++) {
            boolean changed = false;
            for (int[] r : richer) {
                if (quiet[result[r[0]]] < quiet[result[r[1]]]) {
                    result[r[1]] = result[r[0]];
                    changed = true;
                }
            }
            if (!changed) {
                break;
            }
        }
        return result;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).