LeetCode in Kotlin

851. Loud and Rich

Medium

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

class Solution {
    fun loudAndRich(richer: Array<IntArray>, quiet: IntArray): IntArray {
        val result = IntArray(quiet.size)
        for (i in quiet.indices) {
            result[i] = i
        }
        for (k in quiet.indices) {
            var changed = false
            for (r in richer) {
                if (quiet[result[r[0]]] < quiet[result[r[1]]]) {
                    result[r[1]] = result[r[0]]
                    changed = true
                }
            }
            if (!changed) {
                break
            }
        }
        return result
    }
}

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