LeetCode in Kotlin

3332. Maximum Points Tourist Can Earn

Medium

You are given two integers, n and k, along with two 2D integer arrays, stayScore and travelScore.

A tourist is visiting a country with n cities, where each city is directly connected to every other city. The tourist’s journey consists of exactly k 0-indexed days, and they can choose any city as their starting point.

Each day, the tourist has two choices:

• Stay in the current city: If the tourist stays in their current city curr during day i, they will earn stayScore[i][curr] points.
• Move to another city: If the tourist moves from their current city curr to city dest, they will earn travelScore[curr][dest] points.

Return the maximum possible points the tourist can earn.

Example 1:

Input: n = 2, k = 1, stayScore = [[2,3]], travelScore = [[0,2],[1,0]]

Output: 3

Explanation:

The tourist earns the maximum number of points by starting in city 1 and staying in that city.

Example 2:

Input: n = 3, k = 2, stayScore = [[3,4,2],[2,1,2]], travelScore = [[0,2,1],[2,0,4],[3,2,0]]

Output: 8

Explanation:

The tourist earns the maximum number of points by starting in city 1, staying in that city on day 0, and traveling to city 2 on day 1.

Constraints:

• 1 <= n <= 200
• 1 <= k <= 200
• n == travelScore.length == travelScore[i].length == stayScore[i].length
• k == stayScore.length
• 1 <= stayScore[i][j] <= 100
• 0 <= travelScore[i][j] <= 100
• travelScore[i][i] == 0

Solution

import kotlin.math.max

class Solution {
    fun maxScore(n: Int, k: Int, stayScores: Array<IntArray>, travelScores: Array<IntArray>): Int {
        // dp[day][city]
        val dp = Array<IntArray>(k + 1) { IntArray(n) }
        var result = 0
        for (day in k - 1 downTo 0) {
            for (city in 0 until n) {
                val stayScore = stayScores[day][city] + dp[day + 1][city]
                var travelScore = 0
                for (nextCity in 0 until n) {
                    val nextScore = travelScores[city][nextCity] + dp[day + 1][nextCity]
                    travelScore = max(nextScore, travelScore)
                }
                dp[day][city] = max(stayScore, travelScore)
                result = max(dp[day][city], result)
            }
        }
        return result
    }
}

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