LeetCode in Kotlin
3502. Minimum Cost to Reach Every Position
Easy
You are given an integer array cost of size n. You are currently at position n (at the end of the line) in a line of n + 1 people (numbered from 0 to n).
You wish to move forward in the line, but each person in front of you charges a specific amount to swap places. The cost to swap with person i is given by cost[i].
You are allowed to swap places with people as follows:
- If they are in front of you, you must pay them
cost[i]to swap with them. - If they are behind you, they can swap with you for free.
Return an array answer of size n, where answer[i] is the minimum total cost to reach each position i in the line.
Example 1:
Input: cost = [5,3,4,1,3,2]
Output: [5,3,3,1,1,1]
Explanation:
We can get to each position in the following way:
i = 0. We can swap with person 0 for a cost of 5.i = 1. We can swap with person 1 for a cost of 3.i = 2. We can swap with person 1 for a cost of 3, then swap with person 2 for free.i = 3. We can swap with person 3 for a cost of 1.i = 4. We can swap with person 3 for a cost of 1, then swap with person 4 for free.i = 5. We can swap with person 3 for a cost of 1, then swap with person 5 for free.
Example 2:
Input: cost = [1,2,4,6,7]
Output: [1,1,1,1,1]
Explanation:
We can swap with person 0 for a cost of 1, then we will be able to reach any position i for free.
Constraints:
1 <= n == cost.length <= 1001 <= cost[i] <= 100
Solution
import kotlin.math.min
class Solution {
fun minCosts(cost: IntArray): IntArray {
var min = cost[0]
val ans = IntArray(cost.size)
ans[0] = min
for (i in 1..<cost.size) {
min = min(min, cost[i])
ans[i] = min
}
return ans
}
}