LeetCode in Kotlin

3660. Jump Game IX

Medium

You are given an integer array nums.

From any index i, you can jump to another index j under the following rules:

• Jump to index j where j > i is allowed only if nums[j] < nums[i].
• Jump to index j where j < i is allowed only if nums[j] > nums[i].

For each index i, find the maximum value in nums that can be reached by following any sequence of valid jumps starting at i.

Return an array ans where ans[i] is the maximum value reachable starting from index i.

Example 1:

Input: nums = [2,1,3]

Output: [2,2,3]

Explanation:

• For i = 0: No jump increases the value.
• For i = 1: Jump to j = 0 as nums[j] = 2 is greater than nums[i].
• For i = 2: Since nums[2] = 3 is the maximum value in nums, no jump increases the value.

Thus, ans = [2, 2, 3].

Example 2:

Input: nums = [2,3,1]

Output: [3,3,3]

Explanation:

• For i = 0: Jump forward to j = 2 as nums[j] = 1 is less than nums[i] = 2, then from i = 2 jump to j = 1 as nums[j] = 3 is greater than nums[2].
• For i = 1: Since nums[1] = 3 is the maximum value in nums, no jump increases the value.
• For i = 2: Jump to j = 1 as nums[j] = 3 is greater than nums[2] = 1.

Thus, ans = [3, 3, 3].

Constraints:

• 1 <= nums.length <= 105
• 1 <= nums[i] <= 109

Solution

import kotlin.math.max
import kotlin.math.min

class Solution {
    fun maxValue(nums: IntArray): IntArray {
        val f = IntArray(nums.size)
        var cur = 0
        for (i in nums.indices) {
            cur = max(cur, nums[i])
            f[i] = cur
        }
        var min = nums[nums.size - 1]
        for (i in nums.size - 2 downTo 0) {
            if (f[i] > min) {
                f[i] = max(f[i], f[i + 1])
            }
            min = min(min, nums[i])
        }
        return f
    }
}

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