LeetCode in Kotlin

3903. Smallest Stable Index I

Easy

You are given an integer array nums of length n and an integer k.

For each index i, define its instability score as max(nums[0..i]) - min(nums[i..n - 1]).

In other words:

• max(nums[0..i]) is the largest value among the elements from index 0 to index i.
• min(nums[i..n - 1]) is the smallest value among the elements from index i to index n - 1.

An index i is called stable if its instability score is less than or equal to k.

Return the smallest stable index. If no such index exists, return -1.

Example 1:

Input: nums = [5,0,1,4], k = 3

Output: 3

Explanation:

• At index 0: The maximum in [5] is 5, and the minimum in [5, 0, 1, 4] is 0, so the instability score is 5 - 0 = 5.
• At index 1: The maximum in [5, 0] is 5, and the minimum in [0, 1, 4] is 0, so the instability score is 5 - 0 = 5.
• At index 2: The maximum in [5, 0, 1] is 5, and the minimum in [1, 4] is 1, so the instability score is 5 - 1 = 4.
• At index 3: The maximum in [5, 0, 1, 4] is 5, and the minimum in [4] is 4, so the instability score is 5 - 4 = 1.
• This is the first index with an instability score less than or equal to k = 3. Thus, the answer is 3.

Example 2:

Input: nums = [3,2,1], k = 1

Output: -1

Explanation:

• At index 0, the instability score is 3 - 1 = 2.
• At index 1, the instability score is 3 - 1 = 2.
• At index 2, the instability score is 3 - 1 = 2.
• None of these values is less than or equal to k = 1, so the answer is -1.

Example 3:

Input: nums = [0], k = 0

Output: 0

Explanation:

At index 0, the instability score is 0 - 0 = 0, which is less than or equal to k = 0. Therefore, the answer is 0.

Constraints:

• 1 <= nums.length <= 100
• 0 <= nums[i] <= 109
• 0 <= k <= 109

Solution

class Solution {
    fun firstStableIndex(nums: IntArray, k: Int): Int {
        val n = nums.size
        val mini = IntArray(n)
        var mint = Int.MAX_VALUE
        for (i in n - 1 downTo 0) {
            if (nums[i] < mint) {
                mint = nums[i]
            }
            mini[i] = mint
        }
        var maxt = 0
        for (i in 0..<n) {
            if (nums[i] > maxt) {
                maxt = nums[i]
            }
            if (maxt - mini[i] <= k) {
                return i
            }
        }
        return -1
    }
}

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