LeetCode in Kotlin

2012. Sum of Beauty in the Array

Medium

You are given a 0-indexed integer array nums. For each index i (1 <= i <= nums.length - 2) the beauty of nums[i] equals:

• 2, if nums[j] < nums[i] < nums[k], for all 0 <= j < i and for all i < k <= nums.length - 1.
• 1, if nums[i - 1] < nums[i] < nums[i + 1], and the previous condition is not satisfied.
• 0, if none of the previous conditions holds.

Return the sum of beauty of all nums[i] where 1 <= i <= nums.length - 2.

Example 1:

Input: nums = [1,2,3]

Output: 2

Explanation: For each index i in the range 1 <= i <= 1: - The beauty of nums[1] equals 2.

Example 2:

Input: nums = [2,4,6,4]

Output: 1

Explanation: For each index i in the range 1 <= i <= 2:

• The beauty of nums[1] equals 1.

• The beauty of nums[2] equals 0.

Example 3:

Input: nums = [3,2,1]

Output: 0

Explanation: For each index i in the range 1 <= i <= 1: - The beauty of nums[1] equals 0.

Constraints:

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

Solution

class Solution {
    fun sumOfBeauties(nums: IntArray): Int {
        val maxArr = IntArray(nums.size)
        maxArr[0] = nums[0]
        for (i in 1 until nums.size - 1) {
            maxArr[i] = Math.max(maxArr[i - 1], nums[i])
        }
        val minArr = IntArray(nums.size)
        minArr[nums.size - 1] = nums[nums.size - 1]
        for (i in nums.size - 2 downTo 0) {
            minArr[i] = Math.min(minArr[i + 1], nums[i])
        }
        var sum = 0
        for (i in 1 until nums.size - 1) {
            if (nums[i] > maxArr[i - 1] && nums[i] < minArr[i + 1]) {
                sum += 2
            } else if (nums[i] > nums[i - 1] && nums[i] < nums[i + 1]) {
                sum += 1
            }
        }
        return sum
    }
}

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