LeetCode in Kotlin

2659. Make Array Empty

Hard

You are given an integer array nums containing distinct numbers, and you can perform the following operations until the array is empty:

• If the first element has the smallest value, remove it
• Otherwise, put the first element at the end of the array.

Return an integer denoting the number of operations it takes to make nums empty.

Example 1:

Input: nums = [3,4,-1]

Output: 5

Operation   Array
1           [4, -1, 3]
2           [-1, 3, 4]
3           [3, 4]
4           [4]
5           []

Example 2:

Input: nums = [1,2,4,3]

Output: 5

Operation   Array
1           [2, 4, 3]
2           [4, 3]
3           [3, 4]
4           [4]
5           []

Example 3:

Input: nums = [1,2,3]

Output: 3

Operation   Array
1           [2, 3]
2           [3]
3           []

Constraints:

• 1 <= nums.length <= 105
• -109 <= nums[i] <= 109
• All values in nums are distinct.

Solution

class Solution {
    fun countOperationsToEmptyArray(nums: IntArray): Long {
        val sortNums = Array(nums.size) { IntArray(2) }
        for (i in nums.indices) {
            sortNums[i][0] = nums[i]
            sortNums[i][1] = i
        }
        sortNums.sortBy { it[0] }
        var res = 0L + nums.size
        for (i in 1..sortNums.lastIndex) {
            if (sortNums[i - 1][1] > sortNums[i][1]) {
                res += nums.size - i
            }
        }
        return res
    }
}

This site is open source. Improve this page.