LeetCode in Kotlin

2826. Sorting Three Groups

Medium

You are given a 0-indexed integer array nums of length n.

The numbers from 0 to n - 1 are divided into three groups numbered from 1 to 3, where number i belongs to group nums[i]. Notice that some groups may be empty.

You are allowed to perform this operation any number of times:

• Pick number x and change its group. More formally, change nums[x] to any number from 1 to 3.

A new array res is constructed using the following procedure:

1. Sort the numbers in each group independently.
2. Append the elements of groups 1, 2, and 3 to res in this order.

Array nums is called a beautiful array if the constructed array res is sorted in non-decreasing order.

Return the minimum number of operations to make nums a beautiful array.

Example 1:

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

Output: 3

Explanation: It’s optimal to perform three operations:

1. change nums[0] to 1.
2. change nums[2] to 1.
3. change nums[3] to 1.

After performing the operations and sorting the numbers in each group, group 1 becomes equal to [0,1,2,3,4] and group 2 and group 3 become empty. Hence, res is equal to [0,1,2,3,4] which is sorted in non-decreasing order.

It can be proven that there is no valid sequence of less than three operations.

Example 2:

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

Output: 2

Explanation: It’s optimal to perform two operations:

1. change nums[1] to 1.
2. change nums[2] to 1.

After performing the operations and sorting the numbers in each group, group 1 becomes equal to [0,1,2,3], group 2 becomes empty, and group 3 becomes equal to [4,5]. Hence, res is equal to [0,1,2,3,4,5] which is sorted in non-decreasing order.

It can be proven that there is no valid sequence of less than two operations.

Example 3:

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

Output: 0

Explanation: It’s optimal to not perform operations.

After sorting the numbers in each group, group 1 becomes empty, group 2 becomes equal to [0,1,2,3] and group 3 becomes equal to [4,5]. Hence, res is equal to [0,1,2,3,4,5] which is sorted in non-decreasing order.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 3

Solution

import kotlin.math.max

class Solution {
    fun minimumOperations(nums: List<Int>): Int {
        val n = nums.size
        val arr = IntArray(3)
        var max = 0
        for (num in nums) {
            var locMax = 0
            val value = num
            for (j in 0 until value) {
                locMax = max(locMax, arr[j])
            }
            locMax++
            arr[value - 1] = locMax
            if (locMax > max) {
                max = locMax
            }
        }
        return n - max
    }
}

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