2826. Sorting Three Groups

Problem

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:

• Sort the numbers in each group independently.
• 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 (Java)

class Solution {
    public int minimumOperations(List<Integer> A) {
        int n = A.size(), dp[] = new int[]{n, n, n, n};
        for (int a : A) {
            dp[a]--;
            dp[2] = Math.min(dp[2], dp[1]);
            dp[3] = Math.min(dp[3], dp[2]);
        }
        return dp[3];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).