Question

2826. Sorting Three Groups

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Description

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

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum number of operations to turn the first $i$ numbers into a beautiful array, and the $i$th number is changed to $j+1$. The answer is $\min(f[n][0], f[n][1], f[n][2])$.

We can enumerate all cases where the $i$th number is changed to $j+1$, and then take the minimum value. Here, we can use a rolling array to optimize the space complexity.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

class Solution:
  def minimumOperations(self, nums: List[int]) -> int:
    f = g = h = 0
    for x in nums:
      ff = gg = hh = 0
      if x == 1:
        ff = f
        gg = min(f, g) + 1
        hh = min(f, g, h) + 1
      elif x == 2:
        ff = f + 1
        gg = min(f, g)
        hh = min(f, g, h) + 1
      else:
        ff = f + 1
        gg = min(f, g) + 1
        hh = min(f, g, h)
      f, g, h = ff, gg, hh
    return min(f, g, h)

class Solution {
  public int minimumOperations(List<Integer> nums) {
    int[] f = new int[3];
    for (int x : nums) {
      int[] g = new int[3];
      if (x == 1) {
        g[0] = f[0];
        g[1] = Math.min(f[0], f[1]) + 1;
        g[2] = Math.min(f[0], Math.min(f[1], f[2])) + 1;
      } else if (x == 2) {
        g[0] = f[0] + 1;
        g[1] = Math.min(f[0], f[1]);
        g[2] = Math.min(f[0], Math.min(f[1], f[2])) + 1;
      } else {
        g[0] = f[0] + 1;
        g[1] = Math.min(f[0], f[1]) + 1;
        g[2] = Math.min(f[0], Math.min(f[1], f[2]));
      }
      f = g;
    }
    return Math.min(f[0], Math.min(f[1], f[2]));
  }
}

class Solution {
public:
  int minimumOperations(vector<int>& nums) {
    vector<int> f(3);
    for (int x : nums) {
      vector<int> g(3);
      if (x == 1) {
        g[0] = f[0];
        g[1] = min(f[0], f[1]) + 1;
        g[2] = min({f[0], f[1], f[2]}) + 1;
      } else if (x == 2) {
        g[0] = f[0] + 1;
        g[1] = min(f[0], f[1]);
        g[2] = min(f[0], min(f[1], f[2])) + 1;
      } else {
        g[0] = f[0] + 1;
        g[1] = min(f[0], f[1]) + 1;
        g[2] = min(f[0], min(f[1], f[2]));
      }
      f = move(g);
    }
    return min({f[0], f[1], f[2]});
  }
};

func minimumOperations(nums []int) int {
  f := make([]int, 3)
  for _, x := range nums {
    g := make([]int, 3)
    if x == 1 {
      g[0] = f[0]
      g[1] = min(f[0], f[1]) + 1
      g[2] = min(f[0], min(f[1], f[2])) + 1
    } else if x == 2 {
      g[0] = f[0] + 1
      g[1] = min(f[0], f[1])
      g[2] = min(f[0], min(f[1], f[2])) + 1
    } else {
      g[0] = f[0] + 1
      g[1] = min(f[0], f[1]) + 1
      g[2] = min(f[0], min(f[1], f[2]))
    }
    f = g
  }
  return min(f[0], min(f[1], f[2]))
}

function minimumOperations(nums: number[]): number {
  let f: number[] = new Array(3).fill(0);
  for (const x of nums) {
    const g: number[] = new Array(3).fill(0);
    if (x === 1) {
      g[0] = f[0];
      g[1] = Math.min(f[0], f[1]) + 1;
      g[2] = Math.min(f[0], Math.min(f[1], f[2])) + 1;
    } else if (x === 2) {
      g[0] = f[0] + 1;
      g[1] = Math.min(f[0], f[1]);
      g[2] = Math.min(f[0], Math.min(f[1], f[2])) + 1;
    } else {
      g[0] = f[0] + 1;
      g[1] = Math.min(f[0], f[1]) + 1;
      g[2] = Math.min(f[0], Math.min(f[1], f[2]));
    }
    f = g;
  }
  return Math.min(...f);
}

Solution 2

class Solution:
  def minimumOperations(self, nums: List[int]) -> int:
    f = [0] * 3
    for x in nums:
      g = [0] * 3
      if x == 1:
        g[0] = f[0]
        g[1] = min(f[:2]) + 1
        g[2] = min(f) + 1
      elif x == 2:
        g[0] = f[0] + 1
        g[1] = min(f[:2])
        g[2] = min(f) + 1
      else:
        g[0] = f[0] + 1
        g[1] = min(f[:2]) + 1
        g[2] = min(f)
      f = g
    return min(f)