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2826. Sorting Three Groups

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 {
    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]});
    }
};

  • 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)


  • 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);
}

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