Leetcode 2826: Sorting Three Groups

Input: The input consists of an array `nums` of integers, where each element is one of the values 1, 2, or 3.

Example: For example, `nums = [3, 1, 2, 3, 1]`

Constraints:

• 1 <= nums.length <= 100

• 1 <= nums[i] <= 3

Output: Return the minimum number of operations required to make the array `nums` non-decreasing.

Example: For `nums = [2, 1, 3, 2, 3]`, the output is `2`.

Constraints:

Goal: The goal is to find the minimum number of deletions to make the array non-decreasing.

Steps:

• Track how many operations are needed to make the array non-decreasing for each possible value (1, 2, 3).

• Iterate through the array and compare each element with its previous value to determine if it should be removed.

Goal: The input will be a non-empty array of integers.

Steps:

• The length of `nums` will be between 1 and 100.

• Each element in `nums` will be 1, 2, or 3.

Assumptions:

• The array contains only 1, 2, or 3 as its elements.

• The array length is manageable (<= 100).

• Input: For `nums = [3, 1, 2, 3, 1]`

• Explanation: By removing the first, third, and fourth elements, we can make the array non-decreasing: `[1, 2, 3]`.

Approach: The solution involves calculating the minimum number of operations required to make the array non-decreasing by tracking the number of deletions for each element in `nums`.

Observations:

• The elements in `nums` are restricted to 1, 2, and 3, which simplifies the problem.

• We need to track the number of removals for each value and minimize the number of deletions.

Steps:

• Initialize three variables to track the number of deletions for each value (1, 2, and 3).

• Iterate over the array and for each element, update the corresponding variable to reflect the minimum deletions required.

Empty Inputs:

• If the array is empty (which won't happen according to constraints), no deletions are needed.

Large Inputs:

• Handle inputs with the maximum length of 100 efficiently.

Special Values:

• If all elements are already in non-decreasing order, no deletions are required.

Constraints:

• Ensure the solution works within time limits for the maximum input size.

int minimumOperations(vector<int>& A) {
    int a = 0, b = 0, c = 0;
    for (int x: A) {
        a += x != 1;
        b = min(a, b + (x != 2));
        c = min(b, c + (x != 3));
    }
    return c;
}

1 : Function Definition

int minimumOperations(vector<int>& A) {

The function 'minimumOperations' is defined, taking a vector of integers 'A' as input and returning the minimum number of operations needed to transform 'A' into the sequence [1, 2, 3].

2 : Variable Initialization

    int a = 0, b = 0, c = 0;

Variables 'a', 'b', and 'c' are initialized to zero. These variables keep track of the number of operations required to transform elements into 1, 2, and 3, respectively.

3 : Loop Start

    for (int x: A) {

A for loop is started, iterating through each element 'x' of the vector 'A'. This loop is used to process each element in 'A' and update the operation counts.

4 : Update a

        a += x != 1;

The variable 'a' is incremented if the current element 'x' is not equal to 1, indicating that an operation is needed to transform this element to 1.

5 : Update b

        b = min(a, b + (x != 2));

The variable 'b' is updated to the minimum of the current 'a' and 'b + 1' if the current element 'x' is not equal to 2. This step tracks the operations needed to transform elements into 2.

6 : Update c

        c = min(b, c + (x != 3));

The variable 'c' is updated to the minimum of the current 'b' and 'c + 1' if the current element 'x' is not equal to 3. This step tracks the operations needed to transform elements into 3.

7 : Return Statement

    return c;

The function returns the value of 'c', which represents the minimum number of operations needed to transform the entire sequence into [1, 2, 3].

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n), where n is the length of the `nums` array.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is O(1), as only a few variables are used.

Leetcode 2826: Sorting Three Groups

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