Leetcode 2869: Minimum Operations to Collect Elements

Input: The input consists of an array of positive integers nums and an integer k. The goal is to collect elements from 1 to k in the minimum number of operations.

Example: nums = [6, 3, 8, 5, 7, 1, 4, 2], k = 3

Constraints:

• 1 <= nums.length <= 50

• 1 <= nums[i] <= nums.length

• 1 <= k <= nums.length

Output: Return the minimum number of operations needed to collect all elements from 1 to k in the collection.

Example: For input nums = [6, 3, 8, 5, 7, 1, 4, 2], k = 3, the output is 6.

Constraints:

Goal: The goal is to collect all numbers from 1 to k in the least number of operations.

Steps:

• Start removing elements from the end of the array one by one.

• Keep track of the elements you collect in your collection.

• Stop when all elements from 1 to k are collected.

Goal: The solution must handle arrays with up to 50 elements and work for all possible values of k.

Steps:

• 1 <= nums.length <= 50

• 1 <= nums[i] <= nums.length

• 1 <= k <= nums.length

Assumptions:

• The input array nums contains all the elements from 1 to k at least once.

• Input: For input nums = [6, 3, 8, 5, 7, 1, 4, 2], k = 3, the output is 6.

• Explanation: After 6 operations, elements 1, 2, and 3 will be collected. Thus, the total number of operations is 6.

Approach: We will process the array from the last element towards the first, collecting elements until we have all the elements from 1 to k in our collection.

Observations:

• The challenge is to determine how many operations it takes to collect the elements from 1 to k, starting from the end of the array.

• One efficient way to approach this is by using a set or array to keep track of the elements that have been collected.

Steps:

• Initialize a set to keep track of collected elements.

• Iterate over the array from the end, collecting elements.

• Stop when all elements from 1 to k are in the set.

Empty Inputs:

• The input array will always contain at least one element.

Large Inputs:

• The solution must efficiently handle arrays of size up to 50 elements.

Special Values:

• If all elements are in order or all elements from 1 to k are already in the array, fewer operations will be needed.

Constraints:

• The solution must respect the constraints on the number of elements and the values of k.

int minOperations(vector<int>& nums, int k) {
    int cnt[51] = {}, i = nums.size() - 1;
    for (int found = 0; found < k; --i)
        if (nums[i] <= k)
            found += cnt[nums[i]]++ == 0;
    return nums.size() - i - 1;
}

1 : Code Declaration

int minOperations(vector<int>& nums, int k) {

Function declaration that takes a vector of integers and an integer k as input, returning an integer result.

2 : Variable Initialization

    int cnt[51] = {}, i = nums.size() - 1;

Initialize an array to count occurrences of numbers up to 50 and set the index variable i to the last element of nums.

3 : Loop Setup

    for (int found = 0; found < k; --i)

Start a for loop that iterates backwards through the array, with a stopping condition based on found elements.

4 : Condition Check

        if (nums[i] <= k)

Condition that checks if the current element is less than or equal to k.

5 : Frequency Count

            found += cnt[nums[i]]++ == 0;

Increment the 'found' variable if the current number has not been seen before, while updating its count in the cnt array.

6 : Result Calculation

    return nums.size() - i - 1;

Return the difference between the size of the nums vector and the current index, minus one, which represents the result of the operation.

Best Case: O(n)

Average Case: O(n)

Worst Case: O(n)

Description: The time complexity is O(n) because we only need to iterate through the array once.

Best Case: O(k)

Worst Case: O(k)

Description: The space complexity is O(k) for storing the collected elements in the set.

Leetcode 2869: Minimum Operations to Collect Elements

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