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2869. Minimum Operations to Collect Elements

Description

You are given an array nums of positive integers and an integer k.

In one operation, you can remove the last element of the array and add it to your collection.

Return the minimum number of operations needed to collect elements 1, 2, ..., k.

 

Example 1:

Input: nums = [3,1,5,4,2], k = 2
Output: 4
Explanation: After 4 operations, we collect elements 2, 4, 5, and 1, in this order. Our collection contains elements 1 and 2. Hence, the answer is 4.

Example 2:

Input: nums = [3,1,5,4,2], k = 5
Output: 5
Explanation: After 5 operations, we collect elements 2, 4, 5, 1, and 3, in this order. Our collection contains elements 1 through 5. Hence, the answer is 5.

Example 3:

Input: nums = [3,2,5,3,1], k = 3
Output: 4
Explanation: After 4 operations, we collect elements 1, 3, 5, and 2, in this order. Our collection contains elements 1 through 3. Hence, the answer is 4.

 

Constraints:

  • 1 <= nums.length <= 50
  • 1 <= nums[i] <= nums.length
  • 1 <= k <= nums.length
  • The input is generated such that you can collect elements 1, 2, ..., k.

Solutions

Solution 1: Traverse in Reverse Order

We can traverse the array in reverse order. For each element encountered during the traversal that is less than or equal to $k$ and has not been added to the set yet, we add it to the set until the set contains elements from $1$ to $k$.

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

  • class Solution {
    public int minOperations(List<Integer> nums, int k) {
        boolean[] isAdded = new boolean[k];
        int n = nums.size();
        int count = 0;
        for (int i = n - 1;; i--) {
            if (nums.get(i) > k || isAdded[nums.get(i) - 1]) {
                continue;
            }
            isAdded[nums.get(i) - 1] = true;
            count++;
            if (count == k) {
                return n - i;
            }
        }
    }
}


  • class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        int n = nums.size();
        vector<bool> isAdded(n);
        int count = 0;
        for (int i = n - 1;; --i) {
            if (nums[i] > k || isAdded[nums[i] - 1]) {
                continue;
            }
            isAdded[nums[i] - 1] = true;
            if (++count == k) {
                return n - i;
            }
        }
    }
};

  • class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        is_added = [False] * k
        count = 0
        n = len(nums)
        for i in range(n - 1, -1, -1):
            if nums[i] > k or is_added[nums[i] - 1]:
                continue
            is_added[nums[i] - 1] = True
            count += 1
            if count == k:
                return n - i


  • func minOperations(nums []int, k int) int {
	isAdded := make([]bool, k)
	count := 0
	n := len(nums)
	for i := n - 1; ; i-- {
		if nums[i] > k || isAdded[nums[i]-1] {
			continue
		}
		isAdded[nums[i]-1] = true
		count++
		if count == k {
			return n - i
		}
	}
}

  • function minOperations(nums: number[], k: number): number {
    const n = nums.length;
    const isAdded = Array(k).fill(false);
    let count = 0;
    for (let i = n - 1; ; --i) {
        if (nums[i] > k || isAdded[nums[i] - 1]) {
            continue;
        }
        isAdded[nums[i] - 1] = true;
        ++count;
        if (count === k) {
            return n - i;
        }
    }
}

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