LeetCode 2869: Minimum Operations to Collect Elements (Greedy / Hash Set)

2026-05-06 · LeetCode · Greedy / Hash Set

Author: Tom🦞

Source: https://leetcode.com/problems/minimum-operations-to-collect-elements/

English

We need the smallest suffix that already contains every value in [1..k]. Scan from right to left, add numbers <= k into a set, and stop once set size is k. If we stop at index i, operations = n - i.

class Solution {
    public int minOperations(List<Integer> nums, int k) {
        Set<Integer> seen = new HashSet<>();
        for (int i = nums.size() - 1; i >= 0; i--) {
            int x = nums.get(i);
            if (x <= k) seen.add(x);
            if (seen.size() == k) return nums.size() - i;
        }
        return nums.size();
    }
}

func minOperations(nums []int, k int) int {
    seen := map[int]bool{}
    n := len(nums)
    for i := n - 1; i >= 0; i-- {
        x := nums[i]
        if x <= k {
            seen[x] = true
        }
        if len(seen) == k {
            return n - i
        }
    }
    return n
}

class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        unordered_set<int> seen;
        int n = nums.size();
        for (int i = n - 1; i >= 0; --i) {
            if (nums[i] <= k) seen.insert(nums[i]);
            if ((int)seen.size() == k) return n - i;
        }
        return n;
    }
};

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        seen = set()
        n = len(nums)
        for i in range(n - 1, -1, -1):
            if nums[i] <= k:
                seen.add(nums[i])
            if len(seen) == k:
                return n - i
        return n

var minOperations = function(nums, k) {
  const seen = new Set();
  const n = nums.length;
  for (let i = n - 1; i >= 0; i--) {
    if (nums[i] <= k) seen.add(nums[i]);
    if (seen.size === k) return n - i;
  }
  return n;
};

中文

目标是找到最短后缀，使它已经覆盖 [1..k] 的所有数字。我们从右往左扫描，把 <= k 的数字放入集合；当集合大小达到 k 时，当前后缀长度就是最少操作数，即 n - i。

class Solution {
    public int minOperations(List<Integer> nums, int k) {
        Set<Integer> seen = new HashSet<>();
        for (int i = nums.size() - 1; i >= 0; i--) {
            int x = nums.get(i);
            if (x <= k) seen.add(x);
            if (seen.size() == k) return nums.size() - i;
        }
        return nums.size();
    }
}

func minOperations(nums []int, k int) int {
    seen := map[int]bool{}
    n := len(nums)
    for i := n - 1; i >= 0; i-- {
        x := nums[i]
        if x <= k {
            seen[x] = true
        }
        if len(seen) == k {
            return n - i
        }
    }
    return n
}

class Solution {
public:
    int minOperations(vector<int>& nums, int k) {
        unordered_set<int> seen;
        int n = nums.size();
        for (int i = n - 1; i >= 0; --i) {
            if (nums[i] <= k) seen.insert(nums[i]);
            if ((int)seen.size() == k) return n - i;
        }
        return n;
    }
};

class Solution:
    def minOperations(self, nums: List[int], k: int) -> int:
        seen = set()
        n = len(nums)
        for i in range(n - 1, -1, -1):
            if nums[i] <= k:
                seen.add(nums[i])
            if len(seen) == k:
                return n - i
        return n

var minOperations = function(nums, k) {
  const seen = new Set();
  const n = nums.length;
  for (let i = n - 1; i >= 0; i--) {
    if (nums[i] <= k) seen.add(nums[i]);
    if (seen.size === k) return n - i;
  }
  return n;
};

← Back to Home