LeetCode 2870. Minimum Number of Operations to Make Array Empty

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Intuition

  • Mathematics problem.
  • Count the number of occurrences for all numbers in nums.
  • Specify the operation based on the number of occurrences.

Approach

The number of operations required to remove a specific number of occurrences can be found as follows:

Number of Occurrences Number of Operations to Remove this Number
1 -1 (no solution)
2 / 3 1
4 / 5 / 6 2
7 / 8 / 9 3
10 / 11 / 12 4
… …
n-2 / n-1 / n $\lceil n/3 \rceil$
  1. Create an empty map count.
  2. Initialise ans to 0.
  3. Count the number of occurrences for every number in nums, and store it in count.
  4. Use the table above to calculate the number of operations needed.
  5. Return ans.

Pseudocode of minOperations():

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minOperations():
	Input nums an array of integers
	Output minimum number of operations to make array empty
	
	count = empty map
	ans = 0
	for all num in nums do
		if num not in count then
			count[num] = 1
		else
			count[num] += 1
		end if
	end for
	for all (key, value) in map do
		if value == 1 then
			return -1
		end if
		ans += celi(value / 3)
	end for

Complexity

Time complexity: $O(n)$.

  • $n$ iterations to count the number of occurances in the array.

Space complexity: $O(n)$.

  • count will take a space of $n$ if all the numbers in the array are unique.

Code

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class Solution {
public:
  int minOperations(vector<int>& nums) {
    unordered_map<int, int> count;
    for (int num : nums) {
      if (count.find(num) == count.end()) {
        count[num] = 1;
      } else {
        count[num]++;
      }
    }
    int ans = 0;
    for (auto it = count.begin(); it != count.end(); it++) {
      if (it->second == 1) {
        return -1;
      }
      ans += ceil((double)(it->second) / 3);
    }
    return ans;
  }
};