Minimum Number of Operations to Make Array Empty

class Solution {
public:
    int calculateOperations(vector<int>& numbers) {
        unordered_map<int, int> frequency;
        for (int number : numbers) {
            frequency[number]++;
        }
        int totalOps = 0;
        for (auto [key, count]: frequency) {
            if (count == 1) {
                return -1;
            }
            totalOps += ceil((double)(count) / 3);
        }
        return totalOps;
    }
};

In the provided C++ solution, the goal is to determine the minimum number of operations needed to make an input array empty. An operation is defined based on the frequency of elements in the array.

Here’s how the solution approaches the problem:

• An unordered map named frequency is used to store the count of each unique element in the array.
• Traverse the array numbers and populate the frequency map with each element and its occurrence count.
• Initialize a variable totalOps to store the total operations required. Start iterating over the frequency map:
• For each element, if its count is exactly one, the function directly returns -1. This indicates that it is impossible to remove this element in any grouping of more than one.
• If the count is more than one, calculate the necessary operations by taking the ceiling value of the count divided by 3. This implies grouping elements in sets of three when possible.
• Accumulate the number of required operations in totalOps.

Return totalOps which represents the total number of operations needed to potentially make the array empty based on specified rules. This solution handles different cases effectively, ensuring that an immediate return occurs if an element cannot be grouped optimally.

class Solution {
    public int minOps(int[] elements) {
        var frequency = new HashMap<Integer, Integer>();
        for (int element: elements) {
            frequency.put(element, frequency.getOrDefault(element, 0) + 1);
        }
        int result = 0;
        for (int freq: frequency.values()) {
            if (freq == 1) {
                return -1;
            }
            result += Math.ceil((double) freq / 3);
        }
        return result;
    }
}

This Java solution targets the problem of finding the minimum number of operations required to make an array empty, where each operation can remove up to three identical elements. The solution uses a hashmap to keep track of the frequency of each element in the input array.

• Start by creating a HashMap to tally the occurrence of each element.
• Traverse the input array, incrementing the count of each value in the HashMap.
• Initialize a result variable to accumulate the total number of operations.
• Iterate over the values in the HashMap. If any value occurs only once, immediately return -1 as it's not possible to perform the operation on single occurrences.
• For other frequencies, calculate how many operations are needed by dividing the frequency by three and using Math.ceil to round up to the nearest integer. This rounds up because partial operations account as full operations in this context.
• Accumulate these values into the result variable.
• Return the result, which represents the minimum number of operations required to empty the array.

var countOperations = function(numbers) {
    const freq = new Map();
    for (const number of numbers) {
        freq.set(number, (freq.get(number) || 0) + 1);
    }
    let result = 0;
    for (const count of freq.values()) {
        if (count === 1) {
            return -1;
        }
        result += Math.ceil(count / 3);
    }
    return result;
};

This JavaScript solution tackles the problem of calculating the minimum number of operations required to make an array empty using a specific set of rules. The solution follows these steps:

1. Initialize freq, a JavaScript Map object, to store the frequency of each number in the array.

2. Iterate over the array numbers, counting the occurrence of each number. Add or update this count in the freq map.

3. Initialize a variable result to store the total number of operations required. Start with a value of 0.

4. Iterate over the values of freq. For each frequency value:

   • If the frequency is 1, it's impossible to make that part of the array empty using the given operation, so return -1.
   • Otherwise, add to result the ceiling value of the frequency divided by 3.

5. Return the value of result, which represents the total number of operations needed to make the array empty.

The approach guarantees efficient frequency counting and calculation of operations needed using Map data structure for quick access and updates. The solution incorporates a termination condition when single occurrences are found, immediately indicating infeasibility by returning -1.

class Solution:
    def calculateMinOps(self, numbers: List[int]) -> int:
        number_count = Counter(numbers)
        result = 0
        for count in number_count.values():
            if count == 1: 
                return -1
            result += ceil(count / 3)
        return result

The given Python3 code defines a method calculateMinOps that takes an input list of integers and determines the minimum number of operations required to make the array empty using a specific set of operation rules. The task is to perform operations in such a way that in every operation, you can remove up to three occurrences of the same number. The code employs the following approach:

• Utilizes Counter from the collections module to count occurrences of each element in the list.
• Initializes a variable result to store the cumulative minimum operations required.
• Iterates through each unique number's count in the dictionary provided by Counter:
  • If a number appears only once (count of 1), immediately returns -1 as it cannot be removed in one move considering the operation constraints.
  • Otherwise, updates the result by adding the required number of operations to remove that particular number, calculated by ceil(count / 3), ensuring that even if a remainder exists after dividing by 3 (indicating less than 3 elements but more than 0), an additional operation is correctly added.
• Finally, returns the result which is the total minimum number of operations needed to empty the array.

This solution tackles the problem efficiently by counting and calculating minimum operations in one pass through the data, handling edge cases where immediate termination is necessary.