LeetCode 3002: Maximum Size of a Set After Removals Solution

Master LeetCode problem 3002 (Maximum Size of a Set After Removals), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3002. Maximum Size of a Set After Removals

Medium

Array Hash Table Greedy

Problem Explanation

Explanation:

We can approach this problem by iterating through both arrays and counting the occurrences of each element. Then, we can remove the elements with the highest occurrences until we have removed n / 2 elements from each array. Finally, we can insert the remaining elements into a set and return the size of the set.

Here is a step-by-step approach:

Iterate through both arrays and count the occurrences of each element.
Remove the elements with the highest occurrences until n / 2 elements are removed from each array.
Insert the remaining elements into a set.
Return the size of the set as the maximum possible size after removals.

Time complexity:

Counting occurrences: O(n)
Removing elements: O(n)
Inserting into a set: O(n) Overall time complexity: O(n)

Space complexity:

Storing occurrences: O(n)
Storing elements in a set: O(n) Overall space complexity: O(n)

Solution Code

class Solution {
    public int maxSizeSlices(int[] nums1, int[] nums2) {
        Map<Integer, Integer> occurrences = new HashMap<>();
        for (int num : nums1) {
            occurrences.put(num, occurrences.getOrDefault(num, 0) + 1);
        }
        for (int num : nums2) {
            occurrences.put(num, occurrences.getOrDefault(num, 0) + 1);
        }
        
        int targetRemovals = nums1.length / 2;
        Set<Integer> s = new HashSet<>();
        
        while (targetRemovals > 0) {
            int maxOccurrence = 0;
            int maxNum = 0;
            for (int num : occurrences.keySet()) {
                if (occurrences.get(num) > maxOccurrence) {
                    maxOccurrence = occurrences.get(num);
                    maxNum = num;
                }
            }
            if (maxOccurrence > 0) {
                occurrences.put(maxNum, occurrences.get(maxNum) - 1);
                if (occurrences.get(maxNum) == 0) {
                    occurrences.remove(maxNum);
                }
                targetRemovals--;
            }
        }
        
        for (int num : occurrences.keySet()) {
            s.add(num);
        }
        
        return s.size();
    }
}

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