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2856. Minimum Array Length After Pair Removals

Description

Given an integer array num sorted in non-decreasing order.

You can perform the following operation any number of times:

  • Choose two indices, i and j, where nums[i] < nums[j].
  • Then, remove the elements at indices i and j from nums. The remaining elements retain their original order, and the array is re-indexed.

Return the minimum length of nums after applying the operation zero or more times.

 

Example 1:

Input: nums = [1,2,3,4]

Output: 0

Explanation:

Example 2:

Input: nums = [1,1,2,2,3,3]

Output: 0

Explanation:

Example 3:

Input: nums = [1000000000,1000000000]

Output: 2

Explanation:

Since both numbers are equal, they cannot be removed.

Example 4:

Input: nums = [2,3,4,4,4]

Output: 1

Explanation:

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 109
  • nums is sorted in non-decreasing order.

Solutions

Solution 1: Greedy + Priority Queue (Max Heap)

We use a hash table $cnt$ to count the occurrence of each element in the array $nums$, then add each value in $cnt$ to a priority queue (max heap) $pq$. Each time we take out two elements $x$ and $y$ from $pq$, decrease their values by one. If the value after decrement is still greater than $0$, we add the decremented value back to $pq$. Each time we take out two elements from $pq$, it means we delete a pair of numbers from the array, so the length of the array decreases by $2$. When the size of $pq$ is less than $2$, we stop the deletion operation.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

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class Solution:
    def minLengthAfterRemovals(self, nums: List[int]) -> int:
        cnt = Counter(nums)
        pq = [-x for x in cnt.values()]
        heapify(pq)
        ans = len(nums)
        while len(pq) > 1:
            x, y = -heappop(pq), -heappop(pq)
            x -= 1
            y -= 1
            if x > 0:
                heappush(pq, -x)
            if y > 0:
                heappush(pq, -y)
            ans -= 2
        return ans

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class Solution {
    public int minLengthAfterRemovals(List<Integer> nums) {
        Map<Integer, Integer> cnt = new HashMap<>();
        for (int x : nums) {
            cnt.merge(x, 1, Integer::sum);
        }
        PriorityQueue<Integer> pq = new PriorityQueue<>(Comparator.reverseOrder());
        for (int x : cnt.values()) {
            pq.offer(x);
        }
        int ans = nums.size();
        while (pq.size() > 1) {
            int x = pq.poll();
            int y = pq.poll();
            x--;
            y--;
            if (x > 0) {
                pq.offer(x);
            }
            if (y > 0) {
                pq.offer(y);
            }
            ans -= 2;
        }
        return ans;
    }
}

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class Solution {
public:
    int minLengthAfterRemovals(vector<int>& nums) {
        unordered_map<int, int> cnt;
        for (int x : nums) {
            ++cnt[x];
        }
        priority_queue<int> pq;
        for (auto& [_, v] : cnt) {
            pq.push(v);
        }
        int ans = nums.size();
        while (pq.size() > 1) {
            int x = pq.top();
            pq.pop();
            int y = pq.top();
            pq.pop();
            x--;
            y--;
            if (x > 0) {
                pq.push(x);
            }
            if (y > 0) {
                pq.push(y);
            }
            ans -= 2