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1338. Reduce Array Size to The Half

Description

You are given an integer array arr. You can choose a set of integers and remove all the occurrences of these integers in the array.

Return the minimum size of the set so that at least half of the integers of the array are removed.

 

Example 1:

Input: arr = [3,3,3,3,5,5,5,2,2,7]
Output: 2
Explanation: Choosing {3,7} will make the new array [5,5,5,2,2] which has size 5 (i.e equal to half of the size of the old array).
Possible sets of size 2 are {3,5},{3,2},{5,2}.
Choosing set {2,7} is not possible as it will make the new array [3,3,3,3,5,5,5] which has a size greater than half of the size of the old array.

Example 2:

Input: arr = [7,7,7,7,7,7]
Output: 1
Explanation: The only possible set you can choose is {7}. This will make the new array empty.

 

Constraints:

  • 2 <= arr.length <= 105
  • arr.length is even.
  • 1 <= arr[i] <= 105

Solutions

Solution 1: Counting + Sorting

We can use a hash table or an array $\textit{cnt}$ to count the occurrences of each number in the array $\textit{arr}$. Then, we sort the numbers in $\textit{cnt}$ in descending order. We traverse $\textit{cnt}$ from largest to smallest, adding the current number $x$ to the answer and adding $x$ to $m$. If $m \geq \frac{n}{2}$, we return the answer.

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

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class Solution:
    def minSetSize(self, arr: List[int]) -> int:
        cnt = Counter(arr)
        ans = m = 0
        for _, v in cnt.most_common():
            m += v
            ans += 1
            if m * 2 >= len(arr):
                break
        return ans

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class Solution {
    public int minSetSize(int[] arr) {
        int mx = 0;
        for (int x : arr) {
            mx = Math.max(mx, x);
        }
        int[] cnt = new int[mx + 1];
        for (int x : arr) {
            ++cnt[x];
        }
        Arrays.sort(cnt);
        int ans = 0;
        int m = 0;
        for (int i = mx;; --i) {
            if (cnt[i] > 0) {
                m += cnt[i];
                ++ans;
                if (m * 2 >= arr.length) {
                    return ans;
                }
            }
        }
    }
}

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class Solution {
public:
    int minSetSize(vector<int>& arr) {
        int mx = ranges::max(arr);
        int cnt[mx + 1];
        memset(cnt, 0, sizeof(cnt));
        for (int& x : arr) {
            ++cnt[x];
        }
        sort(cnt, cnt + mx + 1, greater<int>());
        int ans = 0;
        int m = 0;
        for (int& x : cnt) {
            if (x) {
                m += x;
                ++ans;
                if (m * 2 >= arr.size()) {
                    break;
                }
            }
        }
        return ans;
    }
};

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func minSetSize(arr []int) (ans int) {
    mx := slices.Max(arr)
    cnt := make([]int, mx+1)
    for _, x := range arr {
        cnt[x]++
    }
    sort.Ints(cnt)
    for i, m := mx, 0; ; i-- {
        if cnt[i] > 0 {
            m += cnt[i]
            ans++
            if m >= len(arr)/2 {
                return
            }
        }
    }
}

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function minSetSize(arr: number[]): number {
    const cnt = new Map<number, number>();
    for (const v of arr) {
        cnt.set(v, (cnt.get(v) ?? 0) + 1);
    }
    let [ans, m] = [0, 0];
    for (const v of Array.from(cnt.values()).sort((a, b) => b - a)) {
        m += v;
        ++ans;
        if (m * 2 >= arr.length) {
            break;
        }
    }
    return ans;
}

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