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# Reduce Array Size to The Half

Given an 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 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.
```

**Example 3:**

```
Input: arr = [1,9]
Output: 1
```

**Example 4:**

```
Input: arr = [1000,1000,3,7]
Output: 1
```

**Example 5:**

```
Input: arr = [1,2,3,4,5,6,7,8,9,10]
Output: 5
```

**Constraints:**

* `1 <= arr.length <= 10^5`
* `arr.length` is even.
* `1 <= arr[i] <= 10^5`

```java
class Solution {
    public int minSetSize(int[] arr) {
        Map<Integer, Integer> map = new HashMap<>();
        ArrayList<Integer>[] list = new ArrayList[arr.length + 1];

        for (int num : arr)
            map.put(num, map.getOrDefault(num, 0) + 1);

        for (int num : map.keySet()) {
            int count = map.get(num);
            if (list[count] == null) {
                list[count] = new ArrayList<Integer>();
            }
            list[count].add(num);
        }

        int steps = 0, res = 0;
        for (int i = arr.length; i > 0; i--) {
            List<Integer> cur = list[i];
            if (cur == null || cur.size() == 0)
                continue;
            for (int num : cur) {
                steps += i;
                res++;
                if (steps >= arr.length / 2)
                    return res;
            }
        }
        return arr.length;
    }
}
```


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