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Formatted question description: https://leetcode.ca/all/1338.html

# 1338. Reduce Array Size to The Half (Medium)

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

Related Topics:
Array, Greedy

## Solution 1.

• class Solution {
public int minSetSize(int[] arr) {
Map<Integer, Integer> numberCountMap = new HashMap<Integer, Integer>();
for (int num : arr) {
int count = numberCountMap.getOrDefault(num, 0);
count++;
numberCountMap.put(num, count);
}
Map<Integer, Integer> countNumbersMap = new TreeMap<Integer, Integer>(new Comparator<Integer>() {
public int compare(Integer key1, Integer key2) {
return key2 - key1;
}
});
Set<Integer> numberSet = numberCountMap.keySet();
for (int num : numberSet) {
int count = numberCountMap.getOrDefault(num, 0);
int nums = countNumbersMap.getOrDefault(count, 0);
nums++;
countNumbersMap.put(count, nums);
}
int halfLength = arr.length / 2;
Iterator<Integer> iterator = countNumbersMap.keySet().iterator();
int size = 0;
int removeCount = 0;
while (iterator.hasNext() && removeCount < halfLength) {
int count = iterator.next();
int nums = countNumbersMap.getOrDefault(count, 0);
for (int i = 0; i < nums; i++) {
removeCount += count;
size++;
if (removeCount >= halfLength)
return size;
}
}
return size;
}
}

############

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;
}
}
}
}
}

• // OJ: https://leetcode.com/problems/reduce-array-size-to-the-half/
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
int minSetSize(vector<int>& arr) {
unordered_map<int, int> m;
for (int n : arr) m[n]++;
vector<int> v;
for (auto p : m) v.push_back(p.second);
sort(v.begin(), v.end(), greater<int>());
int ans = 0, cnt = 0;
for (int n : v) {
cnt += n;
++ans;
if (cnt >= (arr.size() + 1) / 2) break;
}
return ans;
}
};

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


• func minSetSize(arr []int) (ans int) {
mx := 0
for _, x := range arr {
mx = max(mx, x)
}
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
}
}
}
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

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