# Solution For Reduce Array Size To The Half

Problem Statement:
Given an array A of integers, we must reduce the array size to half. We are allowed to remove any set of elements from the array such that the new length of the array becomes half of its original length (rounded down).

Return the minimum size of the set of numbers that need to be removed.

Solution:
To solve this problem, we need to find the frequency of each number in the array A. After that, we need to sort the frequencies of each number in descending order. We will keep removing the numbers with the highest frequencies until we reach half of the array size or we have removed all the elements from the array. The size of the set of numbers removed will be our answer.

Algorithm:

1. Create an empty dictionary freq to store the frequency of each number in A.
2. Traverse the array A and update the frequency of each number in the freq dictionary.
3. Sort the dictionary by values (i.e., frequencies) in descending order.
4. Traverse the sorted dictionary and remove the elements with the highest frequencies until we reach half of the array size or we have removed all the elements from the array.
5. Return the size of the set of numbers removed.

Code:

def minSetSize(A: List[int]) -> int:
freq = {}
for num in A:
freq[num] = freq.get(num, 0) + 1
sorted_freq = sorted(freq.items(), key=lambda x: x[1], reverse=True)
target_size = len(A) // 2
num_removed = 0
for num, count in sorted_freq:
target_size -= count
num_removed += 1
if target_size <= 0:
break
return num_removed

Time Complexity:
The time complexity of this algorithm is O(n log n), where n is the length of the array A. The time complexity of sorting the dictionary is O(n log n) in the worst case.

Space Complexity:
The space complexity of this algorithm is O(n), where n is the length of the array A. We are storing the frequencies of the numbers in a dictionary freq.

Example:
Input: A = [3,3,3,3,5,5,5,2,2,7] Output: 2
Explanation: We can remove [3,3] to get the new array [3,5,5,5,2,2,7] with length half of A.

## Step by Step Implementation For Reduce Array Size To The Half

```public int minSetSize(int[] arr) {
// create a map to store the frequency of each number
Map map = new HashMap<>();
for (int i : arr) {
map.put(i, map.getOrDefault(i, 0) + 1);
}
// create an array of frequencies
int[] freq = new int[map.size()];
int index = 0;
for (int i : map.values()) {
freq[index++] = i;
}
// sort the array in descending order
Arrays.sort(freq);
int count = 0;
int sum = 0;
// add up the frequencies until we reach half of the length of the array
for (int i = freq.length - 1; i >= 0; i--) {
sum += freq[i];
count++;
if (sum >= arr.length / 2) {
break;
}
}
return count;
}```
```def reduceArraySizeToTheHalf(nums):

# create a dictionary to store the count of each number
num_count = {}

# store the count of each number in the dictionary
for i in nums:
if i in num_count:
num_count[i] += 1
else:
num_count[i] = 1

# sort the dictionary in descending order by value
sorted_num_count = sorted(num_count.items(), key = lambda x: x[1], reverse = True)

# keep track of the sum of the counts
count_sum = 0

# iterate through the sorted dictionary
for key, value in sorted_num_count:

# add the count to the sum
count_sum += value

# if the sum is greater than or equal to half the size of the array, return the count
if count_sum >= len(nums) / 2:
return value```
```var reduceArraySizeToTheHalf = function(arr) {
// create a map to store the frequency of each element in the array
let map = new Map();
for (let i = 0; i < arr.length; i++) {
let key = arr[i];
// if the key already exists in the map, increment the value
if (map.has(key)) {
map.set(key, map.get(key) + 1);
// otherwise, set the value to 1
} else {
map.set(key, 1);
}
}
// sort the map in descending order by value
let sortedMap = new Map([...map.entries()].sort((a, b) => b[1] - a[1]));
// create a new array to store the elements that we will keep
let newArr = [];
// iterate through the map
for (let [key, value] of sortedMap) {
// if the length of the new array is less than half the length of the original array
// and the value is greater than 1
if (newArr.length < arr.length / 2 && value > 1) {
// push the key into the new array
newArr.push(key);
}
}
// return the new array
return newArr;
};```
```int minSetSize(vector& arr) {
int n = arr.size();

// Hash to store frequencies
unordered_map hash;

// Traverse array elements and
// count frequencies
for (int i = 0; i < n; i++)
hash[arr[i]]++;

// Traverse the map and return
// the first element whose
// frequency is greater than n/2
for (auto it = hash.begin(); it != hash.end(); it++)
if (it->second > n/2)
return it->second;
}```
```public class Solution {
public int MinSetSize(int[] arr) {
// create a hashmap to store the number of occurrences for each number
var map = new Dictionary();
foreach (var num in arr) {
if (!map.ContainsKey(num)) {
}
else {
map[num]++;
}
}

// sort the hashmap by value in descending order
var list = map.ToList();
list.Sort((a, b) => b.Value.CompareTo(a.Value));

// keep track of the sum of values and the number of items in the list
int sum = 0;
int count = 0;

// iterate through the list and add the values until the sum is greater than or equal to half the length of the array
foreach (var item in list) {
sum += item.Value;
count++;
if (sum >= arr.Length / 2) {
break;
}
}

return count;
}
}```
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