Reduce Array Size To The Half - Leetcode Solution

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.

1