package com.fishercoder.solutions;

import java.util.*;

/**
 * 1342. 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
 * */
public class _1342 {
    public static class Solution1 {
        public int minSetSize(int[] arr) {
            Map<Integer, Integer> map = new HashMap<>();
            for (int i : arr) {
                map.put(i, map.getOrDefault(i, 0) + 1);
            }
            List<Integer> list = new ArrayList<>();
            for (int k : map.keySet()) {
                list.add(map.get(k));
            }
            Collections.sort(list, Collections.reverseOrder());
            int sum = 0;
            int i = 0;
            while (sum < arr.length / 2) {
                sum += list.get(i++);
            }
            return i--;
        }
    }
}

package com.fishercoder.solutions;

import java.util.Collections;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

/**
 * 1342. 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
 * */
public class _1342 {
    public static class Solution1 {
        public int minSetSize(int[] arr) {
            Map<Integer, Integer> map = new HashMap<>();
            for (int i : arr) {
                map.put(i, map.getOrDefault(i, 0) + 1);
            }
            List<Integer> list = new ArrayList<>();
            for (int k : map.keySet()) {
                list.add(map.get(k));
            }
            Collections.sort(list, Collections.reverseOrder());
            int sum = 0;
            int i = 0;
            while (sum < arr.length / 2) {
                sum += list.get(i++);
            }
            return i--;
        }
    }
}

fix build