add a solution for 1338

fishercoder1534 · fishercoder1534 · commit ef904bef901f · 2021-07-06T09:55:21.000-07:00
diff --git a/src/main/java/com/fishercoder/solutions/_1338.java b/src/main/java/com/fishercoder/solutions/_1338.java
@@ -6,41 +6,6 @@
 import java.util.List;
 import java.util.Map;
 
-/**
- * 1338. 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 _1338 {
     public static class Solution1 {
         public int minSetSize(int[] arr) {
@@ -61,4 +26,28 @@ public int minSetSize(int[] arr) {
             return i--;
         }
     }
+
+    public static class Solution2 {
+        public int minSetSize(int[] arr) {
+            Map<Integer, Integer> map = new HashMap<>();
+            for (int num : arr) {
+                map.put(num, map.getOrDefault(num, 0) + 1);
+            }
+            List<int[]> list = new ArrayList<>();
+            for (int key : map.keySet()) {
+                list.add(new int[]{map.get(key), key});
+            }
+            Collections.sort(list, (a, b) -> b[0] - a[0]);
+            int minSet = 0;
+            int count = 0;
+            for (int[] pair : list) {
+                count += pair[0];
+                minSet++;
+                if (count >= arr.length / 2) {
+                    return minSet;
+                }
+            }
+            return minSet;
+        }
+    }
 }
