diff --git a/merge-intervals/hyeri0903.java b/merge-intervals/hyeri0903.java
new file mode 100644
index 0000000000..e14d30afad
--- /dev/null
+++ b/merge-intervals/hyeri0903.java
@@ -0,0 +1,40 @@
+class Solution {
+    public int[][] merge(int[][] intervals) {
+        /**
+        1.prob: overlapping 되는 곳을 merge 해서 return
+        2.constraints
+        - len(intervalse) min = 1, max= 10000
+        - start, end value min = 0, max = 10000
+        3.solution
+        - sorting array -> merge overlapping values
+        time: O(n log n)
+         */
+
+       int n = intervals.length;
+       Arrays.sort(intervals, (a,b) -> Integer.compare(a[0], b[0]));
+
+       List<int[]> answer = new ArrayList<>();
+       int start = intervals[0][0];
+       int end = intervals[0][1];
+
+       for(int i = 1; i < n; i++) {
+        int nextStart = intervals[i][0];
+        int nextEnd = intervals[i][1];
+
+        //overlap
+        if(end >= nextStart) {
+            end = Math.max(end, nextEnd);
+        } else {
+            answer.add(new int[]{start, end});
+
+            start = nextStart;
+            end = nextEnd;
+
+        }
+       }
+
+        //last intervals
+        answer.add(new int[]{start, end});
+        return answer.toArray(new int[answer.size()][]);
+    }
+}
diff --git a/missing-number/hyeri0903.java b/missing-number/hyeri0903.java
new file mode 100644
index 0000000000..14f7efa416
--- /dev/null
+++ b/missing-number/hyeri0903.java
@@ -0,0 +1,30 @@
+class Solution {
+    public int missingNumber(int[] nums) {
+        /**
+        1.distinc number 로 이루어진 0~n 까지의 배열 중 missing number return
+        2.constraints
+        - n 최소 = 1, 최대 = 10000
+        - 배열길이 min=0, max=n
+        3.solutions
+        - sorting 하고 for문으로 missing num check. time: O(n log n), space: O(1)
+        - 전체 sum - 실제 sum
+         */
+
+         Arrays.sort(nums);
+         int n = nums.length;
+         int answer = 0;
+         int i = 0;
+
+         for(i = 0; i < n; i++) {
+            if(nums[i] != i) {
+                answer = i;
+                return answer;
+            }
+         }
+         if(i == n) {
+            answer = n;
+            return answer;
+         }
+         return answer;
+    }
+}