diff --git a/java/088_Merge_Sorted_Array.java b/java/088_Merge_Sorted_Array.java
new file mode 100644
index 0000000..95c7c8e
--- /dev/null
+++ b/java/088_Merge_Sorted_Array.java
@@ -0,0 +1,29 @@
+class Solution {
+    //we are being given two int arrays and two int variables that state the number of elements in each array, respectively
+    public void merge(int[] nums1, int m, int[] nums2, int n) {
+        //I am using a counter so that I can iterate through the second array inside the for loop
+        int counter = 0;
+        //We know that nums1 is of the size n + m, so we can add all the elements to the array and sort them later
+        //For loop adds all values of nums2 to the end of nums1
+        for(int i=m; i<nums1.length; i++){
+            nums1[i] = nums2[counter++];
+        }
+        //Now that all of the elements are in the array, we can begin the sort process
+        //Though we could use the Arrays.sort function to quickly sort the array, I wanted to implement a sorting algorithm myself
+        //I am implementing what's called a "Bubble Sorting Algorithm" to sort this array
+        //This should work best in our scenario as we don't have to work with a large list of numbers
+        for(int i=0; i<nums1.length; i++){
+            for(int j=0; j<nums1.length-i-1; j++){
+                if(nums1[j] > nums1[j+1]){
+                    int temp = nums1[j+1];
+                    nums1[j+1] = nums1[j];
+                    nums1[j] = temp;
+                }
+            }
+        }
+        //The following function simply prints out everything that is contained within our num1 array
+        for(int i=0; i<nums1.length; i++){
+            System.out.println(nums1[i]);
+        }
+    }
+}
\ No newline at end of file
diff --git a/java/442_Find_All_Duplicates_in_an_Array.java b/java/442_Find_All_Duplicates_in_an_Array.java
new file mode 100644
index 0000000..b12d6f5
--- /dev/null
+++ b/java/442_Find_All_Duplicates_in_an_Array.java
@@ -0,0 +1,17 @@
+class Solution {
+    public List<Integer> findDuplicates(int[] nums) {
+        int n = nums.length;
+
+        List<Integer> ans = new ArrayList<>();
+
+        for(int i=0;i<n;i++) {
+            int possibleVal = Math.abs(nums[i]);
+
+            if(nums[possibleVal - 1] < 0) {
+                ans.add(possibleVal);
+            }
+            nums[possibleVal - 1] = -1 * nums[possibleVal - 1];
+        }
+        return ans;
+    }
+}
diff --git a/python/007_Reverse_Integer.py b/python/007_Reverse_Integer.py
index 9f3f20f..19aeba8 100644
--- a/python/007_Reverse_Integer.py
+++ b/python/007_Reverse_Integer.py
@@ -1,17 +1,34 @@
 class Solution:
     def reverse(self, x):
         # https://leetcode.com/problems/reverse-integer/
-        flag = True if x < 0 else False
-        if flag:
+#        flag = True if x < 0 else False
+#       if flag:
+#           x = -x
+#       x = str(x)[::-1]
+
+#       if flag:
+#           x = "-" + x
+
+#       value = 2 ** 31
+#       x = int(x)
+#       if -value <= x < value:
+#           return x
+#       return 0
+        
+        is_neg = False
+        if x < 0:
             x = -x
-        x = str(x)[::-1]
+            is_neg = True
 
-        if flag:
-            x = "-" + x
+        res = 0
+        while x > 0:
+            res *= 10
+            res += x % 10
+            x //= 10
+        if is_neg:
+            res = -res
 
-        value = 2 ** 31
-        x = int(x)
-        if -value <= x < value:
-            return x
-        return 0
-    
+        if res < -2**31 or res > 2**31-1:
+            return 0
+        return res
+    
\ No newline at end of file