Complete Two-Pointer-2

Problem37.java

@@ -0,0 +1,30 @@
+// Time Complexity : O(m+n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+class Solution {
+    public void merge(int[] nums1, int m, int[] nums2, int n) {
+        int p1 = m-1; int p2 = n-1;
+        int idx = m+n-1;
+
+        while(p1 >= 0 && p2 >= 0) {
+            if (nums2[p2] > nums1[p1]) {
+                nums1[idx] = nums2[p2];
+                p2--;
+            } else {
+                nums1[idx] = nums1[p1];
+                p1--;
+            }
+            idx--;
+        }
+
+        while(p2 >= 0) {
+            nums1[idx] = nums2[p2];
+            p2--;
+            idx--;
+        }
+    }
+}

Problem38.java

@@ -0,0 +1,23 @@
+// Time Complexity : O(m+n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+
+        int r = 0; int c = n-1;
+
+        while (r < m && c >= 0) {
+            if (matrix[r][c] == target) return true;
+            else if (matrix[r][c] > target) c--;
+            else r++;
+        }
+
+        return false;
+    }
+}

Problem39.java

@@ -0,0 +1,32 @@
+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No
+
+
+// Your code here along with comments explaining your approach
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        int k = 2;
+        int slow = 0; int fast = 0;
+
+        int count = 0;
+
+        while (fast < nums.length) {
+            if (fast != 0 && nums[fast] == nums[fast-1]) {
+                count++;
+            } else {
+                count = 1;
+            }
+
+            if (count <= k) {
+                nums[slow] = nums[fast];
+                slow++;
+            }
+
+            fast++;
+        }
+
+        return slow;
+    }
+}