Completed Two Pointers 2

src/MergeSortedArray.java

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+/*
+Problem - Merging of 2 arrays
+Approach - We start filling nums1 from the back using two pointers from end of nums1's filled part and nums2.
+At each step, we place the larger of the two elements at the end and move pointers backward.
+After that, if anything is left in nums2, we copy it directly into nums1.
+
+ */
+import java.util.Arrays;
+
+public class MergeSortedArray {
+    public void merge(int[] nums1, int m, int[] nums2, int n) {
+        int p1 = m - 1;
+        int p2 = n - 1;// starting from end of an array because if we start from the beginning of array we will encounter second array goes unsorted
+        int index = m + n - 1;
+        while (p1 >= 0 && p2 >= 0) {
+            if (nums1[p1] > nums2[p2]) {
+                nums1[index] = nums1[p1];
+                p1--;
+            }
+            else {
+                nums1[index] = nums2[p2];
+                p2--;
+            }
+            index--;
+        }
+        while (p2 >= 0) {
+            nums1[index] = nums2[p2];
+            p2--;
+            index--;
+        }
+    }
+    public static void main(String[] args) {
+        int[] nums1 = new int[]{1, 2, 3, 0, 0, 0};
+        int[] nums2 = new int[]{2, 5, 6};
+        MergeSortedArray mergeSortedArray = new MergeSortedArray();
+        mergeSortedArray.merge(nums1, 3, nums2, 3);
+        System.out.println(Arrays.toString(nums1));
+    }
+}

src/RemoveDuplicates.java

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+/*
+Problem - Remove Duplicates from Sorted Array II
+Approach -used two pointers fast(to scan the elements) and slow(to overwrite elements)
+count duplicates and only allow each element to appear at most twice.
+If count is within limit, we place it in the right position using the slow pointer.
+Time Complexity - O(n)
+Space Complexity- O(n)
+ */
+public class RemoveDuplicates {
+    public int removeDuplicates(int[] nums) {
+        int k = 2;
+        int slow = 0, fast = 0;
+
+        int count = 0;
+
+        while(fast < nums.length){
+            if(fast != 0 && nums[fast] == nums[fast - 1]){
+                count++;//duplicate elements increment the count
+            }else{
+                count = 1; // to indicate its the start of new number
+            }
+            if(count <= k){
+                nums[slow] = nums[fast];
+                slow++;
+            }
+            fast++;
+        }
+        return slow;
+    }
+    public  static void main(String[] args) {
+        RemoveDuplicates removeDuplicates = new RemoveDuplicates();
+        System.out.println(removeDuplicates.removeDuplicates(new int[]{1,1,1,2,2,3}));
+        System.out.println(removeDuplicates.removeDuplicates(new int[]{0,0,1,1,1,1,2,3,3}));
+
+    }
+}

src/Search2DSortedMatrix.java

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+/*
+Problem - Search 2D sorted matrix II
+Approach - Start from the bottom-left corner of the matrix or we can start from top-right corner
+If the number is bigger, move up or move left; if it's smaller, move right or move down.
+Keep checking until you find the target or exit the matrix bounds.
+Time Complexity - O(m+n)
+Space Complexity - O(1)
+ */
+
+public class Search2DSortedMatrix {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
+        int m = matrix.length, n = matrix[0].length;
+        int row = 0, column = n - 1; // search using top right corner as a start point
+        while (row <m  && column >=0) {
+            if (matrix[row][column] == target) return true;
+            else if (matrix[row][column] > target) column--;
+            else row++;
+        }
+        return false;
+    }
+
+    public boolean searchMatrix2(int[][] matrix, int target) {
+        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return false;
+        int m = matrix.length, n = matrix[0].length;
+        int row = m-1, column = 0; // search using bottom left corner as a start point
+        while (row >= 0 && column <n) {
+            if (matrix[row][column] == target) return true;
+            else if (matrix[row][column] > target) row--;
+            else column++;
+        }
+        return false;
+    }
+    public static void main(String[] args) {
+        Search2DSortedMatrix s2d = new Search2DSortedMatrix();
+        int[][] values = {
+                {1,4,7,11,15},
+                {2,5,8,12,19},
+                {3,6,9,16,22},
+                {10,13,14,21,23},
+                {18,19,20,23,26}
+        };
+        System.out.println(s2d.searchMatrix(values, 20));
+        System.out.println(s2d.searchMatrix2(values, 15));
+        System.out.println(s2d.searchMatrix2(values, 30));
+    }
+}