Completed Two-Pointers-2

Leetcode240.java

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+// TC - O(m*n)
+// SC - O(1)
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+        int i = m - 1; // row
+        int j = 0; // col
+        // Starting from the left bottom most index so that everything towards right is
+        // in increasing order and everything upwards is in decreasing order. This will
+        // help in eliminating one section while searching.
+        while (i < m && j < n && i >= 0 && j >= 0) { // Keep the row and col values within range
+            if (matrix[i][j] == target)
+                return true;
+            if (matrix[i][j] < target) { //
+                j++;
+            } else {
+                i--;
+            }
+        }
+        return false;
+    }
+}

Leetcode80.java

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+// TC - O(n)
+// SC - O(1)
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        int slow = 0;
+        int fast = 0;
+        int count = 0;
+        int n = nums.length;
+        int k =2;
+        while(fast< n) {
+
+            if(fast != 0 && nums[fast]== nums[fast-1]) {
+                count+=1;
+            } else {
+                count =1;
+            }
+
+            if(count<=k) {
+                nums[slow] = nums[fast];
+                slow++;
+            }
+            fast+=1;
+
+
+        }
+        return slow;
+    }
+}

Leetcode88.java

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+// TC - O(m+n)
+// SC - O(1)
+class Solution {
+    public void merge(int[] nums1, int m, int[] nums2, int n) {
+        int first = m - 1;
+        int second = n - 1;
+        int i = m + n - 1;
+        while (first >= 0 && second >= 0) { // compare the highest element in each array and add them to the right most
+                                            // element on nums1 array
+            if (nums1[first] > nums2[second]) {
+                nums1[i] = nums1[first];
+                first--;
+            } else {
+                nums1[i] = nums2[second];
+                second--;
+            }
+            i -= 1;
+        }
+        while (second >= 0) { // if the nums2 array still has items left in it after the above iteration, we
+                              // can directly copy them into nums1 array via the i'th pointer. We don't need
+                              // to do this for num1 array elements since they are already in place incase
+                              // they have been left out.
+            nums1[i] = nums2[second];
+            second--;
+            i -= 1;
+        }
+    }
+}