Two-Pointers-2 Solutions

DuplicatesFromSortedArray2.java

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+// Time Complexity : O(n).
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Approach :  We have two pointers, one used to swap and other to iterate through the array. We increase the counter if two elements are
+// equal adn based on the condition we swap in the array so that the final resultant array is achieved.
+
+class Solution {
+    public int removeDuplicates(int[] nums) {
+        int count = 1;
+        int j = 1;//pinter for swapping result elements
+        for(int i=1;i<nums.length;i++){
+            if(nums[i] == nums[i-1]){ //if two consecutive elements are equal
+                count++;
+            }else{
+                count = 1;
+            }
+            if(count <= 2){ //as max allowed is twice
+                nums[j] = nums[i];
+                j++;
+            }
+        }
+        return j;//result length
+    }
+}

MergeSortedArray.java

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+// Time Complexity : O(m+n).
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Approach :  We have two pointers, both at end of each array. Basically, we compare two elements from both arrays and place the biggest
+// one at the end of the array. Based on it, we move the pointers accordingly.
+
+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){ //until both arrays wont reach out of bounds
+            if(nums1[p1] <= nums2[p2]){
+                nums1[idx] = nums2[p2];//add the greater one from nums2 to nums1
+                p2--;
+                idx--;
+            } else{
+                nums1[idx] = nums1[p1];//shift the greater one to end pointer in nums1
+                p1--;
+                idx--;
+            }
+        }
+        while(p2>=0){//if still elements are left in nums2, move to nums1
+            nums1[idx] = nums2[p2];
+            p2--;
+            idx--;
+        }
+    }
+}

SearchIn2DArray.java

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+// Time Complexity : O(m+n).
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Approach :  We start looping from [0][n-1] or [m-1][0] as from these points we see a pattern of increasing on one direction and decreasing
+// on another direction. So I picked [0][n-1], compared it with target and moved the pointer lo left if target is greater or moved the
+// pointer down if target is lower.
+
+class Solution {
+    public boolean searchMatrix(int[][] matrix, int target) {
+        int m = matrix.length;
+        int n = matrix[0].length;
+        int i = 0;
+        int j = n-1;
+        while(i<m && j>=0){ // boundary conditions
+            if(matrix[i][j] == target){
+                return true;
+            } else if(target > matrix[i][j]){
+                i++; // move towards bottom in 2d matrix visualization
+            } else{
+                j--;// move towards left in 2d matrix visualization
+            }
+        }
+        return false;
+    }
+}