Done two-pointers-1

problem1.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No, just need to handle mid pointer properly.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// I use three pointers low, mid, high to place 0s to the left and 2s to the right.
+// If nums[mid] is 0, swap with low and move both; if it is 2, swap with high and move high only.
+// If nums[mid] is 1, just move mid, and this finishes sorting in one pass.
+
+class Solution {
+    public void sortColors(int[] nums) {
+
+        int low = 0, mid = 0;
+        int high = nums.length - 1;
+
+        while (mid <= high) {
+
+            if (nums[mid] == 0) {
+                swap(nums, low, mid);
+                low++;
+                mid++;
+            } 
+            else if (nums[mid] == 2) {
+                swap(nums, mid, high);
+                high--;
+            } 
+            else {
+                mid++;
+            }
+        }
+    }
+
+    private void swap(int[] nums, int i, int j) {
+        int temp = nums[i];
+        nums[i] = nums[j];
+        nums[j] = temp;
+    }
+}

problem2.java

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+// Time Complexity : O(n^2)
+// Space Complexity : O(1)  (excluding output list)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : Avoiding duplicates is the main tricky part.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// First I sort the array, then fix one number and use two pointers to find pairs that make sum zero.
+// When the sum is too small I move left pointer, when too large I move right pointer, and when zero I store the triplet.
+// I skip duplicates for i, left, and right so output does not repeat.
+
+import java.util.*;
+
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+
+        Arrays.sort(nums);
+        List<List<Integer>> res = new ArrayList<>();
+
+        for (int i = 0; i < nums.length; i++) {
+
+            if (i > 0 && nums[i] == nums[i - 1]) continue; // skip duplicate i
+
+            int left = i + 1;
+            int right = nums.length - 1;
+
+            while (left < right) {
+
+                int sum = nums[i] + nums[left] + nums[right];
+
+                if (sum == 0) {
+                    res.add(Arrays.asList(nums[i], nums[left], nums[right]));
+                    left++;
+                    right--;
+
+                    // skip duplicates on left
+                    while (left < right && nums[left] == nums[left - 1]) left++;
+
+                    // skip duplicates on right
+                    while (left < right && nums[right] == nums[right + 1]) right--;
+
+                } else if (sum < 0) {
+                    left++;
+                } else {
+                    right--;
+                }
+            }
+        }
+
+        return res;
+    }
+}

problem3.java

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+// Time Complexity : O(n)
+// Space Complexity : O(1)
+// Did this code successfully run on Leetcode : Yes
+// Any problem you faced while coding this : No, just need to move the smaller height pointer.
+
+
+// Your code here along with comments explaining your approach in three sentences only
+// I use two pointers at start and end and calculate area using width and minimum height.
+// To possibly get a bigger area, I always move the pointer with smaller height because that is the limiting factor.
+// I keep updating max area until pointers cross.
+
+class Solution {
+    public int maxArea(int[] height) {
+
+        int left = 0;
+        int right = height.length - 1;
+        int max = 0;
+
+        while (left < right) {
+
+            int h = Math.min(height[left], height[right]);
+            int w = right - left;
+
+            max = Math.max(max, h * w);
+
+            if (height[left] < height[right]) left++;
+            else right--;
+        }
+
+        return max;
+    }
+}