From 997c8433ab0a6a1ede1e9a0926b34c2066496e19 Mon Sep 17 00:00:00 2001
From: Praniksha123 <pvoddinani@gmail.com>
Date: Thu, 2 Jul 2026 21:11:59 +0530
Subject: [PATCH] Create pointers.java

---
 pointers.java | 104 ++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 104 insertions(+)
 create mode 100644 pointers.java

diff --git a/pointers.java b/pointers.java
new file mode 100644
index 00000000..1e022755
--- /dev/null
+++ b/pointers.java
@@ -0,0 +1,104 @@
+
+// Approach:
+// 1. Count the number of 0s, 1s, and 2s in the array.
+// 2. Traverse the array again and place all 0s first.
+// 3. Fill the next positions with 1s.
+// 4. Fill the remaining positions with 2s.
+// 5. This sorts the array using the counting technique.
+
+// Time Complexity: O(n)
+// Space Complexity: O(1)
+class Solution {
+    public void sortColors(int[] nums) {
+        int n=nums.length;
+        int count1=0;
+        int count2=0;
+        int count3=0;
+        for(int i=0;i<n;i++){
+            if(nums[i]==0){
+                count1++;
+            }else if(nums[i]==1){
+                count2++;
+            }else{
+                count3++;
+            }
+        }
+        int z=count1+count2;
+        for(int i=0;i<count1;i++){
+            nums[i]=0;
+        }for(int i=count1;i<z;i++){
+            nums[i]=1;
+        }for(int i=z;i<n;i++){
+            nums[i]=2;
+        }
+        return;
+    }
+}
+//problem 2
+// Approach:
+// 1. Sort the array.
+// 2. Fix one element and use two pointers (left and right) to find two numbers
+//    whose sum with the fixed element equals zero.
+// 3. If the sum is zero, store the triplet and skip duplicate elements.
+// 4. If the sum is less than zero, move the left pointer forward.
+// 5. If the sum is greater than zero, move the right pointer backward.
+// 6. Repeat until all unique triplets are found.
+
+// Time Complexity: O(n^2)
+// Space Complexity: O(1) (excluding the output list)
+class Solution {
+    public List<List<Integer>> threeSum(int[] nums) {
+        Arrays.sort(nums);
+        List<List<Integer>> res=new ArrayList<>();
+        int n=nums.length;
+        for(int i=0;i<n;i++){
+            if(i!=0 && nums[i]==nums[i-1]) continue;
+            int left=i+1;
+            int right=n-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--;
+                    while(left<right && nums[left]==nums[left-1]) left++;
+                    while(left<right && nums[right]==nums[right+1]) right--;
+                }
+                else if(sum>0){
+                    right--;
+                }else{
+                    left++;
+                }
+            }
+        }
+        return res;
+    }
+}
+//problem 3
+// Approach:
+// 1. Place one pointer at the beginning and another at the end of the array.
+// 2. Calculate the area formed by the two lines.
+// 3. Update the maximum area if the current area is larger.
+// 4. Move the pointer with the smaller height inward since it limits the area.
+// 5. Repeat until the two pointers meet.
+
+// Time Complexity: O(n)
+// Space Complexity: O(1)
+class Solution {
+    public int maxArea(int[] height) {
+        int n=height.length;
+        int left=0;
+        int right=n-1;
+        int ans=0;
+        while(left<right){
+            int res=Math.min(height[left],height[right]);
+            ans=Math.max(ans,res*(right-left));
+            if(height[left]<height[right]){
+                left++;
+            }else{
+                right--;
+            }
+        }
+        return ans;
+    }
+}