Merge pull request #276 from pranav-3005/pranav5

added delete-node-in-a-bst.java

Author: kaal-coder

.idea/.gitignore

@@ -0,0 +1,85 @@
+// question link : https://leetcode.com/problems/delete-node-in-a-bst/description/
+class Solution {
+
+    //1.reach the given node using recursion,if not found just return
+    public TreeNode deleteNode(TreeNode root, int val) {
+			if(root==null)
+				return root;
+
+            
+			if(val==root.val)
+			{//2. After found, there are 4 conditions
+                    
+                //2.1 cur node have both left & ryt nodes.  
+				if(root.left!=null && root.right!=null)
+				{
+					
+                    /*concept : take ryt node's left subtree,
+                                connect it to left node's ryt most node
+                                connect the updated left node to ryt node's left (i.e.. ryt.left=left)
+                                and return the updated ryt node. (deleting the cur node)
+                    */ 
+					
+					TreeNode ryt=cleft(root.left,root.right);
+					return ryt;
+				}
+
+                //2.2 have only ryt node
+				else if(root.left==null && root.right!=null)
+				{
+					return root.right;
+				}
+
+                //2.3 have only left node
+				else if(root.left!=null && root.right==null)
+				{
+					return root.left;
+				}
+
+                //2.4 have no child nodes.
+				else
+				{
+					return null;
+				}
+			}
+
+			if(val<root.val)
+			{
+				root.left=deleteNode(root.left,val);
+				return root;
+			}
+			else
+			{
+				root.right=deleteNode(root.right,val);
+				return root;
+			}
+    }
+
+    //cleft() -> does insert() func and connect the updated left to (ryt.left)
+    TreeNode cleft(TreeNode left,TreeNode ryt)
+	{
+		if(ryt.left==null)
+		{
+			ryt.left=left;
+			return ryt;
+		}
+		TreeNode temp=ryt.left;
+		ryt.left=insert(left,temp);
+		return ryt;
+		
+	}
+
+    //insert() -> connect ryt node's left subtree(temp) to rytmost of left node and return the updated left node
+    TreeNode insert(TreeNode root,TreeNode temp)
+	{
+		if(root==null)
+			return temp;
+		
+		root.right=insert(root.right,temp);
+
+		return root;
+	}
+	
+	
+
+}

.idea/hacktoberfest_2023.iml

@@ -0,0 +1,65 @@
+//Question : https://leetcode.com/problems/largest-rectangle-in-histogram/description/
+
+class Solution {
+    public int largestRectangleArea(int[] h) {
+
+        //monotonic stack
+        
+        int n=h.length;
+        int left[]=new int[n];  //left small elem's ind'
+        int ryt[]=new int[n];  //ryt smaller elem's ind'
+        
+        Stack<Integer> lin=new Stack<>(); //storing index in stacks
+        Stack<Integer> rin=new Stack<>();
+        
+        int ans[]=new int[n]; //ans array
+
+        //1.store each elem's ryt side next small elem index in ryt[]
+        for(int i=0;i<n;i++)
+        {
+            //before add cur elem index to stack, 
+            //we check if its the ryt next small of elems in stack
+            while( rin.size()>0 && h[i]< h[rin.peek()] )
+            {
+                ryt[rin.peek()]=i;
+                rin.pop();
+            }
+            rin.push(i);
+        }   
+        //if no small elem present in ryt, assign n.
+        while(rin.size()>0) 
+        {
+            ryt[rin.pop()]=n;
+        }
+
+
+        //2.store each elem's left side next small elem index in left[]
+        for(int i=n-1;i>=0;i--)
+        {
+            //before add cur elem index to stack, 
+            //we check if its the left next small of elems in stack
+            while( lin.size()>0 && h[i]<h[lin.peek()] )
+            {
+                left[lin.pop()]=i;
+            }
+            lin.push(i);
+        }
+        //if no small elem present in left, assign '-1'.
+        while(lin.size()>0)
+        {
+            left[lin.pop()]=-1;
+        }
+
+
+        //3. calculate area of rectangle(i.e area=width*height)
+        //store ans arr in left[]  //height=value at an ind
+        for(int i=0;i<n;i++)
+        {
+            int w=ryt[i]-left[i] - 1; //width
+            ans[i]= w * h[i] ; //area
+        }
+
+        //4. return max of ans arr(i.e...left[])
+        return Arrays.stream(ans).max().getAsInt();
+    }
+}

.idea/misc.xml

@@ -0,0 +1,52 @@
+//Question : https://leetcode.com/problems/minimum-number-of-k-consecutive-bit-flips/description/
+
+//Question exp : convert alll 0 -> 1 by doing k bit flip
+
+//Ans approach : greedy approach
+class Solution {
+    public int minKBitFlips(int[] nums, int k) {
+        int n=nums.length;
+        int flipCount=0; // range [cur - (k-1)elem] will be flipped this much times
+        int totalFlips=0 ; //final ans
+
+        //true will be marked for all range's(cur indx - k-1th indx) end elem.
+        //While traversing, if visited[cur elem]=true,
+        //then we reduce the flipCount of cur_range ,as we move towards nxt range
+        boolean[] visited=new boolean[n];
+
+        for(int i=0;i<n;i++)
+        {
+            //only works if, (which is flipped flipCount no of times) cur_elem's value = 0
+                // 0 flipped even times = 0
+                //1 flipped odd times = 0
+            if( (nums[i]==0 && flipCount%2==0) || (nums[i]==1 && flipCount%2!=0) )
+            {
+                flipCount++;
+                totalFlips++;
+
+                if( i+(k-1) < n )
+                {
+                    //mark end of cur range as true
+                    //so, we can rmv the flipCount of this range, after reaching this endPoint.
+                    visited[i+(k-1)]=true;
+                }
+                else
+                {
+                    //if end elem goes out of range
+                    //hence, all of arr elems cannot be flipped to 1
+                    return -1; 
+                }
+            }
+
+            if(visited[i]==true)
+            {
+                //end point of cur_range is reached.
+                //so reduce the flipCount of cur_range 
+                flipCount--;
+            }
+
+        } 
+
+        return totalFlips;
+    }
+}

.idea/modules.xml

@@ -0,0 +1,53 @@
+//Question : https://leetcode.com/problems/sliding-window-maximum/description/
+
+class Solution {
+
+    //helper class to combine cur_index's value & cur_index
+    public class pair
+    {
+        int val; //value
+        int ind; //index
+
+        pair(int val,int ind)
+        {
+            this.val=val;
+            this.ind=ind;
+        }
+    } 
+
+    public int[] maxSlidingWindow(int[] nums, int k) {
+
+        int n=nums.length;
+
+        //stores ans
+        int ans[]=new int[n-(k-1)];
+
+        //queue of descending order by cur_index's value (max heap)
+        PriorityQueue<pair> pque=new PriorityQueue<>((a,b)->{
+           return b.val-a.val;
+        });
+
+        //start range (1st range)
+        for(int i=0;i<k;i++)
+        {
+            pque.add(new pair(nums[i],i));
+        }
+
+        ans[0]=pque.peek().val;
+
+        //start from k 
+        for(int i=k;i<n;i++)
+        {
+            //adds ans only in range of each subarr
+            //if ans's index not in range, then that elem in rmved from que
+
+            while(pque.size()>0 && pque.peek().ind <= (i-k) )
+                pque.remove();
+
+            pque.add(new pair(nums[i],i));
+            ans[i-k+1]=pque.peek().val; //storing every k window's max
+        }
+        return ans;
+       }   
+
+}

.idea/vcs.xml

@@ -0,0 +1,48 @@
+//Question : https://leetcode.com/problems/trapping-rain-water/description/
+
+class Solution {
+    public int trap(int[] h) {
+        int n=h.length;
+
+        //left and ryt boundary
+        int lb=h[0], rb=h[n-1];
+
+        //traversing range ,start and end within boundary
+        int l=1, r= n-2;
+
+        int tw=0; //total_water
+        
+        //traverse
+        while(l<=r)
+        {
+            if(lb<=rb)
+            {
+                int hw=lb; //theight of water (smallest of(left and ryt boundary))
+                
+                //whether water can be stored at top of building(h[l])
+                if(h[l]<hw)
+                {
+                    int ac= hw - h[l]; //actual height(i.e.. water on top of the building)
+                    tw+= ac; //add to total water
+                }
+                lb= Math.max(lb,h[l]);
+                l++;
+            }
+            else
+            {
+               int hw=rb; //tot height of water
+                
+                //whether water can be stored at top of building
+                if(h[r]<hw)
+                {
+                    int ac= hw - h[r]; //actual height(i.e.. water on top)
+                    tw+= ac;  //add to total water
+                }
+                rb= Math.max(rb,h[r]);
+                r--;
+            }
+        }
+        return tw;
+
+    }
+}