How to solve Leetcode problem 250 in Java

Day 46/100 – #100DaysOfCode 🚀 | #Java #BFS #Graph ✅ Problem Solved: Word Ladder (LeetCode 127) 🧩 Problem Summary: You are given a beginWord, endWord, and a list of words. You must transform beginWord → endWord by changing one character at a time, where each intermediate word must exist in the dictionary. Return the minimum number of transformations. 💡 Approach Used: Used Breadth-First Search (BFS) because: We need the shortest path in terms of transformation steps. Each word is considered a node in a graph. Key Steps: Pre-processed all words into generic pattern groups (e.g., h*t → hot, hit, etc.). Performed BFS from beginWord until reaching endWord. Counted transformation depth as solution. ⚙️ Time Complexity: O(N × L²) 📦 Space Complexity: O(N × L) ✨ Takeaway: This problem is an excellent example of how BFS can be used for shortest transformation sequences in word-graphs. #Java #LeetCode #Graph #BFS #ProblemSolving #100DaysOfCode #CodingChallenge

#100DaysOfCode – Day 77 Delete in a Doubly Linked List Problem: Given a doubly linked list and an integer x, remove the node at position x (1-indexed) and return the head of the updated list. Example: Input: 1 <-> 2 <-> 3 <-> 4, x = 3 Output: 1 <-> 2 <-> 4 My Approach: 1️⃣ Handled edge cases deleting the head or an empty list. 2️⃣ Traversed to the node at position x. 3️⃣ Updated both prev and next pointers to unlink the node cleanly. Time Complexity: O(N) Space Complexity: O(1) Understanding pointer manipulation in linked lists builds the foundation for mastering advanced data structures. Every prev and next link matters! #100DaysOfCode #Java #ProblemSolving #DSA #GeeksforGeeks #LinkedList #Pointers #TakeUForward #CodeNewbie #CodingJourney

🚀 Day 6 of 30 Days of Data Structures Challenge Topics Covered Today: 🔹 Java Collection Framework (JCF) – LinkedList 🔹 Explored in-built LinkedList methods like add(), remove(), get(), addFirst(), addLast() and more. 🔹 Implemented Merge Sort on a LinkedList from Scratch. 📂 Code: https://lnkd.in/gk5-kdKv https://lnkd.in/gfdmpVWf #Java #DataStructures #LinkedList #MergeSort #JCF

🗓 Day 7 / 100 – #100DaysOfLeetCode 🔢 Problem 2536: Increment Submatrices by One Today’s challenge involved processing multiple submatrix increment queries on an n x n matrix, initially filled with zeros. 🧠 My Approach Instead of updating every cell inside each submatrix (which would be too slow for up to 10⁴ queries), I used a row-wise difference array technique. For each query [r1, c1, r2, c2]: Increment prefix[row][c1] Decrement prefix[row][c2+1] (if within bounds) This allows efficient marking of increments. Later, prefix-summing each row reconstructs the final matrix. ⏱ Time Complexity O(q × n) where q = number of queries (We touch r2 - r1 + 1 rows per query) 💾 Space Complexity O(n²) for the prefix matrix #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

Day 53/100 Problem Statement : Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number. Return the decimal value of the number in the linked list. The most significant bit is at the head of the linked list. Input: head = [1,0,1] Output: 5 Solution : https://lnkd.in/g3vuQeUx public int getDecimalValue(ListNode head) { ListNode temp=head; String s=""; while(temp!=null){ s+=temp.val; temp=temp.next; } int ans=Integer.parseInt(s,2); return ans; } #100DaysDSA #100DaysOfCode #Java #Leetcode #Neetcode #Neetcode250 #TUF

Day 43/100 Problem Statement : Given an array nums, you can perform the following operation any number of times: Select the adjacent pair with the minimum sum in nums. If multiple such pairs exist, choose the leftmost one. Replace the pair with their sum. Return the minimum number of operations needed to make the array non-decreasing. An array is said to be non-decreasing if each element is greater than or equal to its previous element (if it exists). Input: nums = [5,2,3,1] Output: 2 Solution : https://lnkd.in/g3Z6NQxN public int minimumPairRemoval(int[] nums) { int length = nums.length - 1; int count = 0; while(length > 0) { boolean increase = true; int minSum = Integer.MAX_VALUE; int minIndex = -1; for(int i = 0; i < length; i++) { if(nums[i] > nums[i + 1]) increase = false; if(nums[i] + nums[i + 1] < minSum) { minSum = nums[i] + nums[i + 1]; minIndex = i; } } if(increase) break; nums[minIndex] = minSum; for(int i = minIndex + 1; i < length; i++) nums[i] = nums[i + 1]; length--; count++; } return count; } #100DaysDSA #100DaysOfCode #Java #Leetcode #Neetcode #Neetcode250 #TUF

📌 LeetCode Day 57 — #107. Binary Tree Level Order Traversal II Problem Description: Given the root of a binary tree, return the bottom-up level order traversal of its nodes' values. That means you need to traverse the tree level by level from left to right, but from the bottom level up to the root. Approach: Use a Queue (BFS): Perform a standard level order traversal using a queue. Store Each Level: For each level, store the node values in a temporary list. Insert at Beginning: Instead of appending levels at the end, insert each level at index 0 to reverse the order. Return Result: The final list represents the bottom-up traversal. Complexity Analysis: Time Complexity: O(n) — every node is visited once. Space Complexity: O(n) — queue and result storage for all nodes. Hashtags: #BinaryTree #BFS #LogicBuilding #LeetCode100Days #Day57 #Java #ProblemSolving

Day 50 of #75DaysDSAChallenge Problem: 7. Reverse Integer Difficulty: 🟠 Medium Platform: LeetCode 🧩 Problem Statement Given a signed 32-bit integer x, return x with its digits reversed. If reversing x causes it to go outside the signed 32-bit integer range [-2³¹, 2³¹ - 1], return 0. Example: Input: x = 123 Output: 321 💡 Approach 1️⃣ Extract the last digit using x % 10. 2️⃣ Remove the last digit using x / 10. 3️⃣ Add the digit to the reversed number. 4️⃣ Before updating, check for overflow conditions. 5️⃣ Continue until all digits are processed. #LeetCode #Java #DSA #CodingChallenge #75DaysDSAChallenge #ProblemSolving #TechLearning #CodingJourney

💻 Day 03/10 of Linked List Series: Inserting a Node at the Beginning of a Linked List Let’s understand how we can add a new node at the start of a linked list 👇 ⚙️ Algorithm: Insert at Beginning Create a new node with the given data (for example, 'A'). Set the next pointer of this new node to point to the current head of the list. Now, make this new node the new head of the linked list. 💡 In simple terms: you’re placing a new link before the first link in the chain. 🧩 Example Explanation: Initial list: B → C → D We want to insert 'A' at the beginning. Steps: Create a new node 'A'. Point 'A' → 'B'. Return node 'A'. ✅ Final list becomes: A → B → C → D ✨ This operation is one of the most common and easiest insert operations in linked lists — it just needs a few pointer adjustments! 📚 Stay tuned for the next part in this #LinkedListSeries! Follow my Brand 👉 #CodeWithLakkojuEswaraSai #CodeWithLakkojuEswaraSai_LinkedList #DSA #Java #10000coders