AMIT vishwakarma’s Post

🚀 Day 10/30 | LeetCode Problem: Reverse Linked List (206) Problem: Given the head of a singly linked list, reverse the list and return the new head. 💡 Approach (Iterative – Three Pointers) To reverse a linked list, we carefully change the direction of pointers. We use three variables: prev → previous node current → current node next_node → temporarily stores next node Steps: Store next node Reverse the current node’s pointer Move prev and current one step forward Repeat until the list ends Finally, prev becomes the new head. ⏱ Complexity Time Complexity: O(n) Space Complexity: O(1) 🧠 Python Code class Solution: def reverseList(self, head): prev = None current = head while current: nxt = current.next current.next = prev prev = current current = nxt return prev 📌 Example Input: [1,2,3,4,5] Output: [5,4,3,2,1] 🎯 Key Takeaway Linked List problems are all about pointer manipulation. Understanding pointer flow is more important than memorizing code. ✅ Accepted 🔖 Hashtags #LeetCode #30DaysOfLeetCode #Day10 #Python #LinkedList #DataStructures #Algorithms #ProblemSolving #CodingJourney #SoftwareEngineering

Day 47/100 – #100DaysOfLeetCode 🚀 🧩 Problem: LeetCode 191 - Number of 1 Bits (Easy) 🧠 Approach 1: Using Brian Kernighan’s Algorithm. 💻 Solution: class Solution:   def hammingWeight(self, n: int) -> int:     count = 0     while n:       n &= (n - 1)        count += 1     return count ⏱ Time | Space: O(k) | O(1) 📌 Key Takeaway: Using n & (n - 1) efficiently removes one set bit at a time, making it faster than checking every bit individually. 🧠 Approach 2: Python’s built-in binary conversion makes counting set bits simple 💻 Solution: class Solution:   def hammingWeight(self, n: int) -> int:     return bin(n).count('1') #leetcode #dsa #development #problemSolving #CodingChallenge

🚀 Day 41 of #100DaysofCode Challenge! Today’s focus was on recursion and backtracking patterns: 🔹 Generate Parentheses Solved this using a backtracking approach by building valid combinations incrementally while maintaining two constraints: The number of opening brackets used must not exceed n The number of closing brackets must not exceed the number of opening brackets This ensures only valid sequences are generated without checking invalid ones afterward. Core idea: Use recursion to explore all valid states Add '(' if openings < n Add ')' if closings < openings Key takeaways: ✔ Backtracking with constraint pruning ✔ Recursive state-space exploration ✔ Understanding how valid combinations grow systematically Time Complexity: O(4ⁿ / √n) (related to Catalan numbers) Space Complexity: O(n) for recursion stack Building stronger intuition for recursion and combinatorial problems every day 🚀 📸 Screenshots of solution are attached! #Day41 #DynamicProgramming #BinaryTree #TreeDP #Algorithms #ProblemSolving #Python #Java #CodingMindset #SoftwareEngineering #AdityaVerma #CodingJourney

🚀 Day 169 of My LeetCode Journey 🚀 1092. Shortest Common Supersequence 🫧 In this problem, we are given two strings, and we need to build the shortest string that contains both of them as subsequences. ▪️ The key idea is related to the Longest Common Subsequence. If both strings share some common characters in order, we should include those characters only once in the final string. ▪️ First, I built a DP table to find the LCS length between the two strings. This helps us understand which characters are common between them. ▪️ After filling the DP table, I traversed it backwards to construct the final string. ▪️ If characters in both strings match, I add that character once to the result and move diagonally in the table. ▪️ If they don’t match, I move in the direction that gave the larger LCS value and add that character to the result. ▪️ If one string finishes before the other, I simply added the remaining characters from the other string. ▪️ In the end, I reversed the collected characters to get the shortest common supersequence, which contains both strings while keeping the length as small as possible. #LeetCode #DynamicProgramming #Strings #Python #CodingJourney #Day169 🔥

🚀 Day 18 of #100DaysOfCode 📌 LeetCode 977 – Squares of a Sorted Array Today’s problem was about transforming a sorted array (with possible negative numbers) into a new array of squared values that remains sorted. 💡 Approach Used: Two Pointers Instead of squaring all elements and sorting again (O(n log n)), I applied the Two Pointer technique to compare absolute values from both ends and filled the result array from the back. 🧠 Key Learnings: ✔️ Importance of optimal thinking over brute force ✔️ How negative numbers affect sorting after squaring ✔️ Efficient use of Two Pointers in array problems ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(n) 🔍 Problem Pattern: Array + Two Pointers Consistency is the key — showing up every day and improving step by step! 💪 #Day18 #100DaysOfCode #LeetCode #DSA #Python #CodingJourney #BTech #DataStructures

Day 43 of 365 days of code Qn 1) simplify path approach used: stack initialize a stack 1) iterate the string path using var i from 0 to len(path) 2) ignore forward slashes 3) initialize ch="" 4) while i<len(path) and path[i] not equal to forward slash: ch+=path[i];i+=1 5) if ch==.. stack.pop elif ch== . continue else: stack.push(ch) 6)i+=1 7) pop all the element of the stack and put them in another stack 8) initialize a string starting with/ as u iterate pop the stack and add the forward slash to the popped string. #365daysOfCode #NeetCode #leetcode #DSA #python #LeetCode #ProblemSolving #Algorithms

🚀 Day 47 of #180DaysOfCode Today I solved LeetCode 921 – Minimum Add to Make Parentheses Valid. This problem focuses on balancing parentheses by determining the minimum number of additions required to make the string valid. A great exercise for improving understanding of stacks and string traversal. 🔹 Problem: LeetCode 921 – Minimum Add to Make Parentheses Valid https://lnkd.in/dVdQdTYZ 📂 GitHub Repo: https://lnkd.in/dR7YYD_m #Python #DSA #LearnInPublic #CodingChallenge #LeetCode #DataStructures #Algorithms #PythonDSA #DailyCoding #Consistency #TechJourney #Upskill #CodeOfTheDay #Day47

✅ Day 74 of 100 Days LeetCode Challenge Problem: 🔹 #653 – Two Sum IV: Input is a BST 🔗 https://lnkd.in/g9vDFKMA Learning Journey: 🔹 Today’s problem required checking whether there exist two nodes in a Binary Search Tree whose values sum to k. 🔹 I performed a Depth-First Search (DFS) traversal using a stack to visit all nodes of the tree. 🔹 While traversing, I stored each node’s value in a list. 🔹 Then I used a hash set to track visited values and checked whether the complement (k - value) already exists. 🔹 If the complement is found, it means two numbers sum to k, so the function returns True. Concepts Used: 🔹 Depth-First Search (DFS) 🔹 Stack-based Tree Traversal 🔹 Hash Set for Fast Lookup 🔹 Two Sum Pattern Key Insight: 🔹 The classic Two Sum approach works well after collecting values from the BST. 🔹 Using a set allows O(1) lookup, making it efficient to check complements during iteration. Complexity: 🔹 Time: O(n) 🔹 Space: O(n) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 64 of 100 Days LeetCode Challenge Problem: 🔹 #1784 – Check if Binary String Has at Most One Segment of Ones 🔗 https://lnkd.in/gpJYed9J Learning Journey: 🔹 Today’s problem focused on determining whether a binary string contains only one contiguous segment of '1's. 🔹 Since the string has no leading zeros, the first character is always '1'. 🔹 If a '1' appears again after a '0', it indicates the start of a second segment. 🔹 A single pass through the string is enough to detect this pattern. Concepts Used: 🔹 String Traversal 🔹 Pattern Detection 🔹 Conditional Logic Key Insight: 🔹 The pattern "01" followed by another '1' reveals multiple segments of ones. 🔹 Detecting this transition allows us to solve the problem efficiently. Complexity: 🔹 Time: O(n) 🔹 Space: O(1) 💬 What’s your favorite trick for solving binary string problems? #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers

✅ Day 69 of 100 Days LeetCode Challenge Problem: 🔹 #1009 – Complement of Base 10 Integer 🔗 https://lnkd.in/geVPugvi Learning Journey: 🔹 Today’s challenge involved finding the bitwise complement of a base-10 integer. 🔹 I first converted the integer into its binary representation using bin(n)[2:] to remove the 0b prefix. 🔹 Then I iterated through each bit and flipped it: • 0 becomes 1 • 1 becomes 0 🔹 After constructing the flipped binary string, I converted it back to a decimal integer using int(s, 2). Concepts Used: 🔹 Binary Representation 🔹 Bit Manipulation 🔹 String Traversal 🔹 Base Conversion (Binary → Decimal) Key Insight: 🔹 Converting the number to binary makes it easy to flip bits directly. 🔹 After inversion, converting the binary string back to base-10 produces the required complement. Complexity: 🔹 Time: O(b) where b is the number of bits in n 🔹 Space: O(b) #LeetCode #Algorithms #DataStructures #CodingInterview #100DaysOfCode #SoftwareEngineering #Python #ProblemSolving #LearningInPublic #TechCareers