Multiplying Strings on LeetCode without Big Integers

🧠 Day 41 / 100 – DSA Practice Solved Count and Say on LeetCode 🔢🗣️✅ 🔹 Problem: Generate the nth term of the count-and-say sequence, where each term is derived by describing the previous term. 🔹 Approach: Used an iterative + string building approach: Start with base case "1" For each iteration, read the previous string Count consecutive characters Append count + character to form next term 🔍 Key Insight: This problem is essentially Run-Length Encoding (RLE) applied repeatedly on strings 🔹 Complexity: ⏱ Time → O(n × m) (m = length of generated string) 📦 Space → O(m) 💯 Result: ✔️ All test cases passed ⚡ Runtime: 8 ms (Beats 62%) Great problem to improve string manipulation & pattern recognition 🚀 #Day41 #100DaysOfCode #LeetCode #Java #DSA #Strings #RLE #Algorithms #CodingJourney

🧠 Day 40 / 100 – DSA Practice Solved String to Integer (atoi) on LeetCode 🔢➡️🧮✅ 🔹 Problem: Convert a string into a 32-bit signed integer while handling whitespace, signs, non-digit characters, and overflow conditions. 🔹 Approach: Followed a step-by-step simulation: Ignore leading whitespaces Handle optional + / - sign Convert digits while checking for overflow Stop at first non-digit character 🔍 Key Insight: Careful handling of edge cases like overflow and invalid input is crucial in this problem 🔹 Complexity: ⏱ Time → O(n) 📦 Space → O(1) 💯 Result: ✔️ All test cases passed ⚡ Runtime: 1 ms (Beats 100%) One of the best problems to master string parsing & edge case handling 🚀 #Day40 #100DaysOfCode #LeetCode #Java #DSA #Strings #Algorithms #CodingJourney

🧠 Day 38 / 100 – DSA Practice Solved Word Search on LeetCode 🔍✅ 🔹 Problem: Given a 2D grid of characters and a word, check if the word exists in the grid by traversing adjacent cells (horizontally or vertically). 🔹 Approach: Used DFS + Backtracking: Start from each cell Explore all 4 directions (up, down, left, right) Mark visited cells to avoid reuse Backtrack if the path doesn’t match 🔍 Key Insight: Backtracking helps explore all possible paths while ensuring each cell is used only once per path 🔹 Complexity: ⏱ Time → O(m × n × 4^L) 📦 Space → O(L) recursion stack 💯 Result: ✔️ All test cases passed ⚡ Runtime: 130 ms (Beats 73%) Great problem to strengthen DFS & Backtracking concepts 🚀 #Day38 #100DaysOfCode #LeetCode #Java #DSA #Backtracking #DFS #Algorithms #CodingJourney

Day 45 of Daily DSA 🚀 Solved LeetCode 867: Transpose Matrix ✅ Problem: Given a 2D matrix, return its transpose (rows become columns and columns become rows). Approach: Created a new matrix and swapped indices while traversing. Steps: Get number of rows and columns Create a new matrix of size cols x rows Traverse original matrix Assign: trans[j][i] = matrix[i][j] Return the new matrix ⏱ Complexity: • Time: O(n × m) • Space: O(n × m) 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 46.46 MB Simple transformations like transpose build strong fundamentals for matrix-based problems 💡 #DSA #LeetCode #Java #Arrays #Matrix #CodingJourney #ProblemSolving

📘 DSA Journey — Day 35 Today’s focus: Binary Search with index patterns. Problem solved: • Single Element in a Sorted Array (LeetCode 540) Concepts used: • Binary Search • Index parity (even/odd pattern) • Search space reduction Key takeaway: The array is sorted and every element appears twice except one. A key observation: Before the single element, pairs start at even indices After the single element, this pattern breaks. Using binary search: • If mid is even and nums[mid] == nums[mid + 1], the single element lies on the right side • Else, it lies on the left side (including mid) By leveraging this pattern, we can find the answer in O(log n) time and O(1) space. Continuing to strengthen binary search intuition and consistency in problem solving. #DSA #Java #LeetCode #CodingJourney

Day 48 of Daily DSA 🚀 Solved LeetCode 48: Rotate Image ✅ Problem: Given an n x n matrix, rotate the image by 90° clockwise — in-place (without using extra space). Approach: Used a two-step transformation: Transpose the matrix Reverse each row Steps: Traverse upper triangle and swap → matrix[i][j] ↔ matrix[j][i] For each row: Use two pointers (left, right) Swap elements to reverse the row Matrix gets rotated in-place ⏱ Complexity: • Time: O(n²) • Space: O(1) 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 43.56 MB In-place transformations are powerful — no extra space, just smart manipulation 💡 #DSA #LeetCode #Java #Matrix #Arrays #CodingJourney #ProblemSolving

Day 47 of Daily DSA 🚀 Solved LeetCode 74: Search a 2D Matrix ✅ Problem: Given a sorted 2D matrix where: • Each row is sorted • First element of each row > last element of previous row Find whether a target exists in the matrix. Approach: Used an optimized staircase search (top-right traversal). Steps: Start from top-right corner If element == target → return true If element > target → move left If element < target → move down Continue until found or out of bounds ⏱ Complexity: • Time: O(n + m) • Space: O(1) 📊 LeetCode Stats: • Runtime: 0 ms (Beats 100%) ⚡ • Memory: 43.84 MB Sometimes choosing the right starting point (top-right) makes the search super efficient 💡 #DSA #LeetCode #Java #Matrix #BinarySearch #CodingJourney #ProblemSolving

#Day362 of #1001DaysOfCode 📘 LeetCode Daily Challenge Problem: XOR After Queries 💡 Approach: Optimized the brute-force simulation using sqrt decomposition. Large step queries were processed directly Small step queries were grouped and batch-processed efficiently Used modular exponentiation + grouped progression updates to reduce time complexity significantly. ⏱ Optimized Time Complexity: ~O(q√n + n√n log MOD) 🧠 Space Complexity: O(n + q) A great problem for learning advanced optimization techniques 🚀 #DSA #Java #LeetCode #ProblemSolving #Algorithms #Coding

🔥 DSA Challenge – Day 136/360 🚀 📌 Topic: Bit Manipulation 🧩 Problem: Number of 1 Bits (Hamming Weight) Problem Statement: Given an integer, return the number of set bits (1’s) present in its binary representation. 🔍 Example: Input: 6 Output: 2 Explanation: Binary of 6 → 110 → 2 set bits 💡 Optimized Approach (Brian Kernighan’s Algorithm): Instead of checking each bit one by one, we use a smart trick: n & (n - 1) removes the rightmost set bit in each iteration This reduces the number of operations to the number of set bits only ⚡ Time Complexity: O(k) (k = number of set bits) ⚡ Space Complexity: O(1) 🚀 Key Takeaway: Using bit manipulation can drastically optimize performance compared to brute force approaches. Always look for patterns in binary operations! #DSA #Coding #Java #BitManipulation #136DaysOfCode #LeetCode #ProblemSolving #SoftwareEngineering

🚀 Day 566 of #750DaysOfCode 🚀 🔍 Problem Solved: Mirror Distance of an Integer Today’s problem was simple but a great reminder of how powerful basic number manipulation can be. 💡 Problem Insight: We define the mirror distance of a number as: 👉 abs(n - reverse(n)) So the task is: Reverse the digits of the number Compute the absolute difference 🧠 Approach: Extract digits using % 10 Build the reversed number Compute absolute difference 📊 Complexity: Time: O(d) → where d = number of digits Space: O(1) 🔥 Key Takeaway: Even the easiest problems reinforce fundamentals like digit extraction, reversal, and absolute difference — building blocks for more complex algorithms. #Day566 #750DaysOfCode #LeetCode #Java #CodingJourney #ProblemSolving #Algorithms #BeginnerFriendly