Implementing strStr() on LeetCode and HackerRank

🚀 Day 51 of #100DaysOfCode — LeetCode + HackerRank Edition Today’s challenge: Spiral Matrix — traverse a matrix in spiral order and return the elements. 🌀🧩 Problem (brief): Given an m x n matrix, return all elements in spiral order (clockwise starting from top-left). 📌 My approach →Use four boundaries: top, bottom, left, right. →Walk: left→right along top, top→bottom along right, right→left along bottom, bottom→top along left. →After each traversal, move the corresponding boundary inward. →Repeat while top <= bottom and left <= right. →Handle edge cases: single row, single column, or empty matrix. 💡 What I learned today: ✅Boundary-driven traversal is a neat pattern — useful for many matrix problems. ✅Always check the top<=bottom and left<=right conditions before traversing opposite sides to avoid duplicates. ✅Dry runs are everything — they reveal off-by-one bugs faster than tests. Let’s share patterns — how do you solve Spiral Matrix? Do you prefer boundary pointers, or recursion (peel the outer layer)? Drop your implementation or edge cases — I’ll compare notes! 👇 #Day51 #100DaysOfCode #LeetCode #Python #DSA #SpiralMatrix #ProblemSolving #CodeNewbie #LearnInPublic

🚀 Day 46 of #100DaysOfCode — Leetcode + HackerRank Edition! Today’s challenge was a twist on yesterday’s recursion puzzle — this time with duplicates in the mix 🔁🧠 🧩 permuteUnique(self, nums: List[int]) -> List[List[int]] — Generate all unique permutations of a list that may contain duplicate integers. 📌 Challenge: → Given a list nums, return all distinct orderings → Each number must appear exactly once per permutation → Example: nums = [1, 1, 2] ✅ Output: [[1,1,2], [1,2,1], [2,1,1]] 🔍 Approach: → Sort the input to group duplicates → Use recursion + backtracking to build paths → Track visited indices with a used[] array → Skip duplicates using: if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]: continue  → Backtrack by popping the last element and resetting used[i] 💡 What made it click: → Visualized the decision tree for [1, 1, 2] — saw how pruning avoids duplicate branches → Practiced dry runs to trace how used[] and sorting work together → Realized how skipping nums[i] when nums[i] == nums[i-1] and used[i-1] == False prevents redundant paths → Appreciated how recursion + pruning = clean, efficient, elegant 📚 What I learned: ✅ How to handle duplicates in permutation problems ✅ How sorting + index tracking helps prune recursion ✅ How to visualize branching decisions and avoid redundant paths ✅ The subtle power of used[] and index-based duplicate checks Have you ever debugged a recursive tree with duplicate values? Let’s swap strategies, dry runs, and visual walkthroughs 💬 #Day46 #Leetcode #Python #Recursion #Backtracking #DryRun #LearnInPublic #CodeNewbie #TechJourney #100DaysOfCode #DSA

🚀 Day 29 of #100DaysOfCode — LeetCode + HackerRank Edition! Today’s challenge was all about precision and pattern matching — and I stuck to the essentials. 🔗 Problem: findSubstring(s, words) (LeetCode) 📌 Challenge: Given a string s and a list of words (all the same length), find all starting indices where the concatenation of all words appears exactly once and without overlap. 🔍 Approach: → Calculated the total length of the target substring using len(words[0]) * len(words) → Built a frequency map using Counter(words) → Used a sliding window of size total_len to scan through s → At each position, extracted word-sized chunks and tracked their frequency → Compared the seen map with the original word_count to validate matches → Appended valid starting indices to the result list 💡 What made it click: → Realized that permutations aren’t necessary — just match word frequencies → Stepping through s in word-sized chunks keeps the logic clean and efficient → Using Counter for both the target and current window made comparison seamless 📚 What I learned: ✅ Frequency maps are a powerful alternative to brute-force approaches ✅ Matching dictionaries directly (seen == word_count) is a validation trick ✅ Clean logic and early breaks make the code both readable and performant Have you solved findSubstring before? Did you go with permutations or lean into frequency maps like I did? Let’s swap strategies 💬 #Day29 #LeetCode #SlidingWindow #StringAlgorithms #Python #ProblemSolving #CleanCode #CodeEveryDay #LearnToCode #CodeNewbie #SoftwareEngineering #TechJourney #100DaysOfCode

🚀 Day 32 of #100DaysOfCode — LeetCode + HackerRank Edition! Today’s challenge was all about frequency mapping and subarray logic — solving the Picking Numbers problem from HackerRank. 🔗 Problem: pickingNumbers(a) (HackerRank) 📌 Challenge: Given an array of integers, find the length of the longest subarray where the absolute difference between any two elements is ≤ 1. 🔍 Approach: → Used Python’s Counter to map frequencies of each number → For each number num, calculated freq[num] + freq[num + 1] → Tracked the maximum such sum to find the longest valid subarray → Avoided brute-force by leveraging frequency patterns instead of scanning all subarrays 💡 What made it click: → Realized that valid subarrays must contain only num and num + 1 → Frequency mapping reveals hidden structure in the data → Clean logic with max() and get() made the solution elegant and readable 📚 What I learned: ✅ Frequency-based thinking can simplify subarray problems ✅ Counter is a powerful tool for pattern detection ✅ Avoiding brute-force leads to scalable, efficient solutions ✅ Sharing dry runs and visual walkthroughs helps others grasp the intuition faster Have you tackled Picking Numbers before? Did you go with sorting, brute-force, or frequency mapping like I did? Let’s swap strategies 💬 #Day31 #HackerRank #SubarrayLogic #FrequencyMapping #Python #ProblemSolving #CleanCode #CodeEveryDay #LearnToCode #CodeNewbie #SoftwareEngineering #TechJourney #100DaysOfCode

🚀 Day 45 of #100DaysOfCode — Leetcode + HackerRank Edition! Today’s challenge was a deep dive into recursion and backtracking 🔁🧠 🧩 permute(self, nums: List[int]) -> List[List[int]] — Generate all possible permutations of a list of distinct integers. 📌 Challenge: → Given a list nums, return all possible orderings → Each number must appear exactly once per permutation → Example: nums = [1, 2, 3] ✅ Output: [[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1]] 🔍 Approach: → Use recursion to build permutations step-by-step → Track current path with curr → Skip numbers already in curr → When len(curr) == len(nums), add a copy to results → Backtrack by popping the last element 💡 What made it click: → Visualized the recursive tree — each level adds one unused number → Practiced dry runs with nums = [1, 2, 3] to see how paths evolve → Realized how curr.pop() resets the state for the next branch → Appreciated how backtracking avoids duplicate paths and keeps memory clean 📚 What I learned: ✅ How to implement recursive backtracking with minimal state ✅ How to use dry runs to trace recursive calls ✅ How to visualize tree-like expansion of choices ✅ The power of copying lists (curr[:]) to preserve snapshots Have you ever visualized recursion like a decision tree? Let’s swap strategies and dry run walkthroughs 💬 #Day45 #Leetcode #Python #Recursion #Backtracking #DryRun #LearnInPublic #CodeNewbie #TechJourney #100DaysOfCode #DSA

🚀 Day 38 of #100DaysOfCode —Leetcode + HackerRank Edition! Today’s challenge was a classic logic puzzle wrapped in a deceptively simple interface: 🧩 isValidSudoku(board: List[List[str]]) — Validate a partially filled Sudoku board. 📌 Challenge: Check whether a 9×9 Sudoku board is valid so far, based on three rules: → Each row must contain digits 1–9 without repetition → Each column must contain digits 1–9 without repetition → Each 3×3 sub-box must contain digits 1–9 without repetition 🔍 Approach: → Used three dictionaries of sets to track digits in rows, columns, and boxes → Iterated through each cell, skipping empty ones ('.') → For each digit, checked if it already exists in the corresponding row, column, or box → If duplicate found → return False; otherwise → add to all three sets 💡 What made it click: → Realized that each 3×3 box can be indexed using (i // 3, j // 3) — a neat trick! → Avoided nested loops or extra flags by using clean set logic → Practiced writing readable validation logic that mirrors real-world constraints 📚 What I learned: ✅ How to track multi-dimensional constraints using dictionaries of sets ✅ The power of clean iteration and early returns for validation problems ✅ Sudoku is a great playground for practicing control flow, indexing, and set operations Have you tackled this one before? Did you use sets, arrays, or something else entirely? Let’s swap strategies 💬 #Day38 #HackerRank #Python #SudokuLogic #ValidationChallenge #ControlFlow #ProblemSolving #LearnInPublic #CodeNewbie #TechJourney #100DaysOfCode#DSA

🚀 Day 35 of #100DaysOfCode — LeetCode + HackerRank Edition! Today’s challenge was all about merging two sorted arrays in-place — no extra space allowed, just clean pointer logic. 📌 Challenge: Merge nums2 into nums1, assuming nums1 has enough trailing space. The final array must be sorted — and the merge must happen in-place. 🔍 Approach: → Used a two-pointer strategy from the back to avoid overwriting values in nums1 → Compared the largest elements from both arrays and placed them at the end → Moved pointers backward (p1, p2, p) until all elements were merged → Handled edge cases like empty nums2 or all elements being smaller than nums1 → Avoided sorting after merge — the logic itself ensures order 💡 What made it click: → Realized that merging from the front causes overwrites — merging from the back preserves data → Visualized the pointer movement with a dry run — that helped me catch off-by-one bugs → Learned that in-place algorithms often rely on reverse traversal and smart indexing 📚 What I learned: ✅ Two-pointer techniques are powerful for in-place operations ✅ Always think about overwrite risks when merging arrays ✅ Dry runs and visual walkthroughs are key to debugging pointer logic Have you tried merging sorted arrays in-place before? Did you use reverse traversal or a different trick? Let’s swap strategies 💬 #Day35 #LeetCode #ArrayProblems #TwoPointerTechnique #ProblemSolving #Python #DebuggingJourney #CleanCode #LearnToCode #CodeNewbie #SoftwareEngineering #TechJourney #100DaysOfCode

🚀 Day 33 of #100DaysOfCode — LeetCode + HackerRank Edition! Today’s challenge was a classic twist on binary search — searching in a rotated sorted array. 🔗 Problem: search(self, nums: List[int], target: int) (LeetCode) 📌 Challenge: Given a rotated sorted array, find the index of a target value in O(log n) time. 🔍 Approach: → Used binary search with a twist: at each step, determined which half of the array is sorted → If the left half is sorted, checked if the target lies within it → If not, searched the right half — and vice versa → Carefully updated low, high, and recalculated mid inside the loop → Avoided brute-force by leveraging the sorted structure of one half at every step 💡 What made it click: → Realized that one half is always sorted, even after rotation → Visualized the array and dry-ran examples like [4,5,6,7,0,1,2] to understand the logic → Fixed a key bug: I was calculating mid only once — moving it inside the loop made everything work! → Debugging this helped me appreciate how small details (like mid placement) can make or break binary search 📚 What I learned: ✅ Binary search can be adapted to rotated arrays with clever logic ✅ Always recalculate mid after updating low and high ✅ Dry runs are powerful for catching logic bugs Have you tackled rotated binary search before? Did you use recursion, iteration, or something else? Let’s swap strategies 💬 #Day33 #LeetCode #BinarySearch #RotatedArray #ProblemSolving #Python #DebuggingJourney #CleanCode #LearnToCode #CodeNewbie #SoftwareEngineering #TechJourney #100DaysOfCode

🚀 Day 31 of #100DaysOfCode — LeetCode + HackerRank Edition! Today’s challenge was all about tracking structure and resilience — decoding the longest valid parentheses substring in a given string. 🔗 Problem: longestValidParentheses(s) (LeetCode) 📌 Challenge: Given a string of '(' and ')', find the length of the longest well-formed (valid) parentheses substring. 🔍 Approach: → Used a stack to track indices of unmatched '(' characters → Introduced a base index to reset when encountering unmatched ')' → On each valid match, calculated the distance to the last unmatched index → Continuously updated the maximum valid length found so far 💡 What made it click: → Realized that tracking indices, not just characters, is key to measuring valid spans → Using stack[-1] after a pop gives the start of the current valid substring → Resetting with a base index ensures recovery after mismatches 📚 What I learned: ✅ Stack-based tracking is powerful for nested structures ✅ Index math reveals patterns that raw character matching misses ✅ Defensive coding (checking if stack is empty) prevents crashes and improves robustness ✅ Visual walkthroughs and guided hints make debugging intuitive and rewarding Have you tackled longestValidParentheses before? Did you go with dynamic programming, two-pass scanning, or stack-based logic like I did? Let’s swap strategies 💬 #Day30 #LeetCode #StackAlgorithms #ParenthesesMatching #Python #ProblemSolving #CleanCode #CodeEveryDay #LearnToCode #CodeNewbie #SoftwareEngineering #TechJourney #100DaysOfCode

🚀 Day 52 of #100DaysOfCode — Leetcode + HackerRank Edition! Today’s challenge was all about bringing order to chaos — merging overlapping intervals into clean, non-overlapping ranges 🗂️✨ 🧩 def merge(self, intervals: List[List[int]]) -> List[List[int]]: — Combine overlapping intervals into a simplified list 📌 Challenge: → Implement a function to merge overlapping intervals → Avoid brute-force approaches that check every pair → Optimize for clarity and performance 🔍 Approach: → First, sort intervals by their start time → Keep track of the last merged interval (prev[0]) → If the current interval overlaps, extend the end boundary → Otherwise, add it as a new interval 💡 What made it click: → Realized sorting upfront makes merging straightforward → Saw how prev[1] = max(prev[1], interval[i][1]) elegantly handles overlaps → Practiced dry runs with examples like: [[1,3],[2,6],[8,10],[15,18]] → [[1,6],[8,10],[15,18]] 📚 What I learned: ✅ Why sorting is the key to simplifying interval problems ✅ How greedy merging avoids unnecessary comparisons ✅ How to handle edge cases like single intervals or fully nested ranges Ever felt like your calendar was a mess of overlapping meetings? This algorithm is the perfect organizer 🗓️😎 Let’s swap dry runs, edge cases, and interval insights 💬 #Day51 #Leetcode #Python #MergeIntervals #GreedyAlgorithm #DryRun #LearnInPublic #CodeNewbie #TechJourney #100DaysOfCode #DSA