Solved LeetCode #1625 with BFS and modular arithmetic

🚀 Day 40 of #100DaysOfDSA Solved LeetCode Problem #69 – Sqrt(x) 🧮 💡 Problem Insight: Given a non-negative integer x, return the square root of x rounded down to the nearest integer. For example: Input: x = 8 Output: 2 (since √8 ≈ 2.828, and floor(2.828) = 2) ✨ Key Learnings: Practiced binary search to find results efficiently without using built-in math functions. Learned how to narrow down search space based on mid-square comparisons. Reinforced understanding of integer division and rounding down behavior. Time Complexity: O(log n) — fast and efficient! Space Complexity: O(1) 💬 Lesson: Binary Search isn’t just for sorted arrays — it’s a mindset for narrowing down possibilities quickly 🚀 #LeetCode #Python #DSA #BinarySearch #ProblemSolving #100DaysOfCode #Day40 #CodingJourney

Day 66: Find Minimum in Rotated Sorted Array II 📉 (Simple Version) I'm tackling an advanced Binary Search problem on Day 66 of #100DaysOfCode: "Find Minimum in Rotated Sorted Array II." The challenge is to find the minimum element in a rotated sorted array that may contain duplicates. My solution uses a modified Binary Search approach. The key difficulty is when the middle element (nums[mid]) equals the rightmost element (nums[right]). In this case, we can't tell if the minimum is on the left or the right. To resolve this ambiguity, I simply discard the rightmost element by setting right -= 1. This adjustment keeps the search on track, achieving an optimal O(log n) complexity in the typical case. My solution achieved 100% runtime efficiency! #Python #DSA #Algorithms #BinarySearch #Arrays #100DaysOfCode #ProblemSolving

Day 16 of #100DaysOfLeetCode Today’s problem focused on substring counting within binary strings and required an efficient approach to handle potentially large input sizes without generating every substring explicitly. 1. Number of Substrings With Only 1s The task was to count the total number of substrings that consist entirely of the character '1', with the final result taken modulo (10^9 + 7). Instead of constructing the substrings, the key insight is that each continuous block of 1s contributes a predictable number of valid substrings. 🔹 My Approach: Iterated through the string while tracking the current streak length of consecutive 1s. Each time a block ended, computed the number of substrings from that block using the formula: k*(k+1)/2 where k is the length of the streak. Added the total from each block to the final answer, applying the modulo constraint throughout. Completed the process with a final update for any trailing block of 1s. What I Learned: This problem reinforces how recognizing mathematical patterns within sequences can transform a brute-force solution into a simple linear scan. Efficient substring counting often comes down to understanding structure rather than enumerating possibilities. 📊 Complexity Analysis: Time Complexity: O(n) — single pass over the string. Space Complexity: O(1) — constant space approach. #day16 #100daysofleetcode #leetcode #DSA #python #leetcodes #striver

🔍 Day 44 DSA Challenge – Problem #33: Search in Rotated Sorted Array 📌 Problem Statement: Given a sorted array nums that may have been rotated at an unknown pivot, find the index of a target value. Return -1 if the target doesn’t exist. The solution must run in O(log n) time. ⚙️ How I Solved It: Applied a modified binary search: Identify which half (left or right) is sorted. Narrow the search to the half where the target could exist. Repeat until the target is found or search space is exhausted. 📊 Performance Stats: ⏱ Runtime: 0 ms (⚡ beats 100%) 💾 Memory: 12.65 MB (beats 42.49%) ✅ Testcases Passed: 196 / 196 🧠 Key Takeaway: Understanding array properties like rotation and leveraging binary search ensures optimal search performance in logarithmic time. #LeetCode #BinarySearch #RotatedArray #Problem33 #Python #DSA #100DaysOfCode #CodingJourney

🧩 Day 54 of #LeetCode365 Problem: 69. Sqrt(x) Category: Binary Search | Math Today’s problem was all about finding square roots… but without using math.sqrt() 😤 — because that’s too easy, right? So I made my own version using binary search. Basically, I played “guess the root” until my code said, “close enough, stop now.” 😎 💻 Approach: 👉 Set l = 0, r = x 👉 Use binary search to narrow down the integer part of √x 👉 Return the best guess (a.k.a. the floor of the real answer) ⚙️ Complexity: ⏱ O(log n) | 💾 O(1) 💡 Lesson: Sometimes you don’t need to know the exact answer — just the closest one that doesn’t crash your code 🤷♂️ #LeetCode #Python #CodingHumor #DSA #BinarySearch #ProgrammerHumor

Day 68: Search in 2D Sorted Matrix (Search Space Reduction) 🔎 I'm continuing the streak on Day 68 of #100DaysOfCode with a challenging matrix search problem! The task is to find a target value in an $m \times n$ matrix where both rows and columns are sorted in ascending order. The key to solving this efficiently is Search Space Reduction. Instead of performing a standard search, my solution uses a smart traversal technique: Starting Point: I begin the search at the top-right corner of the matrix. Decision Logic: If the current value equals the target, we stop. If the current value is greater than the target, the entire current column can be eliminated, so we move left. If the current value is less than the target, the entire current row can be eliminated, so we move down. This strategy eliminates one row or one column in every step, guaranteeing an optimal O(m + n) time complexity and O(1) extra space. #Python #DSA #Algorithms #Matrix #Search #100DaysOfCode #ProblemSolving

✅ Learned to solve “Remove Duplicates from Sorted Array” (in-place, O(n) time, O(1) space)! Sorted input means every duplicate sits next to its twin—perfect setup for the two-pointer pattern: scan once, write uniques forward, and return the count k while keeping the first k positions clean and ordered. What clicked: - Two pointers: one scans, one writes uniques forward - Skip repeats deterministically thanks to sorting - Edge cases covered: empty array, all duplicates, negatives, mixed ranges Level-ups next: “Remove Duplicates II” (allow at most twice) and “Remove Element” to deepen the pattern muscle. What’s your favorite twist on this technique? 🚀 #LeetCode #TwoPointers #Arrays #InPlace #DSA #Algorithms #InterviewPrep #ProblemSolving #TimeComplexity #CodingChallenge #Python #SoftwareEngineering

#96day of #100DaysOfCode 🌱 LeetCode 109: Convert Sorted List to Binary Search Tree Today’s challenge was about building balance — literally! Given a sorted linked list, we need to convert it into a height-balanced BST 🌳 🧠 Key Idea: The middle element of the list becomes the root. Left half forms the left subtree, right half forms the right subtree. Use slow–fast pointers to find the middle efficiently. 📈 Complexity: Time: O(n log n) Space: O(log n) (recursion stack) 💬 Learning: Balance matters — in trees, in code, and in life 🌿 Sometimes, the middle point creates the strongest foundation. #LeetCode #DSA #BinarySearchTree #LinkedList #CodingChallenge #Python #Algorithm #LeetCodeDaily #100DaysOfCode