"Next Permutation in Python: A DSA Challenge"

🚀 DSA Challenge – Day 80 Problem: Maximum Distance Between Valid Pairs 🌊📏 Today’s problem tested the combination of binary search and array monotonicity — finding the farthest valid pair between two non-increasing arrays! 🧠 Problem Summary: We’re given two non-increasing arrays, nums1 and nums2. A pair (i, j) is valid if: i ≤ j, and nums1[i] ≤ nums2[j]. The goal is to find the maximum distance (j - i) among all valid pairs. ⚙️ My Approach: 1️⃣ Iterate through each element in nums1. 2️⃣ Use binary search on nums2 to find the farthest valid index satisfying the condition. 3️⃣ Keep track of the maximum j - i distance encountered. This solution leverages the sorted (non-increasing) property of arrays for logarithmic efficiency. 📈 Complexity: Time: O(n log m) → For each element in nums1, a binary search on nums2. Space: O(1) → Only a few variables used. ✨ Key Takeaway: Sometimes, monotonic properties allow you to blend binary search with iteration — turning what looks like a brute-force problem into a clean and efficient search-based solution. ⚡ 🔖 #DSA #100DaysOfCode #LeetCode #ProblemSolving #Algorithms #BinarySearch #TwoPointers #CodingChallenge #Python #InterviewPrep #TechCommunity #Optimization #EfficientCode #CodeEveryday

⚡ Day 87 of #100DaysOfDSA – 3Sum Closest 🔍 📌 Problem. no 16: Given an integer array nums and an integer target, find the sum of three integers in nums such that the sum is closest to the target. Implemented an optimized Two-Pointer Approach after sorting the array. ⚡ Runtime: 431 ms – Beats 75.40% of submissions 💾 Memory: 12.59 MB – Beats 39.70% of solutions 💻 Language Used: Python ✅ Status: Accepted – All 106 test cases passed successfully! This problem helped me strengthen my skills in pointer manipulation, array traversal, and optimization of nested loops. It was a great exercise in precision and efficiency under constraints. 🔑 Key Learnings: ✔️ Learned to balance between accuracy and performance ✔️ Gained deeper understanding of sorting and two-pointer techniques ✔️ Improved debugging and edge-case analysis for numerical problems 🎯 Day 87 — A logical yet challenging problem that sharpened my optimization mindset and coding discipline! 💪🔥 #100DaysOfCode #100DaysOfDSA #LeetCode #PythonProgramming #DataStructures #AlgorithmPractice #CodeNewbie #DailyCoding #ProblemSolving #TechJourney #WomenWhoCode #CodeLife #CodingChallenge #DSAChallenge #DeveloperLife #PythonDeveloper #LearnToCode #CodingCommunity #CodeEveryday #SoftwareEngineering #KathirCollegeOfEngineering #KathirCollege #BTechLife #AIDS #FutureEngineer #CodingMotivation #ProgrammingSkills

🚀 DSA Challenge – Day 85 Problem: Check if All Integers in a Range Are Covered ✅📏 This problem was an elegant use of the Prefix Sum technique, where I used range updates to efficiently check coverage over an interval. 🧠 Problem Summary: You are given several inclusive integer intervals and a target range [left, right]. You must verify if every integer within [left, right] is covered by at least one of the given intervals. ⚙️ My Approach: 1️⃣ Initialize an array line to track coverage at each integer position. 2️⃣ For every range [a, b], increment line[a] and decrement line[b + 1] — this marks the start and end of coverage. 3️⃣ Convert line into a prefix sum array, so each position reflects how many intervals cover that number. 4️⃣ Finally, iterate through [left, right] to ensure each integer has coverage (> 0). 📈 Complexity: Time: O(n + 52) → Linear scan and prefix sum computation. Space: O(52) → Fixed-size array since ranges are small. ✨ Key Takeaway: Prefix sum is not just for subarray sums — it’s a powerful trick for range marking and coverage problems, offering O(1) updates and O(n) verification. ⚡ 🔖 #DSA #100DaysOfCode #LeetCode #PrefixSum #RangeUpdate #ProblemSolving #Algorithms #CodingChallenge #Python #EfficientCode #Optimization #TechCommunity #InterviewPrep #CodeEveryday #LearningByBuilding

🚀 DSA Challenge – Day 90 Problem: Single Element in a Sorted Array ⚙️🔍 This problem was a clever application of binary search where understanding the index patterns of paired elements was the key to achieving logarithmic efficiency. 🧠 Problem Summary: You are given a sorted array where every element appears exactly twice, except for one element that appears only once. Your task: return that single element in O(log n) time and O(1) space. ⚙️ My Approach: 1️⃣ Used binary search to narrow down the segment containing the single element. 2️⃣ Checked if the middle element breaks the pairing rule — if so, that’s our unique number. 3️⃣ Observed the pattern: Before the single element, pairs start at even indices. After it, pairs start at odd indices. 4️⃣ Used this parity observation to adjust the search boundaries efficiently. 📈 Complexity: Time: O(log n) → Binary search halves the search space each step. Space: O(1) → Constant space used. ✨ Key Takeaway: Sometimes, the structure of sorted pairs reveals more than meets the eye — a small parity trick transforms a linear scan into a logarithmic search. ⚡ 🔖 #DSA #100DaysOfCode #LeetCode #ProblemSolving #BinarySearch #Algorithms #CodingChallenge #Python #Optimization #EfficientCode #TechCommunity #InterviewPrep #LearningByBuilding #CodeEveryday

🚀 DSA Challenge – Day 89 Problem: Subsets II (Handling Duplicates in Power Set) ⚙️✨ This problem was a great exploration of recursion, backtracking, and how to systematically avoid duplicate subsets using careful pruning. 🧠 Problem Summary: You are given an integer array nums that may contain duplicates. Your goal: return all possible subsets (the power set) without including any duplicate subsets. ⚙️ My Approach: 1️⃣ First, sort the array — this step helps group duplicates together, which is crucial for skipping them efficiently. 2️⃣ Use recursive backtracking to explore all possible inclusion/exclusion combinations. 3️⃣ Whenever a duplicate element is found (i.e., nums[i] == nums[i-1]), skip it if it’s at the same recursion depth — ensuring unique subset generation. 4️⃣ Keep track of the current subset and add a copy to the result whenever we reach a new state. 📈 Complexity: Time: O(2ⁿ) → Each element can be either included or excluded. Space: O(n) → For recursion and temporary subset storage. ✨ Key Takeaway: Sorting before recursion is often the simplest way to handle duplicates in backtracking problems. With one small condition check, you can turn exponential chaos into structured exploration. ⚡ 🔖 #DSA #100DaysOfCode #LeetCode #ProblemSolving #Recursion #Backtracking #Algorithms #CodingChallenge #Python #TechCommunity #InterviewPrep #EfficientCode #LearningByBuilding #CodeEveryday

#Day7 of My #100DaysInterviewPrep 🚀 Explored matrix-based problems today — sharpening 2D array handling and traversal techniques. Today's Questions: 1️⃣ Matrix Diagonal Sum (LeetCode #1572) – Summed both primary & secondary diagonals, careful to avoid double-counting the center. ⏱️ Time: O(n), 💾 Space: O(1). 🔑 Reinforced iteration over diagonal patterns. 2️⃣ Search a 2D Matrix (LeetCode #74) – Treated matrix as a sorted 1D array and applied binary search. ⏱️ Time: O(log(m·n)), 💾 Space: O(1). 🔑 Practiced binary search in 2D space. 3️⃣ Transpose Matrix (LeetCode #867) – Flipped rows into columns. ⏱️ Time: O(m·n), 💾 Space: O(m·n). 🔑 Understood row/column swapping patterns. 4️⃣ Toeplitz Matrix (LeetCode #766) – Verified if all diagonals from top-left to bottom-right are constant. ⏱️ Time: O(m·n), 💾 Space: O(1). 🔑 Learned efficient diagonal checks without extra memory. ✨ Key takeaway: Matrix problems build strong intuition for multi-dimensional thinking, efficient traversal, and space optimization. Grateful to Abhishek Kumar and K.R. Mangalam University for their continuous guidance 🙏 #100DaysOfCode #Day7 #LeetCode #Matrices #DSA #Python #CodingChallenge #InterviewPrep #PlacementReady

🧩 Day 42 — Add Digits (LeetCode 258) 📝 Problem Given an integer num, repeatedly add all its digits until the result has only one digit, and return it. 🔁 Approach -Continuously sum the digits of the number until a single-digit result remains. -Alternatively, use the digital root formula: -If num == 0, return 0. -Else, return 1 + (num - 1) % 9. 📊 Complexity -Time Complexity: O(1) -Space Complexity: O(1) 🔑 Concepts Practiced -Mathematical optimization -Modulo operation -Digital root pattern #Leetcode #python #DSA #ProblemSolving #Math #Optimization

🚀 DSA Challenge – Day 97 Problem: Roman to Integer 🔢🏛️ This problem combines string parsing with numerical logic — a classic test of how well you can translate human-readable patterns into computational rules. 🧠 Problem Summary: Given a Roman numeral, convert it into an integer. Roman symbols and their values: Symbol Value I 1 V 5 X 10 L 50 C 100 D 500 M 1000 Special subtraction rules: I before V (5) or X (10) → 4 or 9 X before L (50) or C (100) → 40 or 90 C before D (500) or M (1000) → 400 or 900 ⚙️ My Approach: Maintain a mapping of Roman symbols to integer values. Traverse each symbol and push its value to a stack. If the current value is greater than the previous, subtract the previous from the current before pushing. Finally, sum up all values in the stack. 📈 Complexity Analysis: Time: O(n) — traverse each character once. Space: O(n) — stack stores values temporarily. ✨ Key Takeaway: This problem teaches how to recognize patterns and exceptions while converting symbolic representations into numerical logic — a very common theme in real-world parsing tasks. 🔖 #DSA #100DaysOfCode #LeetCode #RomanToInteger #ProblemSolving #Python #Algorithms #CodingChallenge #TechCommunity #Programming #InterviewPrep #LearningEveryday