Next Permutation Algorithm Explanation

🚀 Day 31 / 100 | Check Digitorial Permutation -Intuition: Permutation does not change the factorial sum. So instead of checking all permutations, compute the factorial sum once and check if its digits are the same as the original number’s digits. If both have same digits ,permutation exists and answer is true. -Approach: O(n) Check factorial for each digit. Traverse the number and calculate sum of factorial of digits. Store original number separately. Now compare digit frequency of original number and factorial sum If frequencies match ,return true Else, return false -Complexity: Time Complexity: O(n) n->number of digits in num. Space Complexity: O(1) #100DaysOfCode #Java #DSA #LeetCode

Day 7 of Daily DSA 🚀 Solved LeetCode 167: Two Sum II – Input Array Is Sorted Approach: Used the two-pointer technique leveraging the sorted nature of the array. Moved pointers inward based on the comparison with the target to find the answer in one pass. Complexity: • Time: O(n) • Space: O(1) (constant extra space) LeetCode Stats: • Runtime: 2 ms (Beats 96.10%) • Memory: 48.54 MB (Beats 38.87%) This problem highlights how understanding input constraints can turn a brute-force solution into an optimal one. #DSA #LeetCode #Java #TwoPointers #ProblemSolving #Consistency

📅 Day 90 of #100DaysOfLeetCode Problem: 3634. Minimum Removals to Balance Array Difficulty: Medium Key Insight: Instead of deciding which elements to remove, focus on keeping the largest possible balanced subarray where max ≤ min × k. Minimum removals = total elements − size of this subarray. Approach: Sort the array to make min/max tracking easy Use a sliding window Expand the right pointer Shrink the left pointer whenever nums[j] > k × nums[i] Track the maximum valid window length Complexity: Time: O(n log n) Space: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

Day 69/100 – LeetCode Challenge ✅ Problem: #229 Majority Element II Difficulty: Medium Language: Java Approach: Extended Boyer-Moore Voting Algorithm Time Complexity: O(n) Space Complexity: O(1) Key Insight: At most two elements can appear more than ⌊n/3⌋ times. Track two candidates and their counts simultaneously. When a new element matches neither candidate, decrement both counts. Solution Brief: Phase 1: Find two potential candidates: Maintain two candidates and their counts Update based on matches or count resets Phase 2: Verify both candidates actually appear > n/3 times Add valid candidates to result list. #LeetCode #Day69 #100DaysOfCode #Array #Java #Algorithm #CodingChallenge #ProblemSolving #MajorityElementII #MediumProblem #BoyerMoore #VotingAlgorithm #DSA

🚀 Day 16 of #100DaysOfCode Solved Remove Nth Node from End of List on LeetCode 🔗✂️ 🧠 Key insight: To remove the Nth node from the end, we need to carefully handle edge cases—especially when the node to remove is the head or when n equals the list length. ⚙️ Approach: 🔹First determine the length of the linked list 🔹Calculate the position from the start 🔹Traverse to the node just before the target 🔹Adjust pointers to remove the node safely ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 67/100 – LeetCode Challenge ✅ Problem: #228 Summary Ranges Difficulty: Easy Language: Java Approach: Single Pass with Range Detection Time Complexity: O(n) Space Complexity: O(1) excluding output Key Insight: Track start of current consecutive range. When gap found (next ≠ previous + 1), close current range: Single number if start == previous Range "start → previous" otherwise Solution Brief: Iterated through array comparing consecutive elements. Detected breaks in consecutive sequences. Formatted ranges appropriately with arrow for multi-number ranges. Added final range after loop. #LeetCode #Day67 #100DaysOfCode #Array #Java #Algorithm #CodingChallenge #ProblemSolving #SummaryRanges #EasyProblem #RangeDetection #SinglePass #DSA

🚀 Day 18 of #100DaysOfCode Solved Remove Duplicates from Sorted List II on LeetCode 🔗 🧠 Key insight: When duplicates appear in a sorted linked list, we need to remove all occurrences, not just one. Using a dummy (sentinel) node makes it easier to handle cases where duplicates start at the head. ⚙️ Approach: 🔹Create a dummy node pointing to the head 🔹Traverse the list with two pointers 🔹If a duplicate sequence is found, skip the entire block 🔹Otherwise, move forward normally ⏱️ Time Complexity: O(n) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #LinkedList #Java #ProblemSolving #LearningInPublic #CodingJourney

🚀 Day 9 of #100DaysOfCode Solved Search in Rotated Sorted Array II on LeetCode using Modified Binary Search (Handling Duplicates) 🔎 🧠 Key insight: With duplicates present, it’s sometimes impossible to identify the sorted half directly. If nums[left] == nums[mid], we safely move left++ to shrink the search space. ⚙️ Approach: 🔹Apply binary search 🔹Handle duplicates by skipping equal boundary values 🔹Identify sorted half and check if target lies within it 🔹Continue search accordingly ⏱️ Time Complexity: O(log n) average, O(n) worst case (due to duplicates) 📦 Space Complexity: O(1) #100DaysOfCode #LeetCode #DSA #BinarySearch #Java #ProblemSolving #LearningInPublic #CodingJourney

Day 11 – LeetCode 567 | Permutation in String 🔁 Problem: Check whether one string’s permutation exists as a substring in another string. Approach Used: 👉 Sliding Window + Frequency Difference Array Core Insight (most people miss this): Window size must be exactly s1.length() When a character enters the window → decrement frequency When a character leaves the window → increment frequency If all frequencies become zero, we found a permutation #Day11 #LeetCode567 #SlidingWindow #DSA #Java #ProblemSolving #InterviewPrep #CodingJourney