Combination Sum II Problem Solution with Backtracking

🔹 Day 98 – LeetCode Practice 📌 Problem: Path Sum II (LeetCode #113) 📊 Difficulty: Medium 🧠 Problem Overview: Given the root of a binary tree and a target sum, the task is to find all root-to-leaf paths where the sum of node values equals the target. Each valid path should be returned as a list of values. ✅ Approach Used: Traversed the tree using depth-first search. Maintained the current path and cumulative sum while moving down the tree. When a leaf node was reached, checked if the sum matched the target. Used backtracking to explore all possible paths correctly. 📈 Submission Results: Status: Accepted ✅ Runtime: 2 ms Memory Usage: 45.54 MB 💡 Reflection: This problem is a great example of how recursion and backtracking work together in tree problems. Managing state carefully is key to collecting all valid paths without duplication. #LeetCode #BinaryTree #DFS #Backtracking #Java #DSA #CodingPractice

🔹 Day 85 – #100DaysOfLeetCode Problem: 3010. Divide an Array Into Subarrays With Minimum Cost I Difficulty: Easy Key Insight: The cost of a subarray depends only on its first element. Since the first subarray always starts at index 0, the problem reduces to selecting the two smallest possible starting elements from the remaining array. Approach: Fix the first subarray cost as nums[0] Find the smallest and second smallest values in nums[1…n-1] Add them to get the minimum total cost Time Complexity: O(n) Space Complexity: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

#200DaysOfCode – Day 108 Combination Sum III Problem:- Combination Sum III Task:- Find all valid combinations of k numbers that sum up to n, using only numbers from 1 to 9, where: Each number is used at most once No duplicate combinations are allowed Example: Input: k = 3, n = 7 Output: [[1, 2, 4]] My Approach:- Used Backtracking (DFS) to explore all possible combinations. Started from a given number to avoid duplicates. Stopped recursion when: The combination size exceeded k The sum became negative Added the combination to the result only when: Exactly k numbers were chosen The sum became 0 Time Complexity:- Exponential (bounded due to range 1–9) Space Complexity:- O(k) (recursive stack + temporary list) Backtracking may look complex at first, but with clear base conditions and pruning, it becomes a powerful and elegant tool for solving combination problems efficiently. #takeUforward #200DaysOfCode #Java #ProblemSolving #LeetCode #Backtracking #Recursion #DSA #CodingJourney #CodeNewbie

🔹 Day 91 – LeetCode Practice 📌 Problem: Three Consecutive Odds (LeetCode #1550) 📊 Difficulty: Easy 🧠 Problem Overview: Given an integer array, determine whether there are three consecutive odd numbers appearing next to each other. Return true if such a sequence exists; otherwise, return false. ✅ My Approach: Traversed the array while keeping track of consecutive odd numbers. Increased a counter whenever an odd number appeared and reset it when an even number was found. As soon as the counter reached three, the condition was satisfied. 📈 Complexity Analysis: Time Complexity: O(n) Space Complexity: O(1) 🚀 Submission Results: Status: Accepted ✅ Runtime: 0 ms (Beats 100%) 🚀 Memory: 44.18 MB (Beats 38.42%) 💡 Reflection: This problem highlights how a simple counter-based logic can efficiently solve pattern-detection tasks. Clean logic and early exits make solutions both fast and readable. #LeetCode #DSA #Java #ProblemSolving #CodingPractice #LearningJourney

🚀 Day 77 of #100DaysOfLeetCode 📌 Problem: 1877. Minimize Maximum Pair Sum in Array 📊 Difficulty: Medium 🧠 Key Insight: To minimize the maximum pair sum, pair the smallest element with the largest, the second smallest with the second largest, and so on. This balances the sums and avoids any single pair becoming too large. ⚙️ Approach: Sort the array Use two pointers: i starting from the beginning j starting from the end Pair nums[i] + nums[j] and track the maximum sum Move both pointers inward until all pairs are formed ⏱️ Complexity: Time: O(n log n) Space: O(1) #LeetCode #Java #ProblemSolving #CodingChallenge #100DaysOfCode #DSA #LearningEveryday

 Day 10/100 Q 12 – Remove Duplicates from Sorted Array (DSA) Today’s problem was all about in-place array manipulation and two-pointer technique. 📌 Problem Statement Given a sorted integer array, remove duplicates in-place so that each unique element appears only once and return the count of unique elements. 🧠 Key Insight Since the array is already sorted, duplicates appear next to each other. Using two pointers, we can overwrite duplicates and keep all unique elements at the beginning of the array without using extra space. ⚙️ Approach Use one pointer to track the last unique element Traverse the array with another pointer Update the array when a new unique element is found ⏱ Complexity Time: O(n) Space: O(1) 💡 What I Learned How sorting simplifies duplicate problems Practical use of the two-pointer technique Writing space-optimized solutions 🔁 On to the next challenge! Consistency > Motivation 💪 #100DaysOfDSA #DSA #Arrays #TwoPointers #Java #ProblemSolving #LeetCode #CodingJourney #DailyLearning

🔹 Day 97 – LeetCode Practice 📌 Problem: Divide Array Into Equal Pairs (LeetCode #2206) 📊 Difficulty: Easy 🧠 Problem Overview: You’re given an integer array containing 2n elements. The goal is to check whether the array can be divided into n pairs such that: Every element is used exactly once Both elements in each pair are equal ✅ Approach Used: Sorted the array to bring identical elements together. Traversed the array while counting occurrences of each number. Verified that every number appears an even number of times, ensuring valid pairs. 📈 Submission Results: Status: Accepted ✅ Runtime: 8 ms Memory Usage: 46.94 MB 💡 Reflection: This problem is a great reminder that sorting can simplify pairing logic significantly. Once the array is ordered, validating pairs becomes straightforward and efficient. #LeetCode #ProblemSolving #Arrays #Java #DSA #CodingPractice #Consistency

Day 29/100 ✅ LeetCode 138: Copy List with Random Pointer Solved this problem by creating a deep copy of a linked list where each node contains an additional random pointer. The approach ensures that both next and random pointers are copied correctly without modifying the original list. Key steps involved: Create copied nodes Link copied nodes with original nodes Assign random pointers efficiently Separate the original and cloned lists This problem helped me strengthen my understanding of linked list manipulation and pointer mapping. Solution:-https://lnkd.in/gD9TWWsg #LeetCode #LeetCode138 #CopyListWithRandomPointer #LinkedList #DSA #Java #ProblemSolving #CodingPractice #DeepCopy #DataStructures #Day29

Day 2/25 – LeetCode Challenge 🚀 🔸Problem : Merge Sorted Array 🔸Difficulty : Easy 🔸Topic : Arrays, Three Pointers 🔸Language : Java Approach 🛠️ : ▫️Used a three-pointer technique starting from the end of both arrays. ▫️Compared elements from nums1 and nums2 and placed the larger one at the correct position. ▫️By filling nums1 from the back, avoided overwriting existing values. ▫️Continued until all elements from nums2 were merged. ▫️This ensures an in-place solution with optimal time complexity. Key Learnings 📚: 🔹Efficient use of two pointers for sorted arrays 🔹Importance of reverse traversal in in-place problems 🔹How to avoid extra space while merging arrays 🔹Reinforced understanding of array manipulation #25DaysOfLeetCode #LeetCode #DSA #Java #ProblemSolving #Coding #Consistency

#200DaysOfCode – Day 107 Subsets with Duplicates Problem:- Subsets II (LeetCode 90) Task: Given an integer array nums that may contain duplicates, return all possible unique subsets (power set). Example: Input: nums = [1,2,2] Output: [[], [1], [1,2], [1,2,2], [2], [2,2]] My Approach: Sorted the array to bring duplicate elements together. Used Backtracking to generate all subsets. Skipped duplicate elements at the same recursion level to avoid repeating subsets. Complexity Analysis: Time Complexity: O(2^N) Space Complexity: O(N) (recursion stack, excluding output) Handling duplicates becomes much easier when the input is sorted and recursion levels are controlled carefully. Backtracking problems look complex at first but the logic becomes clean once broken into steps. #takeUforward #100DaysOfCode #Java #LeetCode #ProblemSolving #Backtracking #Recursion #DSA #CodingJourney #CleanCode #DeveloperLife