Optimizing Min Stack for O(1) Retrieval with Auxiliary MinStack

Day 27 of my #30DayCodeChallenge: The Efficiency of Binary Exponentiation! The Problem: Pow(x, n). Implementing the power function to calculate x". While it sounds simple, the challenge lies in handling large exponents (up to 231 - 1) and negative powers without hitting time limits or overflow. The Logic: This problem is a classic example of Divide and Conquer optimized through Binary Exponentiation (also known as Exponentiation by Squaring): 1. Bitwise Breakdown: Instead of multiplying x by itself n times (O(n)), we decompose n into powers of 2. For example, x13 is x8. x4. x¹. This brings our complexity down to O(log n). 2. The Iterative Jump: In every iteration of the loop, we square the current base (x = x x). If the current bit of n is 1 (checked via n & 1), we multiply our result by the current base. 3. Handling the Edge Cases: * Negative Exponents: If n is negative, we calculate xI" and then take the reciprocal (1/result). Overflow: We use a long for n during calculation to avoid overflow when converting -2, 147, 483, 648 to a positive value. The Calculation: By halving the power at each step, we transform a task that could take 2 billion operations into one that takes just 31. One step closer to mastery. Onward to Day 28! #Java #Algorithms #DataStructures #BinaryExponentiation #ProblemSolving #150DaysOfCode #SoftwareEngineer

Modern businesses need scalable and efficient systems to grow. At QuibiXlab LLC, we design and implement cloud-native architectures using technologies like Kubernetes, Go, and Python. Our focus: ✔️ Scalable infrastructure ✔️ Optimized backend systems ✔️ Automation-driven workflows With experience in enterprise environments, including exposure to companies like Coforge, we aim to bring reliable and efficient solutions to growing businesses. Connect with us to explore how the right architecture can support your growth.

Continuing my DSA journey, * Problem: Number of Atoms (726) * Approach: -I solved this using a stack-based parsing approach. -I used a stack of maps to keep track of atom counts at different levels. While iterating through the string: -On '(' → push a new map onto the stack -On ')' → pop the top map, multiply counts, and merge back -For elements → parse the name and count, then update the current map * Key Insight: -The tricky part was handling nested parentheses and applying multipliers correctly. -Using a stack helps manage different scopes efficiently. * What I Learned: This problem improved my understanding of string parsing with stacks. #LeetCode #DSA #Java #Stack #Strings #ProblemSolving

🔥 Today’s DSA Update— #Day58 #MonotonicStack Today I worked on finding Next Greater Element I (Leetcode 496). 💡 What I understood This pattern is useful for problems where we need to find things like: 1. Next greater element 2. Next smaller element Instead of using nested loops (which take O(n²)), we can solve it in O(n) using a stack. 💡 The key idea We use a stack to keep elements that are still “waiting” for their answer. When we find a bigger element: 1. We pop all smaller elements from the stack 2. For each popped element, we now know its next greater value 3. Each element is pushed and popped only once. 🧠 What I learned today Today’s lesson was that stacks are not just for push and pop operations — they can be used to store state while processing elements. This pattern feels different from normal array traversal because we are managing a “waiting list” of elements. #Java #DSA #Stack #DataStructures #LeetCode #ConsistencyCurve

Day 101: Clean Code is Scalable Code 🚀 Problem 3741: Minimum Distance Between Three Equal Elements II They say if you solve a problem correctly once, the "Hard" version becomes easy. Today was proof of that. The Strategy: • Reusable Logic: My solution from yesterday handled the transition to Part II perfectly. The core logic of tracking indices in a HashMap and applying a sliding window of three was already optimal. • The "Part II" Test: Even with increased constraints or complexity in the problem description, O(N) frequency tracking and index-gap calculations proved to be the robust way to go. • Efficiency: By isolating only the elements that appeared 3+ times, I kept the runtime minimal and the memory footprint low. It's a great feeling when the architecture you built yesterday is strong enough to handle today's challenge without a single line of refactoring. Day 101 and the momentum is only increasing. ⚡ #LeetCode #Java #Algorithms #DataStructures #ProblemSolving #DailyCode

💡 Day 57 of LeetCode Problem Solved! 🔧 🌟 Maximum Distance Between a Pair of Values 🌟 🔗 Solution Code: https://lnkd.in/gSP_QePE 🧠 Approach:  • Two-Pointer Strategy: Maintain two indices (i for nums1 and j for nums2) to scan both non-increasing arrays in a single efficient pass. • Greedy Search: If nums1[i] <= nums2[j], calculate j - i and expand j to find the maximum possible distance. If the value in nums1 is too large, move i forward. ⚡ Key Learning:  • Leveraging the non-increasing nature of arrays with two pointers eliminates the need for nested loops, reducing complexity from O(N^2) to linear O(N+M). ⏱️ Complexity:  • Time: O(n + m) • Space: O(1) #LeetCode #Java #DSA #ProblemSolving #Consistency #100DaysOfCode #CodingJourney #Algorithms #TwoPointers

🚀 Day 567 of #750DaysOfCode 🚀 🔍 Problem Solved: Maximum Distance Between a Pair of Values Today’s challenge was about finding the maximum distance (j - i) such that: ✔️ i ≤ j ✔️ nums1[i] ≤ nums2[j] ✔️ Both arrays are non-increasing 💡 Key Insight: Since both arrays are sorted in descending order, we can avoid brute force and use a Two Pointer approach to achieve optimal performance. 🧠 Approach: Initialize two pointers i and j at 0 If nums1[i] ≤ nums2[j] → valid pair → update distance & move j Else → move i forward Maintain j ≥ i at all times 📊 Complexity: Time: O(n + m) Space: O(1) 🔥 Takeaway: Whenever arrays are sorted, always think of two pointers or binary search before jumping to brute force. This simple shift can reduce complexity from O(n²) → O(n)! #Day567 #750DaysOfCode #LeetCode #Java #DataStructures #Algorithms #TwoPointers #CodingJourney #ProblemSolving

Day 100: Consistency, Growth, and a Milestone 💯 Problem 3740: Minimum Distance Between Three Equal Elements I Today marks 100 days of continuous problem-solving. While the number is just a marker, the real value has been the daily discipline of opening the IDE and tackling whatever challenge LeetCode throws my way. The Strategy: • Frequency Grouping: I used a HashMap to store a list of indices for every unique number in the array. This allowed me to isolate potential triplets instantly. • Sliding Window Logic: For any number appearing three or more times, I looked at a sliding window of three consecutive indices (i,i+1,i+2). • The Formula: Through testing, I identified that the minimum distance between three equal elements can be derived from the span between the first and third occurrence: (index of 3rd - index of 1st) * 2. • Optimization: By iterating through the pre-grouped index lists, I kept the solution efficient and clean. Reflecting on these 100 days, I’ve moved from basic simulations to complex optimizations like Square Root Decomposition and 3D DP. The goal isn't to stop here—it's to keep building, keep optimizing, and keep growing. 🚀 #LeetCode #Java #Algorithms #DataStructures #100DaysOfCode #Consistency #ProblemSolving #DailyCode

Day 49 of Daily DSA 🚀 Solved LeetCode 234: Palindrome Linked List ✅ Problem: Given the head of a singly linked list, return true if it is a palindrome, otherwise false. Approach: Used a stack-based method to compare linked list values from both directions. Steps: Traverse the linked list and push all node values into a stack Start again from head Pop values one by one from stack Compare current node value with popped value If mismatch found → return false If all match → return true ⏱ Complexity: • Time: O(n) • Space: O(n) 📊 LeetCode Stats: • Runtime: 15 ms • Memory: 101.66 MB Sometimes using extra space can make the logic cleaner and easier to implement. #DSA #LeetCode #Java #LinkedList #Stack #CodingJourney #ProblemSolving

🔢 Day 47/75: Optimizing with Brian Kernighan’s Algorithm! Today’s challenge was counting the number of "1" bits in an integer. While the naive approach is to check all 32 bits one by one, I implemented a much more elegant solution known as Brian Kernighan’s Algorithm. The Logic: The expression n = n & (n - 1) has a fascinating property: it always flips the least significant set bit (the rightmost 1) to 0. By running this in a loop until $n$ becomes 0, the number of iterations equals exactly the number of set bits. If a number has only three "1" bits, the loop runs 3 times, not 32. The Result: ✅ Runtime: 0 ms (Beats 100.00% of Java users) ✅ Efficiency: $O(k)$ time complexity, where $k$ is the number of set bits. Mastering these bitwise shortcuts is what separates a good solution from a great one! 🚀 #75DaysOfCode #LeetCode #Java #BitManipulation #Algorithms #ComputerScience #Optimization #SoftwareEngineering