How to Rotate an Array in O(n) Time and O(1) Space

Day 17/30 — Circular Array Rotation 🔁 Rotate an array K times to the right and answer queries asking what value sits at a given index after rotation. Instead of doing K physical rotations (which is wasteful when K is large), compute where each queried index came from in the original array using modular arithmetic. It’s simple, efficient, and perfect for interview explanations. Why it’s neat: for a rotated index m, the element there originally lived at origin = (m - (K % n) + n) % n So every query is O(1). Reduce K modulo n to handle large K. This yields overall time O(n + q) and O(1) extra space (ignoring input).

Topic - Arrays 📈 Day 21 🧠 Questions - 4 Sum Problem Statement : Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that: 0 <= a, b, c, d < n a, b, c, and d are distinct. nums[a] + nums[b] + nums[c] + nums[d] == target You may return the answer in any order. Input: nums = [1,0,-1,0,-2,2], target = 0 Output: [[-2,-1,1,2],[-2,0,0,2],[-1,0,0,1]] TC: O(N3) SC: O(No. of quads) -> this space is used to store and return the answer. No extra space is used to solve the problem

You know that feeling when you react just a bit too late? Well, it's not just you! Traditional vibration controllers, while great for damping, often struggle with attenuating the first cycle peak after an impulse disturbance, which can be critical if certain displacement limits must not be exceeded. To tackle this, I trained a machine learning model using random forest to predict the first cycle peak and zero-crossing time very early in the vibratory response. I used the predictions to generate a counter pulse that significantly reduces the first cycle peak. Finally, I integrated a PID controller to handle residual vibrations and improve settling time. The result is a control system that mitigates the initial peak and settles efficiently. While I achieved this with synthetic, noise-free data, and a real-life system would be more complex, this approach demonstrates the potential of early detection. By using early-stage output parameters to build a regression model mapping them to later-stage outputs, we can predict a system's response to disturbance and mitigate the effects of the disturbances without access to measurements of the disturbance. This approach could prove invaluable in solving many engineering challenges. #VibrationControl #MachineLearning #ControlSystems #PredictiveControl #ActiveControl #SystemDynamics #RandomForest #EngineeringInnovation https://lnkd.in/gdP4Ejqe

GFG=>Tree boundary traversal 🌳 medium level solved with O(n) time complexity approach ->do the traversal(obviously) ->create 3 functions ,function 1=travelling towards left node and pushing node->data into an array(answer),function 2=find the leaf nodes and push them into answer, function 3= same as 1st one but in reverse manner.. ->return the answer; by following these steps you will get one error....try and resolve that ;) #aimless

In this Video you will learn SPATIAL ANALYSIS TOOLS: These tools help us analyze relationships between spatial features for example, identifying high elevation areas, calculating distance from rivers, or combining different raster datasets. BUFFER: The Buffer tool creates a zone around a feature at a specific distance CLIP: It works like a cookie cutter, it cuts one layer using the shape of another layer. INTERSECT: The Intersect tool finds the overlapping area between two or more layers. UNION: The Union tool merges all polygons from multiple layers into a single layer,keeping all attributes. https://lnkd.in/dxBH69i7

#CODETOBER — Day 12, Problem 3 📌 Problem: Product of Array Except Self 🔗 [https://lnkd.in/g6YwjJQB) My Approach: • Created an ans array initialized with 1s to store the result. • First loop: computed the left product for each index and stored it in ans. • Second loop (backward): maintained a variable right for the right product and multiplied it with ans[i] on the go. • This avoided using extra space for separate left/right arrays. • Time Complexity: O(n) • Space Complexity: O(1) (excluding output array)

This program takes an integer array, computes the square of each element, and then sorts the resulting values in non-decreasing order. It ensures that all numbers, including negatives, are transformed into their squared values and arranged in a properly ordered sequence for further processing.

611. Valid Triangle Number 611. Valid Triangle Number Difficulty: Medium Topics: Array, Two Pointers, Binary Search, Greedy, Sorting Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle. Example 1: Input: nums = [2,2,3,4] Output: 3 Explanation: Valid combinations are: 2,3,4 (using the first 2) 2,3,4 (using the second 2) 2,2,3 Example 2: Input: nums = [4,2,3,4] Output: 4 Constraints: 1 <= nums.length <= 1000 0 <= nums[i] <= 1000 Solution: We need to find the number of triplets in an array that can form a triangle. The key property of a triangle is that the sum of any two sides must be greater than the third side. Sort the Array: Sorting helps in efficiently checking the triangle condition. After sorting, we can fix the largest side and then check for pairs of smaller sides that satisfy the triangle inequality. Two Pointers Technique: For each element as the potential largest side (starting from the end of the sorte https://lnkd.in/g2y3sj-f

611. Valid Triangle Number 611. Valid Triangle Number Difficulty: Medium Topics: Array, Two Pointers, Binary Search, Greedy, Sorting Given an integer array nums, return the number of triplets chosen from the array that can make triangles if we take them as side lengths of a triangle. Example 1: Input: nums = [2,2,3,4] Output: 3 Explanation: Valid combinations are: 2,3,4 (using the first 2) 2,3,4 (using the second 2) 2,2,3 Example 2: Input: nums = [4,2,3,4] Output: 4 Constraints: 1 <= nums.length <= 1000 0 <= nums[i] <= 1000 Solution: We need to find the number of triplets in an array that can form a triangle. The key property of a triangle is that the sum of any two sides must be greater than the third side. Sort the Array: Sorting helps in efficiently checking the triangle condition. After sorting, we can fix the largest side and then check for pairs of smaller sides that satisfy the triangle inequality. Two Pointers Technique: For each element as the potential largest side (starting from the end of the sorte https://lnkd.in/g2y3sj-f

🔺 LeetCode 2221: Find Triangular Sum of an Array 📌 Problem: Starting with an integer array nums, repeatedly replace it with a new array where each element is (nums[i] + nums[i+1]) % 10, until only one element remains → return that as the triangular sum. 🧠 Key Idea At each step, the array size reduces by 1. Continue this until only one element remains. This can be solved in-place by updating the array row by row. No need for extra space → O(1) space. ⚡ Complexity Time: O(n²) (triangular reductions) Space: O(1) ✅ Result Accepted in 0 ms (Beats 100%) 🚀