#PERMUTATIONS LEETCODE UPDATE#
🔗 The Approach: Slide & Update → Maintain Deque → Maximums Unveiled! 🔄🌠 🎉 Drumroll, please! The Deque held the sliding window's maximums, as revealed by the front element. It's like curating the perfect view! 🌆🔝
🔄 With each slide, I confidently popped elements from the back if they were smaller, ensuring the front held the max.
➡ As I slid through the array, the Deque gracefully maintained the window's integrity and the maximums within. Think of it as a window that lets you peek both ways! 🪟👀
🔄 The secret weapon: Deque (Double-Ended Queue). 💡 Problem detected! Task at hand: finding maximums in sliding windows. 💡 My Strategy: Hey #CodeCrafters, time to unveil my winning technique using the Deque approach! 🔄🔝 🚀 Just crushed the LeetCode challenge of the day! 🎉 "Sliding Window Maximum" – consider yourself conquered! 💪🧠 This version adds a touch of playfulness to the explanation, inviting your LinkedIn audience to join you in the world of algorithms and coding. Let's keep learning and coding! 🎉💻 #leetcode #BreadthFirstSearch #ProblemSolving #AlgorithmMagic #leetcode2023 #datastructuresandalgorithms #dsa #complexity #logicalthinkingĬustomize the emojis and style as you see fit.
#PERMUTATIONS LEETCODE FREE#
🤖 Feel free to raise your hand if you have questions or want to dive deeper into this ocean of coding artistry. 🧙♂️ An algorithmic symphony with infinite potential in solving puzzles that involve graphs and networks. This C++ wizardry taps into the power of BFS, elegantly conquering the "01 Matrix" challenge. 🎟️ The cell with its newfound fame joins the queue, bringing along its updated distance, and 'vis' recording is set accordingly.ħ️⃣ **Curtain Call:** As the final act, the queue empties, and our masterpiece is ready! The 'vis' vector contains the golden map of distances. If it's within bounds, uncharted territory in 'vis', and holds the value '1', it gets a VIP pass. 🕺 We dance through four directions using a loop, checking each cell's eligibility. The distance gets its own spotlight as 'step'. 🎭 Grab the front element from the queue and label its coordinates as 'row' and 'col'. Every iteration follows these choreographed steps: For each '0', we append its details to 'q', setting distance to '0', and update the 'vis' vector.ĥ️⃣ **Directional Dance:** Enter 'delrow' and 'delcol', arrays that map the moves needed to explore adjacent cells.Ħ️⃣ **BFS Ballet:** The heart of the action begins! BFS unfolds within a loop that operates while the queue 'q' isn't empty. Each pair bundles cell indices and distance from the closest '0'.Ĥ️⃣ **Seeding BFS:** Next, we loop through the grid, identifying '0' cells. 🚀 Here's a step-by-step guide to demystify the code:ġ️⃣ **Initialization:** We kick things off by defining 'n' and 'm' as the grid's rows and columns, respectively.Ģ️⃣ **Visualizing Vis:** The next step involves creating a 2D vector 'vis' of size 'n x m', acting as our GPS to track cell distances from the nearest '0'.ģ️⃣ **Queue Quest:** We introduce an empty queue 'q' that stores pairs. Hey LinkedIn crew! 👋 Let's unravel the magic behind a C++ implementation using Breadth-First Search (BFS) to conquer the "01 Matrix" challenge. 🔍 **Decoding BFS in C++: Navigating the "01 Matrix"** #leetcode #leetcode2023 #coding #share #logicalthinking #problemsolving #dsa If you want to try out ,the link to the problem is in the comments 💭. This is because the function uses a recursive call stack with a maximum depth of n, and each call to the function takes constant space.
#PERMUTATIONS LEETCODE CODE#
✨Space Complexity: The space complexity of this code is O(n), where n is the size of the input vector nums. This is because there are n! permutations of a vector of size n, and the function generates each permutation by recursively swapping elements, taking O(n) time for each permutation. ✨Time Complexity: The time complexity of this code is O(n*n!), where n is the size of the input vector nums. I hope this helps you understand how this code works! This generates all possible permutations by recursively swapping elements. The function then iterates over all elements from index idx to the end of the vector, swapping the element at index i with the element at index idx, calling itself recursively with idx+1, and then swapping the elements back. The base case for the recursion is when idx is equal to the size of nums, in which case the current permutation of nums is added to ans. It takes as input the vector nums, the vector of vectors ans, and an index idx. The function getPermuation is a recursive helper function that generates the permutations. The function permute takes as input a vector of integers nums and returns a vector of vectors ans containing all the permutations of nums.
Here's a step-by-step explanation of the approach: Hey everyone Today I solved problem 46. Permutations on LeetCode using C++.
0 kommentar(er)
