Turn weights file to executable
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✨ DAY 15:) #codingjourney 35. Search Insert Position ✨ Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You must write an algorithm with O(log n) runtime complexity. Example 1: Input: nums = [1,3,5,6], target = 5 Output: 2 Example 2: Input: nums = [1,3,5,6], target = 2 Output: 1 Example 3: Input: nums = [1,3,5,6], target = 7 Output: 4 Constraints: 1 <= nums.length <= 104 -104 <= nums[i] <= 104 nums contains distinct values sorted in ascending order. -104 <= target <= 104
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Leetcode problem of the day : 2 Keys Keyboard The problem that we are tackling for today is that to find the minimum number of steps required to get exactly n 'A' characters on a text editor starting with a single 'A'. The available operations are "copy all" and "paste". The challenge is to determine how to optimally reach n characters with the fewest steps. Approach used: 1. Initialization: - We start with 1 'A' on the text editor. - We initialize curr to represent the current number of 'A's, ans to track the number of steps, and curr_copy to store the most recent copy. 2. Loop Until Reaching n: - While the current number of 'A's (curr) is less than n, we check if n is divisible by the current number of 'A's. If it is, it indicates that this is a good point to "copy all" and then "paste" repeatedly to reach n. - When the current number of 'A's divides n, we perform a "copy all" operation (increasing the step count by 1) and store the copied value in curr_copy. - We then increase the current number of 'A's by pasting the copied characters and update the step count again. 3. Continue Until curr Equals n: - We repeat the process until we reach exactly n characters. 4. Return the Result: - Finally, return ans, which contains the minimum number of steps taken to reach n characters. Time Complexity : O(sqrt(N)) Space Complexity : O(1) Link : https://lnkd.in/ge8Z8zsF Feel free to connect for more insights on algorithm optimization and problem-solving strategies! 😉
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I currently have 2 models which are related by a 1:1 relationship which will always be displayed to Check it out: https://lnkd.in/d382Z_tP Join the conversation! #flask #flaskmarshmallow #flasksqlalchemy #marshmallow #sqlalchemy
combine 2 models with 11 relation into one scheme without nesting?
https://querifyquestion.com
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Leetcode problem of the day : Crawler Log Folder The problem that we are tackling for today is that we need to determine the minimum number of operations required to navigate to the main folder given a series of operations in the form of log entries. The operations include: -Moving up one level ("../") -Staying in the current folder ("./") -Moving into a subfolder (any other string) Approach used: 1. Initialize Variable: - minOperation: This variable will track the current depth of the folder structure. 2. Iterate Over Each Log Entry: - For each log entry: - If the log entry is "../", it indicates moving up one level. Decrease the minOperation by 1 if it's greater than 0 (to ensure we don't go above the root level). - If the log entry is "./", it indicates staying in the current folder. Do nothing in this case. - For any other log entry, it indicates moving into a subfolder. Increase the minOperation by 1. 3. Return the Result: - The final value of minOperation will be the minimum number of operations required to reach the main folder. Time Complexity : O(N) Space Complexity : O(1) Link : https://lnkd.in/gNqFiPu9 Feel free to connect for more insights on algorithm optimization and problem-solving strategies! 😉
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Today ,I solved a problem called "Count and Say"..... Problem Description: The count-and-say sequence is a sequence of digit strings defined by the recursive formula: countAndSay(1) = "1" countAndSay(n) is the run-length encoding of countAndSay(n - 1). Run-length encoding (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "3322251" we replace "33" with "23", replace "222" with "32", replace "5" with "15" and replace "1" with "11". Thus the compressed string becomes "23321511". Given a positive integer n, return the nth element of the count-and-say sequence. Algorithm: This Algorithm constructs the count-and-say sequence by iteratively building each term from the previous term. The inner loop counts consecutive characters and appends the count followed by the character to the new term. This process repeats until the desired term is reached. Consistency.....
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** read the problem at the bottom and try it out before reading the rest For this problem, it’s good to keep in mind that when you have a torch being taken, it must also come back. Also, note that once a person has crossed it doesn’t mean that they can’t come back hypothetically. To solve this problem, I would use a greedy algorithmic approach. In problems seeking the fastest time, this approach uses the biggest resources available to solve the problem. It’s reasonable that the faster people should go first. (2 and 1) and send back 1. From there, 10 and 5 should go, and send back 2. Lastly, 2 and 1 should go. That is 2 + 1 + 10 + 2 + 2 = 17 minutes. The overall approach is to take the minimum value(s) at each point, except the 2nd time going to, in which case you should take the max. Pseudo code: Dict = {A: 10, B: 5, C: 2, D: 1} min(A, B, C, D) = C, D (2 min) min(C,D) = D(1 min) max(A,B, D) = A, B (10 min) min(A,B,C) = C(2 min) min(C,D) = C,D(2 min) Taking a greedy approach is better if you want near-optimal solutions, but due to the small size of the problem space, we can assume this is an optimal solution. You could verify by calculating the permutations(Here's the code) from itertools import permutations times = [1, 2, 5, 10] # Generate all permutations crossing_orders = permutations(times) # Initialize variables shortest_time = float('inf') best_order = None # Iterate through each permutation to calculate the total time for order in crossing_orders: total_time = max(sum(order[:2]), order[0]) + max(order[2:]) if total_time < shortest_time: shortest_time = total_time best_order = order
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Let's improve our algorithm skills. The most common error with algorithms is unnecessarily using loops. In this post, I'll try to help you improve your knowledge about complexities. https://lnkd.in/duyNiSs7
Two Sum – LeetCode
http://guooconde.wordpress.com
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Leetcode problem of the day : Minimum Cost to Convert Source String to Target String The problem that we are tackling for today is that given two strings, source and target, and a set of character transformation rules with associated costs, the goal is to find the minimum cost to convert the source string into the target string using the given transformations. If it's not possible to make the conversion for any character, return -1. Approach used: 1. Initialize Distance Matrix: - Create a 2D distance array where dist[i][j] represents the minimum cost to transform character i to character j. - Initialize the distances with Integer.MAX_VALUE to represent initially no direct transformations. 2. Populate Initial Distances: - Fill the distance array with the given transformation costs. 3. Floyd-Warshall Algorithm: - Compute the shortest paths for all pairs of characters using the Floyd-Warshall algorithm. This will help in finding the minimum cost to transform one character into another considering all possible intermediate transformations. 4. Calculate Minimum Cost: - Iterate through the source and target strings, and for each character pair, add the corresponding transformation cost to the total cost. - If a transformation is not possible (i.e., cost is still Integer.MAX_VALUE), return -1. Time Complexity : O(N+M) Space Complexity : O(1) Link : https://lnkd.in/gw_37zrP Feel free to connect for more insights on algorithm optimization and problem-solving strategies! 😉
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Want to visualise a histogram from a Polars column? You could use the .hist method...but the easiest way is to to go straight to the plot.hist method. This gives a nice interactive Bokeh plot. See the API docs here for more options:
Hist — hvPlot 0.9.2 documentation
hvplot.holoviz.org
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