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Day 1 of #geekstreak2024 Problem - Minimum Cost of Ropes Today I have taken the challenge "geekstreak2024" and today is the beginning of my 21-day POTD Challenge Powered By Deutsche Bank! The first challenge was solving the "Minimum Cost of Ropes" problem. Problem - https://lnkd.in/guZ45Tgr Approach - To find the minimum cost of connecting ropes, insert all rope lengths into a min-heap (priority queue). Repeatedly extract the two shortest ropes, connect them, and add the combined length to the total cost. Then, insert the new rope back into the heap. Continue this process until only one rope remains. The total cost at the end will be the minimum cost of connecting all ropes. #geekstreak2024 #deutschebank #CodingChallenge #ProblemSolving #POTD #GeeksForGeeks

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🌟 Day 1 of the 21 days #geekstreak2024 Today's Problem: Minimum Cost of Ropes Problem Statement: Given an array representing the lengths of different ropes, the task is to connect the ropes into one single rope. The cost to connect two ropes is equal to the sum of their lengths, and the goal is to minimize the total cost of connecting all the ropes. Approach: I tackled this problem using a Greedy Algorithm combined with a Min-Heap (Priority Queue). The core idea is to repeatedly connect the two shortest ropes first, that helps to keep the total connection cost as low as possible at every step. Once two ropes are connected, their combined length is inserted back into the heap, allowing the process to continue efficiently. This method ensures that each local decision contributes to an overall optimal solution. Explanation: - Initialized a Min-Heap to store all the rope lengths. - At each iteration, we extract the two smallest ropes, calculate the cost of connecting them, and add their combined length back to the heap. - Above process is repeated until only one rope left. ⏳ Time Complexity: O(n log n) Space Complexity: O(n) GeeksforGeeks Deutsche Bank #GeeksforGeeks #CPP #geekstreak2024 #deutschebank #Potd #geekstreakchallenge2024

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Day 15 /21 #geekstreak2024 Problem - Roof Top Day 15 of my 21-day POTD Challenge Powered By Deutsche Bank Today’s challenge was an exciting one! Given the heights of consecutive buildings, the goal was to find the maximum number of consecutive steps I could take where each step results in a gain in altitude! A great problem to understand array traversal and step counting. Approach - I initialized two variables: ans to store the maximum number of consecutive steps, and curr to count the current streak of altitude gains. By comparing consecutive building heights, I updated the streak and kept track of the longest sequence #geekstreak2024 #deutschebank #CodingChallenge #ProblemSolving #POTD GeeksforGeeks

Day 06 of the #geekstreak2024 Challenge! 21 Days POTD Challenge by Deutsche Bank and GeeksforGeeks! 💠 Today's problem: Minimize the Heights II 💠 Approach: 🔹 Sort the array to handle tower heights in increasing order, simplifying height modifications. 🔹 Modify heights by either adding or subtracting K for each tower, and compare only the new smallest and largest values after each modification. 🔹Track the minimum difference between the tallest and shortest towers based on these boundary values to find the optimal result. #geekstreak2024 #deutschebank #gfg #potd

🎯 Day 15 of #geekstreak2024 – 21 Days POTD Challenge powered by Deutsche Bank on GeeksforGeeks 💡 Problem: Roof Top You are given the heights of consecutive buildings. You can move from the roof of a building to the roof of the next adjacent building. You need to find the maximum number of consecutive steps you can put forward such that you gain an increase in altitude with each step. 🔍 Approach: 1. Loop through the array and count consecutive steps where the altitude increases. 2. If the altitude doesn't increase, reset the count. 3. Track the maximum count of consecutive steps. 🕒 Time Complexity: O(n) Space Complexity: O(1) #geekstreak2024 #DeutscheBank #CodingChallenge #ProblemSolving #POTD

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Day 1/21 of #geekstreak2024 #21dayofgfgchallenge 🚀 In the question we have been given an array of length of ropes and the goal is to combine all ropes(2 at a time) into one single rope , and every time we do so the cost gets updated to the sum of the lengths. We need to minimize this cost. I used the concept of min-heap(priority queue) to store the lengths of the rope, which ensures the smallest ropes are accessible. The lengths of the top two ropes are added and is pushed back in the heap. After all the ropes are combined the cost is returned which is the minimum cost. #geekstreak2024 #21dayofgfgchallenge #deutchebank GeeksforGeeks Deutsche Bank