Sanket Disale’s Post

Hello Everyone, I participated in 𝐋𝐞𝐞𝐭𝐂𝐨𝐝𝐞 𝐰𝐞𝐞𝐤𝐥𝐲 𝐂𝐨𝐧𝐭𝐞𝐬𝐭 407 and successfully solved 𝟯/𝟰 𝐩𝐫𝐨𝐛𝐥𝐞𝐦𝐬. 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟭: Iterate every bit of n,k. - Increment count when n => '1' and k => '0' in a bit. - Impossible condition: n => '0' and k => '1' in a bit. - Return count 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟮: - If the total count of vowels is 0 then Alice will lose, otherwise she will always win. 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟯: -   - Traverse the binary string and keep track of the number of '1's in the string for every i. - If s[i] == '0' and s[i-1] == '1', increase operations by the count of '1's. - Return operations Why increase the operations by the count of '1's: - Assume the string: "001100." - At the 4th index, s[i]=='0' and s[i-1]=='1' satisfies the condition. - After the first operation, the string will look like: 001001. - After the second operation, the string will look like: 000011. - Total operations possible for 'i' is the count of '1's. 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟰: Will upsolve I'm excited about these challenges and looking forward to more contests! #leetcode #learning #contest #ProgrammingSkills #Algorithm #CodingJourney

Let's talk about today's LeetCode Weekly Contest 382. Question 1 - Convert the string into uppercase, then just count the changing characters. Submission Code - https://lnkd.in/g7NgiAFj Question 2 - We can just count the frequencies of every element, then we will assume every element as the smallest element of the subset and try to find the maximum length of the subset. as kth power of 1 is 1 itself so we have to count the ferquency of 1 and take the maximum odd freq. and max of both Submission Code - https://lnkd.in/gGS9NAf5 Questions 3 - Sum of x + y should be odd Submission Code - https://lnkd.in/gyJc4cVe Question 4 - Not able to solve during the contest. Connect 🤝 : Amir Khan #leetcode #datastructures #coding #algorithms  #contest #globralRank #weekly #rank #linkedIn #learning

Day 80 of the #100DaysOfCode Challenge 🌟 -->Solving the "Building an Aquarium" Problem 🐠🌊 Today, I tackled an interesting problem called "Building an Aquarium" where the task is to determine the largest possible height of a tank built around a piece of coral, while ensuring the total amount of water used does not exceed a given limit. Here’s a summary of my approach: ->Problem Overview: Objective: Find the largest height h for the tank, while limiting the total water usage to x units. Link to the problem 👉 https://lnkd.in/dcQxU-zE ->My Approach: 1.Binary Search Algorithm: Utilized binary search to efficiently find the optimal height h.Initialized a large array to store coral column heights and set up standard input/output optimizations. 2.Input Reading: Read the number of test cases and iterate over each one.For each test case, read the number of columns n and the maximum water usage x. 3.Binary Search Execution: Set the initial range for the binary search with low as 1 and high as the maximum integer value. Within the binary search loop, calculated the middle point mid and determined the amount of water required to fill the tank to height mid. 4.Water Calculation: For each column, calculated the difference if mid is greater than the column height, and summed it up to get the total water required. 5.Condition Checking: If the total water required is within the allowed limit x, updated the answer answer to mid and moved the lower bound up. Otherwise, adjusted the upper bound down. 6.Output: After exiting the loop, printed the maximum possible height h for each test case. Here's a link to my solution: https://lnkd.in/dvWFXKtz -->Key Takeaways: ->Efficient Computation: Binary search method ensures finding the largest height h while staying within the water usage limit x. ->Optimized Solution: By using a binary search algorithm, the solution is able to handle large test cases efficiently. I'm always open to learning from others and welcome any insights or alternative approaches to this problem. Thank you for following along with my #100DaysOfCode journey! 💻🚀 #100DaysOfCode #CodingChallenge #PersonalGrowth #Consistency #Discipline #Programming #TechJourney #Coding #Algorithm #Optimization #CodeforcesStreak

🌟 Day 92 of my 100 Days of Code Challenge 🌟 Today, I tackled an intriguing problem called "T-primes." A T-prime is defined as a positive integer that has exactly three distinct positive divisors. Given an array of nnn positive integers, the challenge was to determine whether each number is a T-prime or not. Link to the problem 👉 https://lnkd.in/dzm2sirE 🔍 My Approach: To solve this problem, I leveraged the properties of prime numbers and their squares: 1. Identifying T-primes: A number is a T-prime if and only if its square root is a prime number. 2. Sieve of Eratosthenes: I constructed a Boolean array of length 1,000,001 where prime indices are marked as true and others as false. This was achieved using the Sieve of Eratosthenes. 3. Checking Each Number: For each number in the input array, I checked if its square root is a prime number and if the square of that square root equals the number. If both conditions are met, the number is a T-prime; otherwise, it is not. Here's a link to my solution: https://lnkd.in/djCYrfjp 📈 Outcome: This approach efficiently determined whether each number is a T-prime, ensuring accurate results for all input cases. If you have any suggestions for a better approach or if you have any questions about my method, feel free to reach out! Let's keep learning and growing together! 🚀 #100DaysOfCode #CodingChallenge #PersonalGrowth #Consistency #Discipline #Programming #TechJourney #Coding #Algorithm #Optimization #CodeforcesStreak #CodingJourney #ProblemSolving

🌟 Day 74 of #100DaysOfLeetCode Challenge! 🚀 Today I solved the problem no. 1442 on LeetCode: Count Triplets That Can Form Two Arrays of Equal XOR APPROACH : 1. Create a prefix XOR array prefixXor where prefixXor[i] represents the XOR of elements from the start of the array up to the (i−1)-th element. 2. Initialize prefixXor[0] to 0 to handle XOR calculations correctly. 3. Use two nested loops to iterate over all possible pairs of indices (𝑖,𝑘) where i<k. 4. For each pair, check if prefixXor[i] is equal to prefixXor[k + 1]. This condition implies that the XOR of the subarray from 𝑖 to k is zero, i.e., arr[i...k] XOR is 0. 5. If the condition prefixXor[i] == prefixXor[k + 1] holds true, it means there are (k−i) valid values for j such that i<j≤k. 6. Increment the count by (k−i) for each valid pair (i,k). 7. After iterating through all pairs (i,k), return the total count of valid triplets. TIME COMPLEXITY: O(n^2) SPACE COMPLEXITY: O(n) Feel free to contribute your approaches as well. #100DaysOfLeetCode #CodingChallenge #Algorithms #Programming  #TechSkills #CodingAdventure #GeekModeActivated  #dsachallenge #codingjourney #problemsolving #leetcodechallenge

🌟 Day 18 of #100DaysOfLeetCode Challenge! 🚀 Today I solved the problem no. 1614 on LeetCode: Maximum Nesting Depth of the Parantheses. APPROACH: 1. Initialize variables count and maxdepth to keep track of the current depth and maximum depth encountered so far, respectively. 2. If the input string s is empty, return 0, as there are no parentheses to evaluate. 3. Check if the input string s has a length of 1. If it does, ensure that it contains either '(' or ')'. If it doesn't, return 0, indicating that the string is invalid. Otherwise, continue with the loop. 4. For each character in s if it's an opening parenthesis '(': Increment the count to reflect the increase in depth. Update maxdepth to be the maximum of the current maxdepth and the count. 5. If it's a closing parenthesis ')': Decrement the count to reflect the decrease in depth. 6. Return the maxdepth value, which represents the maximum depth of nested parentheses encountered in the input string. TIME COMPLEXITY : O(n) SPACE COMPLEXITY : O(1) Feel free to contribute your approaches as well! #100DaysOfLeetCode #CodingChallenge #Algorithms #Programming #TechSkills #CodingAdventure #GeekModeActivated #dsachallenge #codingjourney #problemsolving #leetcodechallenge

🌟 Day 54 of #100DaysOfLeetCode Challenge! 🚀 Today I solved the problem no. 786 on LeetCode: K-th Smallest Prime Fraction. APPROACH (MIN-HEAP) : 1. Initialize a priority queue pq to store pairs of floats and pairs of integers. 2. Iterate over the array using two nested loops to generate all possible fractions. The outer loop iterates over elements starting from the beginning (i), and the inner loop iterates over elements starting from the end (j). This ensures that fractions are generated in non-decreasing order. 3. For each pair (i, j), calculate the fraction as arr[i] / arr[j], cast it to a float, and push it into the priority queue along with the pair (arr[i], arr[j]). 4. Pop elements from the priority queue k-1 times. This effectively discards the k-1 smallest fractions, leaving the k-th smallest at the top of the priority queue. 5. Extract the k-th smallest prime fraction from the top of the priority queue. 6. Return the result vector containing the indices of the elements that form the k-th smallest prime fraction. TIME COMPLEXITY: O(N*N + k*logN) SPACE COMPLEXITY: O(N*N) Feel free to contribute your approaches as well. #100DaysOfLeetCode #CodingChallenge #Algorithms #Programming  #TechSkills #CodingAdventure #GeekModeActivated  #dsachallenge #codingjourney #problemsolving #leetcodechallenge

🌟 Day 73 of #100DaysOfLeetCode Challenge! 🚀 Today I solved the problem no. 1404 on LeetCode: Number of Steps to Reduce a Number in Binary Representation to One APPROACH : 1. Initialize the variable steps to 0, which will keep track of the number of steps taken. 2. Loop while the length of the string s is greater than 1, indicating that the number is not yet reduced to 1. 3. If the last character of the string s is '0', it indicates that the number is even. In this case, remove the last character from the string (s.pop_back()), which is equivalent to dividing the number by 2. 4. If the last character of the string s is '1', it indicates that the number is odd. In this case, iterate from the end of the string towards the beginning to find the rightmost '1'. Flip all the '1's encountered to '0' until a '0' is found. If no '0' is found (i.e., all digits are '1'), prepend a '1' to the string, indicating addition by 1. 5. Increment the steps counter after each operation. 6. Return the total number of steps taken. TIME COMPLEXITY: O(n^2) SPACE COMPLEXITY: O(n) Feel free to contribute your approaches as well. #100DaysOfLeetCode #CodingChallenge #Algorithms #Programming  #TechSkills #CodingAdventure #GeekModeActivated  #dsachallenge #codingjourney #problemsolving #leetcodechallenge

Hello #Connections I have participated Weekly Contest 399 and Biweekly Contest 131 in LeetCode . Weekly Contest 399 here is my approach - 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟭👨💻: check all pairs (i, j), if (nums1[i] % (nums2[j] * k)) == 0) increase answer by 1. 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟮👨💻: single character c repeating greater than 9 times then we add 9'ch' and start from 0 again. It's simple just count and add to answer. 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟯👨💻: hint(math) nums1[i] % (nums2[j] * k) == 0 -> nums1[i] % k == 0 && nums1[i] % nums2[j] == 0. 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 4👨💻:Solving after contest Biweekly Contest 131 here is my approach - 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 𝟭👨💻: Brute Force and apply XOR 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 2👨💻: Using map for occurence and there index track and according to query fill the answer array 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 3👨💻: Using two maps 1 for bolls and another for colors according to query fill and remove the element from the map 𝗣𝗿𝗼𝗯𝗹𝗲𝗺 4👨💻:Solving after contest #contest #dsa #coding #leetcode #learning

Let's talk about today's LeetCode Bi-Weekly Contest 136. Question 1 - Just store the frequency and check accordingly. Submission Link: https://lnkd.in/gs6fmftK Question 2 - Just brute force, count for both cases, and return the minimum. Submission Link: https://lnkd.in/gTy5u8FF Question 3 - For this, first we can try for all four pairs for every cell, and add the difference to the total answer, But we will be left with the mid if we have an odd count of rows or columns, for that we need to handle them separately. For that, we can store the count of 0,1 and 2. For 0 we don't need to do anything, but for 2, if we have an odd count then we will try to compensate that with the count of 1s, or if we don't have a count of 1s then we will be be removing the extra count of 2. Submission Link: https://lnkd.in/gAR2AmN3 Question 4 - Started 50 minutes late so I didn't get time to solve it. Follow 🤝 : Prateek Goel #leetcodechallenge #globralRank #weekly #rank #learning #linkedin #coding #learning #contest #biweekly