Leetcode with dani

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🚀 Explore Palindrome Challenges with the Two-Pointer Technique!🧑‍💻 Palindromes are a classic problem type that can be efficiently tackled using the two-pointer technique. Whether you're just starting or looking to refine your skills, these questions will help you master this approach. Check out these palindrome problems: 1.Valid Palindrome 🔗 [Link](https://leetcode.com/problems/valid-palindrome/) Description: Determine if a string is a palindrome, considering only alphanumeric characters and ignoring cases. 2. Valid Palindrome II 🔗 [Link](https://leetcode.com/problems/valid-palindrome-ii/) Description: Can the string become a palindrome by deleting just one character? 3. Palindrome Linked List 🔗 [Link](https://leetcode.com/problems/palindrome-linked-list/) Description: Check if a singly linked list is a palindrome using two pointers and a bit of list manipulation. 4. Longest Palindromic Substring 🔗 [Link](https://leetcode.com/problems/longest-palindromic-substring/) Description: Find the longest palindromic substring in a given string, with pointers expanding from each character. 5. Palindromic Substrings 🔗 [Link](https://leetcode.com/problems/palindromic-substrings/) Description: Count how many palindromic substrings are present in the string using two pointers. 6. Shortest Palindrome. 🔗 [Link](https://leetcode.com/problems/shortest-palindrome/) Description: Find the shortest palindrome by adding characters to the start of the string. 7. Palindrome Partitioning 🔗 [Link](https://leetcode.com/problems/palindrome-partitioning/) Description: Partition a string into all possible palindromic substrings. 8. Palindrome Pairs 🔗 [Link](https://leetcode.com/problems/palindrome-pairs/) Description: Given a list of words, find all pairs of distinct indices that form palindromes. Ready to boost your problem-solving skills? Dive into these problems, practice your two-pointer technique, and ace those palindrome challenges! 💥 Happy Coding! 💻💡

▎1. 3Sum def threeSum(nums):     nums.sort()     result = []     for i in range(len(nums) - 2):         if i > 0 and nums[i] == nums[i - 1]:             continue         left, right = i + 1, len(nums) - 1         while left < right:             total = nums[i] + nums[left] + nums[right]             if total < 0:                 left += 1             elif total > 0:                 right -= 1             else:                 result.append([nums[i], nums[left], nums[right]])                 while left < right and nums[left] == nums[left + 1]:                     left += 1                 while left < right and nums[right] == nums[right - 1]:                     right -= 1                 left += 1                 right -= 1     return result ▎2. Minimum Window Substring from collections import Counter def minWindow(s, t):     if not t or not s:         return ""         dict_t = Counter(t)     required = len(dict_t)         l, r = 0, 0     formed = 0     window_counts = {}         ans = float("inf"), None, None         while r < len(s):         character = s[r]         window_counts[character] = window_counts.get(character, 0) + 1                 if character in dict_t and window_counts[character] == dict_t[character]:             formed += 1                 while l <= r and formed == required:             character = s[l]                         if r - l + 1 < ans[0]:                 ans = (r - l + 1, l, r)                         window_counts[character] -= 1             if character in dict_t and window_counts[character] < dict_t[character]:                 formed -= 1                         l += 1                 r += 1         return "" if ans[0] == float("inf") else s[ans[1]: ans[2] + 1] ▎3. Fruit Into Baskets def totalFruits(fruits):     left, right = 0, 0     basket = {}     max_fruits = 0         while right < len(fruits):         basket[fruits[right]] = basket.get(fruits[right], 0) + 1                 while len(basket) > 2:             basket[fruits[left]] -= 1             if basket[fruits[left]] == 0:                 del basket[fruits[left]]             left += 1                 max_fruits = max(max_fruits, right - left + 1)         right += 1         return max_fruits

1. 3Sum (Hard)     Answer: The solution involves sorting the array and using a two-pointer technique to find triplets that sum to zero. The time complexity is O(n^2). 2. Minimum Window Substring (Hard)     Answer: Use a sliding window approach to maintain a count of characters and expand/shrink the window until the minimum substring is found. The time complexity is O(n). 3. Fruit Into Baskets (Easy)     Answer: Utilize a sliding window to track the types of fruits and their counts, ensuring you only have two types at any time. The maximum length of the window gives the answer. The time complexity is O(n).

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Understanding the constraints, time complexity is key to solving LeetCode problems effectively. If you find these concepts confusing, don’t worry , you’re not alone!. this might help you to undestand the constraints

3. Fruit Into Baskets (Easy) Description: You are visiting a farm and have a basket that can hold at most two types of fruits. You want to maximize the number of fruits you collect. Given an array of integers representing the types of fruits in a row, find the maximum number of fruits you can collect in one basket. - Example: For input [1,2,1], the maximum number of fruits collected is 3.