Question: Find the majority element in an array, which is defined as the element that appears more than n/2 times. Use an efficient algorithm to solve this in linear time O(n) and constant space O(1).

Can you solve this? With in 20 min. Challenge ur self

🌟 Finding the Majority Element in an Array 🌟

Have you ever wondered how to identify the element that appears more than half the time in an array? 🤔 Let’s dive into a clever algorithm that does just that: The Boyer-Moore Voting Algorithm! 🗳

▎🔍 How It Works:

1. Initialization:
- Start with two variables:
- candidate = None
- count = 0

2. Candidate Selection:
- Loop through each element in the array:
- If count is 0, set the current element as the new candidate.
- If the current element matches the candidate, increase count.
- If it doesn’t match, decrease count.

3. Verification (Optional):
- After the first pass, you can run through the array again to confirm that your candidate is indeed the majority element!

▎📈 Why Use This Algorithm?
- Time Complexity: \\( O(n) \\) – Efficiently processes the array in linear time!
- Space Complexity: \\( O(1) \\) – Uses only a constant amount of extra space!

▎🚀 Example:
Consider the array: [2, 2, 1, 1, 1, 2, 2].
- The algorithm will help you find that 2 is the majority element! 🎉

This method is not only efficient but also elegant.

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Python Tips: Two Confusing methods.

strip() and split() Methods

strip(): This method removes any leading or trailing spaces or specific characters from a string.

  text = "  Hello, World!  "
  print(text.strip())  # Output: "Hello, World!"
  

split(): This method splits a string into a list of substrings based on a specified separator.

  data = "apple,banana,cherry"
  print(data.split(','))  # Output: ['apple', 'banana', 'cherry']
  

To see split() in action, check out the [Compare Version Numbers](https://leetcode.com/problems/compare-version-numbers/) problem on LeetCode, where you'll split version strings using a dot (.) as the separator.

1. 3Sum (Hard)
Denoscription: Given an integer array nums, return all unique triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.
- Example: For input nums = [-1,0,1,2,-1,-4], the output should be [[-1,-1,2],[-1,0,1]].

2. Minimum Window Substring (Hard)
Denoscription: Given two strings s and t, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return an empty string.
- Example: For s = "ADOBECODEBANC" and t = "ABC", the minimum window is "BANC".

3. Fruit Into Baskets (Easy)
Denoscription: 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.

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

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).

▎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

🚀 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/)
Denoscription: 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/)
Denoscription: Can the string become a palindrome by deleting just one character?

3. Palindrome Linked List
🔗 [Link](https://leetcode.com/problems/palindrome-linked-list/)
Denoscription: 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/)
Denoscription: 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/)
Denoscription: Count how many palindromic substrings are present in the string using two pointers.

6. Shortest Palindrome.
🔗 [Link](https://leetcode.com/problems/shortest-palindrome/)
Denoscription: Find the shortest palindrome by adding characters to the start of the string.

7. Palindrome Partitioning
🔗 [Link](https://leetcode.com/problems/palindrome-partitioning/)
Denoscription: Partition a string into all possible palindromic substrings.

8. Palindrome Pairs
🔗 [Link](https://leetcode.com/problems/palindrome-pairs/)
Denoscription: 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! 💻💡

#Q20 #leet_codeQ15 Medium noscript. 3Sum
Given an integer array nums, return all the triplets [nums[i], nums[j], nums[k]] such that i != j, i != k, and j != k, and nums[i] + nums[j] + nums[k] == 0.

Notice that the solution set must not contain duplicate triplets.



Example 1:

Input: nums = [-1,0,1,2,-1,-4]
Output: [[-1,-1,2],[-1,0,1]]
Explanation:
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
Example 2:

Input: nums = [0,1,1]
Output: []
Explanation: The only possible triplet does not sum up to 0.
Example 3:

Input: nums = [0,0,0]
Output: [[0,0,0]]
Explanation: The only possible triplet sums up to 0.

In my opinion, if you're preparing for an interview and have no enough time left, it’s best not to spend more than 45 minutes on a single problem. Instead, I believe you should focus on understanding different patterns and concepts. This approach can help you become more versatile and better equipped to tackle various questions during the interview. what do you think?

did you solve this problem?

https://news.1rj.ru/str/+m835nLa3A0c2MjBk join our discussion group

✨ Master Python Division: int(x / y) vs. x // y ✨

Ever wonder about the difference between int(x / y) and x // y in Python? Let’s break it down! 🧠👇

int(x / y):
- Performs regular division first (x / y), giving a floating-point result.
- Then, truncates the result to an integer using int(), removing the decimal part.
- Example:
- int(7 / 3) ➡️ 2 (since 7 / 3 ≈ 2.3333, truncates to 2).
- int(-7 / 3) ➡️ -2 (since -7 / 3 ≈ -2.3333, truncates to -2).

x // y (Floor Division):
- Directly performs floor division, rounding down to the nearest integer.
- Example:
- 7 // 3 ➡️ 2 (largest integer ≤ 2.3333).
- -7 // 3 ➡️ -3 (largest integer ≤ -2.3333).

Key Differences:
- Rounding:
- int(x / y) truncates towards zero.
- x // y rounds down towards negative infinity.
- Negative numbers:
- For positive results, both are the same.
- For negative results, int(x / y) truncates, but x // y rounds down.

Example with Negative Numbers:
- int(-7 / 3) ➡️ -2
- -7 // 3 ➡️ -3

after solving 3 sum try to solve the next question