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Leetcode with dani

Leetcode with dani

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Join us and let's tackle leet code questions together: improve your problem-solving skills Preparing for coding interviews learning new algorithms and data structures connect with other coding enthusiasts

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"Can You Solve the 100 Locker Riddle?"

Your rich, eccentric uncle has left behind a mysterious will. You and 99 relatives are called to his mansion for the reading. Here’s the twist:

There are 100 lockers, each hiding a secret word. Each of you is assigned a number from 1 to 100. The rules are:

Heir 1 opens every locker.

Heir 2 closes every 2nd locker.

Heir 3 changes the state of every 3rd locker (opens if closed, closes if open).

This continues until Heir 100 changes the state of the 100th locker.

At the end, only the lockers that remain open will reveal the code to the safe.

Here’s the challenge:
Without going through all 100 steps, can you figure out which lockers will stay open?

If you solved it differently or have any questions, drop your approach or thoughts in the comments! Let’s learn and grow together. 💡

def maxSubArray(nums):
    max_current = max_global = nums[0]
    for i in range(1, len(nums)):
        max_current = max(nums[i], max_current + nums[i])
        max_global = max(max_global, max_current)
    return max_global

Solution: Kadane’s Algorithm The most efficient way to solve this problem is using Kadane’s Algorithm, which runs in O(n) time and uses O(1) space. Here’s how it works: Initialize two variables: max_current: Tracks the maximum sum of the subarray ending at the current position. max_global: Tracks the overall maximum sum found so far. Iterate through the array: For each element, update max_current to be the maximum of the current element itself or the sum of max_current and the current element. Update max_global to be the maximum of max_global and max_current. Return max_global as the result.

can u Write a Python function to check if a word is a palindrome?… but you can’t use loops or reversed().

🔥 Problem of the Day: "Maximum Subarray" (Medium) LeetCode #53 | Topic: Dynamic Programming / Greedy 📝 Problem Statement Given an integer array nums, find the contiguous subarray with the largest sum. Return the sum. Example: Input: nums = [-2,1,-3,4,-1,2,1,-5,4] Output: 6 (Because [4,-1,2,1] sums to 6) 💡 Hints to Get Started Should you track the current subarray or just its sum? Pro Tip: Kadane’s Algorithm can solve this in O(n) time! ⏳ Time to Solve: 40 minutes! Drop your solution in the comments 💬. I’ll post the optimized answer tomorrow

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Approach Brute Force: Use two nested loops to check every possible pair of lines. Time Complexity: O(n²) Space Complexity: O(1) Optimized Approach: Use two pointers to traverse the array from both ends. Time Complexity: O(n) Space Complexity: O(1) Solution Code
def maxArea(height):
    left, right = 0, len(height) - 1
    max_area = 0
    
    while left < right:
        width = right - left
        current_height = min(height[left], height[right])
        current_area = width * current_height
        max_area = max(max_area, current_area)
        
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    
    return max_area

# Example usage:
height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
print(maxArea(height))  # Output: 49
Explanation We use two pointers, left and right, to traverse the array from both ends. The area between the two lines is calculated as width * height, where width = right - left and height = min(height[left], height[right]). To maximize the area, we move the pointer pointing to the shorter line inward. This approach ensures we only traverse the array once, making it efficient with a time complexity of O(n). Additional Resource LeetCode Container With Most Water Discussion

Approach Brute Force: Use two nested loops to check every possible pair of lines. Time Complexity: O(n²) Space Complexity: O(1) Optimized Approach: Use two pointers to traverse the array from both ends. Time Complexity: O(n) Space Complexity: O(1) Solution Code
def maxArea(height):
    left, right = 0, len(height) - 1
    max_area = 0
    
    while left < right:
        width = right - left
        current_height = min(height[left], height[right])
        current_area = width * current_height
        max_area = max(max_area, current_area)
        
        if height[left] < height[right]:
            left += 1
        else:
            right -= 1
    
    return max_area

# Example usage:
height = [1, 8, 6, 2, 5, 4, 8, 3, 7]
print(maxArea(height))  # Output: 49
Explanation We use two pointers, left and right, to traverse the array from both ends. The area between the two lines is calculated as width * height, where width = right - left and height = min(height[left], height[right]). To maximize the area, we move the pointer pointing to the shorter line inward. This approach ensures we only traverse the array once, making it efficient with a time complexity of O(n). Additional Resource LeetCode Container With Most Water Discussion

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🚀 Daily LeetCode Challenge: Container With Most Water 🚀 Post Content Hey everyone! 👋 Here’s today’s LeetCode problem to sharpen your problem-solving skills. Let’s dive in! 💻 Problem: Container With Most Water Difficulty: Medium Link: Container With Most Water - LeetCode Problem Statement You are given an integer array height of length n. There are n vertical lines drawn such that the two endpoints of the i-th line are (i, 0) and (i, height[i]). Find two lines that, together with the x-axis, form a container that holds the most water. Return the maximum amount of water the container can store. Example Input: python Copy height = [1, 8, 6, 2, 5, 4, 8, 3, 7] Output: python Copy 49 Explanation: The container is formed by the lines at indices 1 (height = 8) and 8 (height = 7). The width of the container is 8 - 1 = 7. The height of the container is min(8, 7) = 7. The total area (water held) is width * height = 7 * 7 = 49.

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can u solve it with out hash?and with only one loop and constant O(1) space?

share ur answer

409. Longest Palindrome Given a string s which consists of lowercase or uppercase letters, return the length of the longest palindrome that can be built with those letters. Letters are case sensitive, for example, "Aa" is not considered a palindrome. Example 1: Input: s = "abccccdd" Output: 7 Explanation: One longest palindrome that can be built is "dccaccd", whose length is 7. Example 2: Input: s = "a" Output: 1 Explanation: The longest palindrome that can be built is "a", whose length is 1. Constraints: 1 <= s.length <= 2000 s consists of lowercase and/or uppercase English letters only. Submit

🎄✨ Merry Christmas! 🎄✨ Wishing you all joy, love, and peace this holiday season. As for the New Year… umm, is it appropriate to say "Happy New Year" now? 🤔 I mean, we’re rocking the Ethiopian calendar here! 🗓😎

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