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Leetcode with dani

Leetcode with dani

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Approach Brute Force: Use two nested loops to check every possible pair of lines. Time Complexity: O(nΒ²) 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(nΒ²) 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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πŸš€ Daily LeetCode Challenge: Container With Most Water πŸš€ Post Content Hey everyone! πŸ‘‹ Here’s today’s LeetCode problem to sharpen your problem-solving skills. Let’s dive in! πŸ’» 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?

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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

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If you were interviewed, how was your technical interview?
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Mistakes are proof that you’re trying. Embrace every error as a lesson and keep moving forward

The is the interview question I just had: Given a binary array nums, return the maximum number of consecutive 1's in the array if you can flip at most one 0. Example 1: Input: nums = [1,0,1,1,0] Output: 4 Explanation: - If we flip the first zero, nums becomes [1,1,1,1,0] and we have 4 consecutive ones. - If we flip the second zero, nums becomes [1,0,1,1,1] and we have 3 consecutive ones. The max number of consecutive ones is 4. max_cont= 4 max_count= 3 max_count =4 [1,0,1,1,0] Example 2: Input: nums = [1,0,1,1,0,1] Output: 4 Explanation: - If we flip the first zero, nums becomes [1,1,1,1,0,1] and we have 4 consecutive ones. - If we flip the second zero, nums becomes [1,0,1,1,1,1] and we have 4 consecutive ones. The max number of consecutive ones is 4. Constraints: 1 <= nums.length <= 105 nums[i] is either 0 or 1.

Have you been interviewed?
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Given a string s consisting of lowercase letters only, return the length of the longest alphabetical continuous substring. def longestContinuousSubstring(s: str) -> int: Example 1: Input: s = "abacaba" Output: 2 Explanation: There are 4 distinct continuous substrings: "a", "b", "c" and "ab". "ab" is the longest continuous substring. Example 2: Input: s = "abcde" Output: 5 Explanation: "abcde" is the longest continuous substring. Constraints: 1 <= s.length <= 10^5 s consists of only English lowercase letters. """

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""" Given a binary array nums, return the maximum number of consecutive 1's in the array. Example 1: Input: nums = [1,1,0,1,1,1] Output: 3 Explanation: The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3. Example 2: Input: nums = [1,0,1,1,0,1] Output: 2 Constraints: 1 <= nums.length <= 105 nums[i] is either 0 or 1. """

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