LeetCode Day-3/365 Coding Journey

Today I solved two problems:

1. Find Minimum in Rotated Sorted Array (LeetCode #153)

Python Solution:

class Solution:
    def findMin(self, nums):
        l, r = 0, len(nums) - 1
        while l < r:
            if nums[l] < nums[r]:
                return nums[l]
            mid = l + (r - l) // 2
            if nums[mid] < nums[l]:
                r = mid
            else:
                l = mid + 1
        return nums[l]

Approach:

• Use binary search to locate the smallest element in the rotated sorted array.
• Compare nums[l] and nums[r] to determine if the current range is sorted.
• Adjust the search range based on comparisons with the mid element.

Learnings:

• Binary search is a powerful tool for logarithmic time complexity.
• Handling edge cases like no rotation or single-element arrays is crucial.

Other Approaches:

• A simple linear scan to find the minimum value, but it takes O(n)O(n) time.

2. Container with Most Water (LeetCode #11)

Python Solution:

class Solution:
    def maxArea(self, height):
        l, r = 0, len(height) - 1
        maxWater = 0
        while l < r:
            maxWater = max(maxWater, min(height[l], height[r]) * (r - l))
            if height[l] < height[r]:
                l += 1
            else:
                r -= 1
        return maxWater

Approach:

• Use two pointers to maximize the area between the container walls.
• Calculate the area using the shorter wall height and width (r - l).
• Move the pointer pointing to the shorter wall inward to potentially increase the area.

Learnings:

• Two-pointer techniques effectively solve problems involving pairs of elements.
• Always consider edge cases, such as arrays with minimal size.

Other Approaches:

• A brute-force approach checking all possible pairs, which takes O(n2)O(n²) time.

Thanks to iCodeGuru for their help and support in my coding journey. Their guidance motivates me to keep learning and solving problems. Let’s grow together!