Challenges and Reflections

Facing Rejections and Demotivation

Today brought another round of job application rejections, adding to a tally of over 100 attempts. The continuous setbacks have taken a toll on motivation. The day unfolded with a sense of demotivation, leading to a retreat into a Chinese drama series, “Qin Dynasty Epic” on Amazon Video Prime. The desire to engage in any productive activity seemed elusive, resulting in a day primarily spent on passive activities like watching shows and eating.

Reflecting on the current state of the economy and the impending Christmas season, the slowdown in job opportunities was anticipated. Despite knowing this, the struggle persists. I find myself in a challenging phase, trying to navigate through tough times.

Adding to the frustration, a job assessment for a data engineer/data architect position turned out to be an unexpected hurdle. The assessments, filled with brain teasers, logic questions, and irrelevant English reading challenges, felt like a misuse of time and resources. The disconnect between the assessment content and the technical nature of the job left me baffled and frustrated.

Navigating the Day with LeetCode

In an attempt to maintain a sense of routine and skill development, I turned to LeetCode for coding practice. However, the fatigue from job rejections lingered, making it difficult to fully engage. Despite this, the commitment to daily coding practice remained intact.

Here is a LeetCode solution I found in the editorial section, serving as a reference for future study:

# Longest Increasing Subsequence
class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        dp = [1] * len(nums)
        for i in range(1, len(nums)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)

        return max(dp)

# other solution with O(nlogn)
class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        sub = [nums[0]]
        
        for num in nums[1:]:
            if num > sub[-1]:
                sub.append(num)
            else:
                # Find the first element in sub that is greater than or equal to num
                i = 0
                while num > sub[i]:
                    i += 1
                sub[i] = num

        return len(sub)

Encouragement for the Future

To my future self: This challenging phase will pass. In times of adversity, stay strong and resilient. Smile and appreciate the positives, maintaining both physical and mental well-being. Remember, growth often emerges from the toughest experiences.