Top Tweets for #CodeBreakdown
When βLet Me Refactorβ Turns Into a Horror Story π
Everything was perfect, until you touched it.
#Nimblechapps #NimbleMeme #NimbleMemes #RefactorDisaster #CodeBreakdown #DevPain #BugLife #DebuggingMess #CleanCodeTrap #CorporateComedy #FunnyAtWork #RelatableWorkMemes
Complete Code Crash is Always Disheartening π₯²
Visit us: https://t.co/7A6IEPvxgJ
#DeveloperLife #BugFixing #CodingHumor #MemeLife #TechStruggles #ProgrammingProblems #FunnyMemes #CodeBreakdown #DevMemes #SoftwareDevelopment #TechJokes #DebuggingLife #ClashOfClansMeme
When the manager steps in to 'fix' the code freshers wrote, and things just get more confusing. We've all been there! π
Tag a colleague who knows this struggle!
#CorporateLife #CodingStruggles #ManagerFixes #FreshersCode #DeveloperLife #OfficeHumor #CodeBreakdown #TechLife
Unlocking the Code: Full Forms of Popular Programming Languages! π»β¨ #Coding #Programming #Tech ππ #CodeBreakdown #TechEducation #ProgrammingFun #hartronambalacity #learnmorewithhartron
Lazy JavaScript developers rejoice! Sharavana's "Compiler In One Byte" explains how a compiler breaks code into small pieces, saving us from writing assembly language. #LazyDev #Compiler #CodeBreakdown https://t.co/iQlkU4Qclu
Which application effectively manages user traffic: @Swiggy or @zomato ? ππ½οΈπ₯
#TervPro #ZomatoVsSwiggy #CodingBattle #AppComparison #FrontendVsBackend #CodeBreakdown
π Unlocking Python's mysteries! π What's the value of 4 + 3 % 5?
Let's break it down: First, we tackle the modulo operator (%), then addition (+).
After simplification, it's 4 + 3 = 7. Voila! π
#PythonTips #CodeBreakdown #LearningPython #takeoffupskill
Trying to explain coding to non-tech friends...
#techthursday #TechForNonTechies #TechTalks #CodeBreakdown
In the coding world, facing a gigantic problem is like staring at a mountain from the base β overwhelming, right? But here's a trick: break that mountain into manageable hills.
#ProblemSolving101 #CodeBreakDown #SmallWinsBigResults
I did something new today π Instead of my usual way of sharing what I learned from a #PreppinData challenge by putting it into a tweet or a Word document; I wrote a quick @tableau Prep #CodeBreakDown on @Medium. Check out 'Custom Split in Tableau Prep': https://t.co/REQepndpsc
πΌοΈ Posting the code snippet below! πΈ Take a peek into my solution for the "Assign Cookies" problem on LeetCode. π Let's break it down together! #CodeBreakdown #LeetCodeChallenge
π§ Intuition:
Rows, columns, and a dance of ones and zeros! πΊπ My strategy involves counting ones in rows/columns, then building the difference matrix. β¨ Simplicity meets elegance. π‘ #CodeBreakdown #Programming
@clcoding In the list k = [2,1,0,3,0,2,1], k.index(0) finds the first zero's index, which is 2. Now, guess how many times '2' appears in the list? Bingo! Twice. That's why k.count(k.index(0)) prints 2! π―π‘ #PythonPuzzles #CodeBreakdown #PythonGiants
#CodeBreakdown π
Step 4: Traverse the input list once more, and use the prefix sum list to find the count of numbers smaller than the current number.
for i in range(len(nums)):
if nums[i] == 0:
continue
res[i] = list1[nums[i]-1]
#CodeBreakdown π
Step 3: Create a prefix sum list from the frequency list, which will help in determining the count of smaller numbers efficiently.
for i in range(1, len(list1)):
list1[i] += list1[i-1]
#CodeBreakdown π
Step 2: Traverse through the input list and update the frequency list accordingly.
for i in range(len(nums)):
list1[nums[i]] +=1
#CodeBreakdown π
trick here is to look out for the constraint in the problem:
Step 1: Initialize two lists, one to keep track of the frequency of each number (0-100) and another to store the results.
list1 = [0]*101
res = [0] * len(nums)
#CodeBreakdownβοΈ
Within the loop, we shift elements to the right based on the number of zeros counted. If we encounter a zero, we duplicate it by assigning it to the next position as well.
if (i + zeros) <= n:
arr[i + zeros] = arr[i]
if arr[i] == 0:
zeros -= 1
..
#CodeBreakdownβοΈ
We then start a reverse loop, traversing the array from the end to the start. This way, we ensure the duplication process does not affect the upcoming elements.
for i in range(n, -1, -1):
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