Ever wondered what makes up a code? Here's the breakdown: ✔️ Writing codes (30%) ✔️ Applying logic (35%) ✔️ Learning from courses (10%) ✔️ Debugging syntax errors (15%) ✔️ Compilation process (10%) #CodeBreakdown #LogicInCode #LearningJourney #DebuggingSkills #CompilationProcess

🐍 Unlocking Python's mysteries! 🔍 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! 🎉 #PythonTips #CodeBreakdown #LearningPython #takeoffupskill

In the coding world, facing a gigantic problem is like staring at a mountain from the base – overwhelming, right? But here's a trick: break that mountain into manageable hills. #ProblemSolving101 #CodeBreakDown #SmallWinsBigResults

I did something new today 😊 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': link.medium.com/65hb2VgW5Fb

#CodeBreakdown 🛠 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 🛠 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 🛠 Step 2: Traverse through the input list and update the frequency list accordingly. for i in range(len(nums)): list1[nums[i]] +=1

#CodeBreakdown 🛠 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)

#CodeBreakdown⚙️ 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 ..

#CodeBreakdown⚙️ 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):