Each day I solve several coding challenges and puzzles from Codr's ranked mode. The goal is to reach genius rank, along the way I explain how I solve them. You do not need any programming background to get started, and you will learn a ton of new and interesting things as you go.
let S = 0; for (let i = 0; i < 192; i++) { let h = Math.floor(i/2) if (h > 0) S += i % h } // S = ? (number) Today's coding challenge is a very interesting one, because we need to play smarter than usual to solve it. It's only a few lines of code but we need a strategy to solve it.
The for-loop iterates 192 times which is quite a lot, so doing it in our heads, paper or excel will be a huge overkill. Let's start by briefly analyzing what the code does in pseudo-code:
S = 0 for i in [0 to 192]: h = floor(i/2) S += i % h return S The variable h is half of i. The remainder from dividing i by h is added to S. Let's do a few iterations to illustrate this:
i:0 S+=0 i:1 S+=0 i:2 S+=0 i:3 S+=0 i:4 S+=0 i:5 S+=1 i:6 S+=0 i:7 S+=1 i:8 S+=0 i:9 S+=1 i:10 S+=0 i:11 S+=1 i:12 S+=0 i:13 S+=1 i:14 S+=0 i:15 S+=1 i:16 S+=0 i:17 S+=1 i:18 S+=0 i:19 S+=1 Notice that the numbers [0, 9] produce three ones. But everything else, that is [10, 19] produce five ones. The same is true for [20, 29], and so on... In a nutshell this algorithm is related to the number of odd numbers within the range, with the exception of the first ten numbers:
The first 10 numbers (0 to 9) produce 3 odd numbers. The next 90 numbers (10 to 99) produce 9*5=45 odd numbers. The next 90 numbers produce once again 45 odd numbers. The last 2 numbers (190 to 192) produce 1 odd number. 3 + 45 + 45 + 1 = 94 
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