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Cover image for Road to Genius: advanced #44

Road to Genius: advanced #44

codr profile image Ilya Nevolin ・2 min read

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.

function maxa(arr) {
  let 💰 = 0;
  for (let i = 0; i < arr.length; i++) {
    for (let j = i + 1; j < arr.🚀; j++) {
      let cA = Math.abs(i - j);
      cA *= Math.min(☃️[i], arr[j]);
      if (cA > max)
        max = cA;
    }
  }
  return max;
}
let A = maxa([5, 💧, 3, 7, 1, 4]);

// ☃️ = ? (identifier)
// 🚀 = ? (identifier)
// 💧 = ? (number)
// 💰 = ? (identifier)
// such that A = 20 (number)
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Today's challenge is relatively simple but does require some brain power.

The first bug 💰 is a variable declaration, after quickly skimming all lines the only variable that's used but hasn't been declared yet is max.

The second bug 🚀 is very likely to be the propery length of an array.

The third bug ☃️ is a variable which is being used in an array like fashion, just like its neighbor, so my best bet is that it should be arr.

The final bug 💧 should be a number, and it's part of the input for the function maxa; we must ensure that A = 20 to complete the challenge. To determine this number we have to analyze what the function does.

The funcion maxa begins with a for-loop starting at i = 0 over all numbers, then a second for-loop that starts at j = i+1. The variable cA determines the distance between i and j, it's then multiplied by the minimum of the values at i and j; finally the largest recorded value for cA is stored in (and returns) max.

The goal is to find i and j such that max = 20. Let's write in pseudo-code to help us out:

max = cA * min(arr_i, arr_j)

factors of 20 are:
1 * 20
2 * 10
4 * 5

-----

let Y = abs(i - j),        Y must be either 4 or 5
let Z = min(arr_i, arr_j), Z must be either 5 or 4
then cA = Y * Z = 20

-----

the largest value for Y is abs(0 - 5) = 5
then Z = min(5, 4) = 4
then cA = 5 * 4 = 20  --> requirement met
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Since the position of 💧 in the array is neither 0 or 5, its value doesn't matter as long as it won't result in a cA value larger than 20. So we can pick the smallest value such as 1:

coding challenge answer

By solving these challenges you train yourself to be a better programmer. You'll learn newer and better ways of analyzing, debugging and improving code. As a result you'll be more productive and valuable in business. Get started and become a certified Codr today at https://nevolin.be/codr/

Discussion (1)

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Farhan Yahya

Same here. everyday I solve a competitive programming question but on Code Chef.

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