Road to Genius: superior #52

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 intersected(a, b) {
  if (a[0] > b[1] || a[1] < b[0])
    return false;
  return true;
}
function mergeTwo(a, b) {
  return [Math.min(🚀[0], b[0]), Math.max(a[1], b[1])];
}
function merge(VLS) {
  VLS.😈((a, b) => a[0] - b[🍎]);
  for (let i = 0; i < VLS.length - 1; i++) {
    const cur = VLS[i];
    const next = VLS[i + 1];
    if (intersected(cur, next)) {
      VLS[i] = undefined;
      VLS[i + 1] = mergeTwo(cur, next);
    }
  }
  return VLS.filter(q => q);
}
let arr = [[1, 10], [10, 15], [15, 15]];
let A = 🐼(arr);
A = A[0][1];

// 🐼 = ? (identifier)
// 🍎 = ? (number)
// 😈 = ? (identifier)
// 🚀 = ? (identifier)
// such that A = 15 (number)

Today's challenge seems to be some specific algorithm related to merging arrays. To solve the challenge we have to fix four bugs.

The easiest bug is 🐼 which should be a call to the function merge.

The next bug 🍎 is likely going to be 0, because the arrow function seems to be comparing a[0] with b[0]. Consequently 🍎 is likely going to be sort function. Similarly the bug 🚀 should be a.

After a brief analysis, this algorithm is designed to merge the given intervals. Notice that the input arr is an array of intervals in the form of [start, end]. When two intervals overlap (using intersected function) then the two intervals are merged into one.

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/