Daily JavaScript Challenge #JS-56: Find Missing Number in Arithmetic Sequence

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Daily JavaScript Challenge: Find Missing Number in Arithmetic Sequence

Hey fellow developers! 👋 Welcome to today's JavaScript coding challenge. Let's keep those programming skills sharp!

The Challenge

Difficulty: Medium

Topic: Numbers and Sequences

Description

Given an array that represents an arithmetic sequence with exactly one number missing, find the missing number in the sequence. An arithmetic sequence is a sequence of numbers with a constant difference between consecutive terms.

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https://www.dpcdev.com/

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Write your solution
Test it against the provided test cases
Share your approach in the comments below!

Want to Learn More?

Check out the documentation about this topic here: https://en.wikipedia.org/wiki/Arithmetic_progression

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Did you find any interesting edge cases?
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This is part of our Daily JavaScript Challenge series. Follow me for daily programming challenges and let's grow together! 🚀

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Top comments (5)

raskinfe • Jan 3 • Edited

Here is my solution for the challenge. Addressing more edge cases and finding multiple missing elements in the sequence. I have not tested it but I assume it should work fine :)

`const findMissingSequence = (arg) => {
if (!Array.isArray(arg)) {
throw new Error("Provided argument is not an array");
}
const length = arg.length;
if (length <= 1) {
return [];
}

const diffs = [];
for (let i = 0; i < length - 1; i++) {
    diffs.push(arg[i + 1] - arg[i]);
}

diffs.sort((a, b) => a - b)

const diff = diffs.shift();

if (diffs.some(d => d % diff !== 0)) {
    throw new Error('Provided argument is not an arithmetic sequence');
}

const missing = [];
let nextSequence = -Infinity;
const sequence = new Set(arg);
let i = 1;
while (true) {
    nextSequence = arg[0] + diff*(i - 1);
    if (nextSequence > arg[length - 1]) { 
      break;
    }

    if (!sequence.has(nextSequence)) {
        missing.push(nextSequence)
    }

    i++
}

return missing;