Problem

The four adjacent digits in the 1000-digit number that has the greatest product is 9 × 9 × 8 × 9 = 5832.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

Find the thirteen adjacent digits in the 1000-digit number that has the greatest product. What is the value of this product?

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Problem Description

Well, a 1000-digit number is given to us.

The 4 adjacent digits with the greatest product are 9 × 9 × 8 × 9 = 5832.

We need to find the product of a 13 adjacent digits in this number and to top it off, it is the greatest product of any 13 adjacent digits in this 1000-digit number.

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Approach

To solve this problem, first, we are going to convert this 1000-digit number to an array of length 1000. We use String.prototype.split() function to convert the number to an array.

Then, we loop through the array till we reach the last 13 digits.

Slice the array for every 13 digits using the Array slice() method and then calculate the product using the Array reduce() method.

Push this product to a new array.

Once all the digits are handled, then using Math.max(), return the greatest of the calculated product.

Slicing the array and calculating its product is a much better solution compared to looping through every digit and calculating its product in terms of time complexity.

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Solution

const givenNumber = '7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450';

const gvnNumArr = givenNumber.trim().split('').map(Number);

const largestProduct = n => {
    let prod = 0,
        prodArray = [];

    for (let i = 0; i <= gvnNumArr.length - n; i++) {
        const last = i + n;

        prod = gvnNumArr.slice(i, last).reduce((acc, curr) => acc * curr, 1);
        prodArray.push(prod);
    }

    return Math.max(...prodArray);
}

console.log(largestProduct(13));

You can find my solution on GitHub 08

This particular solution was executed with a better complexity in Hacker rank. If you have another solution, please leave it in the comments below.

For the other Project Euler Solutions, please follow the series Project Euler Solutions in JS.

Thank you!